# Babbage problem

Babbage problem
You are encouraged to solve this task according to the task description, using any language you may know.

Charles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:

What is the smallest positive integer whose square ends in the digits 269,696?
— Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.

He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.

The task is to find out if Babbage had the right answer — and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand. As Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.

For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].

Motivation

The aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.

## 11l

V n = 1
L n ^ 2 % 1000000 != 269696
n++
print(n)

## 360 Assembly

An assembler program always seems a bit tricky for non system engineer because it deals directly with the operating system and with the hardware instructions. Here we have a 32-bit computer with 16 32-bit registers. The caller (the operating system to keep it simple) is calling you giving your location address stored in register-15 and has stored in register-14 his return address. To save each program context, register-13 points to a 18 word save area. Do not spend time in understanding the context saving and restoring in the prologue and epilogue part of the program. What you have to know, “360” architecture uses 32-bit signed binary arithmetic, so here the maximum integer value is 2^31-1 (2147483647). Therefore the solution must be less than 2147483647. The multiplication and the division use a pair of registers; coding “MR 4,2” means multiply register-5 by register-2 and place result in the (register-4,register-5) pair; the same way “DR 4,2” means divide the (register-4,register-5) pair by register-2 and place the quotient in register-5 and the reminder in register-4. We use in the below program this intermediate 64-bit integers to find a solution with a value up to 2^31-1 even when we have to compute the square of this value.

*        Find the lowest positive integer whose square ends in 269696
*        The logic of the assembler program is simple :
*        loop for i=524 step 2
*          if (i*i modulo 1000000)=269696 then leave loop
*        next i
*        output 'Solution is: i=' i '  (i*i=' i*i ')'
BABBAGE  CSECT                     beginning of the control section
USING BABBAGE,13          define the base register
B     72(15)              skip savearea (72=18*4)
DC    17F'0'              savearea (18 full words (17+1))
STM   14,12,12(13)        prolog: save the caller registers
LA    6,524               let register6 be i and load 524
LOOP     LR    5,6                 load register5 with i
MR    4,6                 multiply register5 with i
LR    7,5                 load register7 with the result i*i
D     4,=F'1000000'       divide register5 with 1000000
C     4,=F'269696'        compare the reminder with 269696
BE    ENDLOOP             if equal branch to ENDLOOP
LA    6,2(6)              load register6 (i) with value i+2
B     LOOP                branch to LOOP
ENDLOOP  XDECO 6,BUFFER+15         edit registrer6 (i)
XDECO 7,BUFFER+34         edit registrer7 (i squared)
XPRNT BUFFER,L'BUFFER     print buffer
L     13,4(0,13)          epilog: restore the caller savearea
LM    14,12,12(13)        epilog: restore the caller registers
XR    15,15               epilog: set return code to 0
BR    14                  epilog: branch to caller
BUFFER   DC    CL80'Solution is: i=............  (i*i=............)'
END   BABBAGE             end of the control section
Output:
Solution is: i=       25264  (i*i=   638269696)


-- The program is written in the programming language Ada. The name "Ada"
-- has been chosen in honour of your friend,
--      Augusta Ada King-Noel, Countess of Lovelace (née Byron).
--
-- This is an program to search for the smallest integer X, such that
-- (X*X) mod 1_000_000 = 269_696.
--
-- In the Ada language, "*" represents the multiplication symbol, "mod" the
-- modulo reduction, and the underscore "_" after every third digit in
-- literals is supposed to simplify reading numbers for humans.
-- Everything written after "--" in a line is a comment for the human,
-- and will be ignored by the computer.

-- We need this to tell the computer how it will later output its result.

procedure Babbage_Problem is

-- We know that 99_736*99_736 is 9_947_269_696. This implies:
-- 1. The smallest X with X*X mod 1_000_000 = 269_696 is at most 99_736.
-- 2. The largest square X*X, which the program may have to deal with,
--    will be at most 9_947_269_69.

type Number is range 1 .. 99_736*99_736;
X: Number := 1;
-- X can store numbers between 1 and 99_736*99_736. Computations
-- involving X can handle intermediate results in that range.
-- Initially the value stored at X is 1.
-- When running the program, the value will become 2, 3, 4, etc.

begin
-- The program starts running.

-- The computer first squares X, then it truncates the square, such
-- that the result is a six-digit number.
-- Finally, the computer checks if this number is 269_696.
while not (((X*X) mod 1_000_000) = 269_696) loop

-- When the computer goes here, the number was not 269_696.
X := X+1;
-- So we replace X by X+1, and then go back and try again.

end loop;

-- When the computer eventually goes here, the number is 269_696.
-- E.e., the value stored at X is the value we are searching for.
-- We still have to print out this value.

-- Number'Image(X) converts the value stored at X into a string of
-- printable characters (more specifically, of digits).
-- I did already run the program, and it did print out 25264.
end Babbage_Problem;


## Aime

integer i;

i = sqrt(269696);
while (i * i % 1000000 != 269696) {
i += 1;
}

o_(i, "\n");

## ALGOL 68

As with other samples, we use "simple" forms such as "a := a + 1" instead of "a +:= 1".

COMMENT text between pairs of words 'comment' in capitals are
for the human reader's information and are ignored by the machine
COMMENT

COMMENT Define s to be the integer value 269 696              COMMENT
INT s = 269 696;

COMMENT Name a location in the machine's storage area that will be
used to hold integer values.
The value stored in the location will change during the
calculations.
Note, "*" is used to represent the multiplication operator.
":=" causes the location named to the left of ":=" to
assume the value computed by the expression to the right.
"sqrt" computes an approximation to the square root
of the supplied parameter
"MOD" is an operator that computes the modulus of its
left operand with respect to its right operand
"ENTIER" is a unary operator that yields the largest
integer that is at most its operand.
COMMENT
INT v := ENTIER sqrt( s );

COMMENT the construct: WHILE...DO...OD repeatedly executes the
instructions between DO and OD, the execution stops when
the instructions between WHILE and DO yield the value FALSE.
COMMENT
WHILE ( v * v ) MOD 1 000 000 /= s DO v := v + 1 OD;

COMMENT print displays the values of its parameters
COMMENT
print( ( v, " when squared is: ", v * v, newline ) )
Output:
     +25264 when squared is:  +638269696


## APL

If at all possible, I would sit down at a terminal with Babbage and invite him to experiment with the various functions used in the program.

      ⍝ We know that 99,736 is a valid answer, so we only need to test the positive integers from 1 up to there:
N←⍳99736
⍝ The SQUARE OF omega is omega times omega:
SQUAREOF←{⍵×⍵}
⍝ To say that alpha ENDS IN the six-digit number omega means that alpha divided by 1,000,000 leaves remainder omega:
ENDSIN←{(1000000|⍺)=⍵}
⍝ The SMALLEST number WHERE some condition is met is found by taking the first number from a list of attempts, after rearranging the list so that numbers satisfying the condition come before those that fail to satisfy it:
SMALLESTWHERE←{1↑⍒⍵}
SMALLESTWHERE (SQUAREOF N) ENDSIN 269696

Output:
25264

## AppleScript

AppleScript's number types are at their limits here, but we can just get to the first Babbage number, after 638 integer root tests on suffixed numbers:

-- BABBAGE -------------------------------------------------------------------

-- babbage :: Int -> [Int]
on babbage(intTests)

script test
on toSquare(x)
(x * 1000000) + 269696
end toSquare

on |λ|(x)
hasIntRoot(toSquare(x))
end |λ|
end script

script toRoot
on |λ|(x)
((x * 1000000) + 269696) ^ (1 / 2)
end |λ|
end script

set xs to filter(test, enumFromTo(1, intTests))
zip(map(toRoot, xs), map(test's toSquare, xs))
end babbage

-- TEST ----------------------------------------------------------------------
on run
-- Try 1000 candidates

unlines(map(curry(intercalate)'s |λ|("  ->  "), babbage(1000)))

--> "2.5264E+4 -> 6.38269696E+8"
end run

-- GENERIC FUNCTIONS ---------------------------------------------------------

-- curry :: (Script|Handler) -> Script
on curry(f)
script
on |λ|(a)
script
on |λ|(b)
|λ|(a, b) of mReturn(f)
end |λ|
end script
end |λ|
end script
end curry

-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m > n then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo

-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter

-- hasIntRoot :: Int -> Bool
on hasIntRoot(n)
set r to n ^ 0.5
r = (r as integer)
end hasIntRoot

-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn

-- unlines :: [String] -> String
on unlines(xs)
intercalate(linefeed, xs)
end unlines

-- zip :: [a] -> [b] -> [(a, b)]
on zip(xs, ys)
set lng to min(length of xs, length of ys)
set lst to {}
repeat with i from 1 to lng
set end of lst to {item i of xs, item i of ys}
end repeat
return lst
end zip

Output:
2.5264E+4  ->  6.38269696E+8

## ARM Assembly

Works with: as version Raspberry Pi
/* ARM assembly Raspberry PI  */
/*  program babbage.s   */

/************************************/
/* Constantes                       */
/************************************/
.equ STDOUT, 1     @ Linux output console
.equ EXIT,   1     @ Linux syscall
.equ WRITE,  4     @ Linux syscall

/*********************************/
/* Initialized data              */
/*********************************/
.data
sMessResult:           .ascii "Result = "
sMessValeur:           .fill 11, 1, ' '            @ size => 11
szCarriageReturn:      .asciz "\n"

/*********************************/
/* UnInitialized data            */
/*********************************/
.bss
/*********************************/
/*  code section                 */
/*********************************/
.text
.global main
main:                                             @ entry of program

ldr r4,iNbStart                               @ start number = 269696
mov r5,#0                                     @ counter multiply
ldr r2,iNbMult                                @ value multiply = 1 000 000
mov r6,r4
1:
mov r0,r6
bl squareRoot                                 @ compute square root
umull r1,r3,r0,r0
cmp r3,#0                                     @ overflow ?
bne 100f                                      @ yes -> end
cmp r1,r6                                     @ perfect square
bne 2f                                        @ no -> loop
bl conversion10                               @ call conversion decimal
bl affichageMess                              @ display message
b 100f                                        @ end
2:
mul r3,r5,r2                                  @ multiply by 1 000 000
b 1b

100:                                              @ standard end of the program
mov r0, #0                                    @ return code
mov r7, #EXIT                                 @ request to exit program
svc #0                                        @ perform the system call

iNbStart:                 .int 269696
iNbMult:                  .int 1000000
/******************************************************************/
/*     compute squareRoot                                       */
/******************************************************************/
/* r0 contains n          */
/* r0 return result or -1 */
squareRoot:
push {r1-r5,lr}                     @ save  registers
cmp r0,#0
beq 100f                            @ if zero -> end
movlt r0,#-1                        @ if negatif return - 1
blt 100f
cmp r0,#4                           @ if <  4 return 1
movlt r0,#1
blt 100f
@ start
clz r2,r0                           @ number of zeros on the left
rsb r2,#32                          @ so many useful numbers right
bic r2,#1                           @ to have an even number of digits
mov r3,#0b11                        @ mask for extract 2 bits
lsl r3,r2
mov r1,#0                           @ init résult with 0
mov r4,#0                           @ raz remainder area

1:                                      @ begin loop
and r5,r0,r3                        @ extract 2 bits with mask
lsl r5,r1,#1                        @ multiplication by 2
lsl r5,#1                           @ shift left one bit
orr r5,#1                           @ bit right = 1
lsl r1,#1                           @ shift left one bit
subs r4,r5                          @ sub remainder
addmi r4,r4,r5                      @ if negative restaur register
subs r2,#2                          @ decrement number bits
movmi r0,r1                         @ if end return result
bmi 100f
lsl r4,#2                           @ no -> shift left remainder 2 bits
lsr r3,#2                           @ and shift right mask 2 bits
b 1b                                @ and loop

100:
pop {r1-r5,lr}                      @ restaur registers
bx lr                               @return
/******************************************************************/
/*     display text with size calculation                         */
/******************************************************************/
/* r0 contains the address of the message */
affichageMess:
push {r0,r1,r2,r7,lr}                          @ save  registres
mov r2,#0                                      @ counter length
1:                                                 @ loop length calculation
ldrb r1,[r0,r2]                                @ read octet start position + index
cmp r1,#0                                      @ if 0 its over
bne 1b                                         @ and loop
@ so here r2 contains the length of the message
mov r1,r0                                      @ address message in r1
mov r0,#STDOUT                                 @ code to write to the standard output Linux
mov r7, #WRITE                                 @ code call system "write"
svc #0                                         @ call systeme
pop {r0,r1,r2,r7,lr}                           @ restaur des  2 registres */
bx lr                                          @ return
/******************************************************************/
/*     Converting a register to a decimal unsigned                */
/******************************************************************/
/* r0 contains value and r1 address area   */
/* r0 return size of result (no zero final in area) */
/* area size => 11 bytes          */
.equ LGZONECAL,   10
conversion10:
push {r1-r4,lr}                                 @ save registers
mov r3,r1
mov r2,#LGZONECAL
1:                                                  @ start loop
bl divisionpar10U                               @ unsigned  r0 <- dividende. quotient ->r0 reste -> r1
strb r1,[r3,r2]                                 @ store digit on area
cmp r0,#0                                       @ stop if quotient = 0
subne r2,#1                                     @ else previous position
bne 1b                                          @ and loop
@ and move digit from left of area
mov r4,#0
2:
ldrb r1,[r3,r2]
strb r1,[r3,r4]
cmp r2,#LGZONECAL
ble 2b
@ and move spaces in end on area
mov r0,r4                                         @ result length
mov r1,#' '                                       @ space
3:
strb r1,[r3,r4]                                   @ store space in area
cmp r4,#LGZONECAL
ble 3b                                            @ loop if r4 <= area size

100:
pop {r1-r4,lr}                                    @ restaur registres
bx lr                                             @return

/***************************************************/
/*   division par 10   unsigned                    */
/***************************************************/
/* r0 dividende   */
/* r0 quotient    */
/* r1 remainder   */
divisionpar10U:
push {r2,r3,r4, lr}
mov r4,r0                                          @ save value
ldr r3,iMagicNumber                                @ r3 <- magic_number    raspberry 1 2
umull r1, r2, r3, r0                               @ r1<- Lower32Bits(r1*r0) r2<- Upper32Bits(r1*r0)
mov r0, r2, LSR #3                                 @ r2 <- r2 >> shift 3
add r2,r0,r0, lsl #2                               @ r2 <- r0 * 5
sub r1,r4,r2, lsl #1                               @ r1 <- r4 - (r2 * 2)  = r4 - (r0 * 10)
pop {r2,r3,r4,lr}
bx lr                                              @ leave function
iMagicNumber:  	.int 0xCCCCCCCD
Output:
Result = 25264

## Arturo

n: new 0
while [269696 <> (n^2) % 1000000]
-> inc 'n

print n

Output:
25264

## AutoHotkey

; Give n an initial value
n = 519

; Loop this action while condition is not satisfied
while (Mod(n*n, 1000000) != 269696) {
; Increment n
n++
}

; Display n as value
msgbox, %n%
Output:
25264

## AWK

# A comment starts with a "#" and are ignored by the machine.  They can be on a
# line by themselves or at the end of an executable line.
#
# A program consists of multiple lines or statements.  This program tests
# positive integers starting at 1 and terminates when one is found whose square
# ends in 269696.
#
# The next line shows how to run the program.
# syntax: GAWK -f BABBAGE_PROBLEM.AWK
#
BEGIN { # start of program
# this declares a variable named "n" and assigns it a value of zero
n = 0
# do what's inside the "{}" until n times n ends in 269696
do {
n = n + 1 # add 1 to n
} while (n*n !~ /269696$/) # print the answer print("The smallest number whose square ends in 269696 is " n) print("Its square is " n*n) # terminate program exit(0) } # end of program  Output: The smallest number whose square ends in 269696 is 25264 Its square is 638269696  ## BASIC ### Applesoft BASIC This is an implementation based on the alternative solution for the BBC BASIC. We know that 269696 is not a perfect square, so we can safely start with N=1269696 and add 1000000 each time N is not a perfect square. The ST function returns the remainder of the difference between N and the square of the integer part of the root (R). There are a couple of quirks in AppleSoft BASIC, the largest integer is 32767, but the largest number not displayed on scientific notation is 999999999, which explains the strange IF statement in line 170.  100 : 110 REM BABBAGE PROBLEM 120 : 130 DEF FN ST(A) = N - INT (A) * INT (A) 140 N = 269696 150 N = N + 1000000 160 R = SQR (N) 170 IF FN ST(R) < > 0 AND N < 999999999 THEN GOTO 150 180 IF N > 999999999 THEN GOTO 210 190 PRINT "SMALLESt NUMBER WHOSE SQUARE ENDS IN"; CHR$ (13);
"269696 IS ";R;", AND THE
SQUARE IS"; CHR$(13);N 200 END 210 PRINT "THERE IS NO SOLUTION FOR VALUES SMALLER"; CHR$(13);
"THAN 999999999."

Output:
]RUN
SMALLEST NUMBER WHOSE SQUARE ENDS IN
269696 IS 25264, AND THE SQUARE IS
638269696

### Commodore BASIC

Based on the C language implementation

10 rem This code is an implementation of Babbage Problem
20 num = 100 : rem We can safely start at 100
30 s = num*num
40 r = s - int(s/1000000)*1000000 : rem remainder when divided by 1,000,000
50 if r = 269696 then goto 100    : rem compare with 269,696
60 print "n="num"sq="s"rem="r
70 num = num+1
80 goto 30
90 rem Print out the result
100 print:print "The smallest number whose square ends in 269696 is:"
110 print num;"....";num;"squared = ";s
120 end


### BASIC256

#This code is an implementation of Babbage Problem
number = 2
DO
number += 2
UNTIL ((number^2) % 1000000) = 269696
PRINT "The smallest number whose square ends in 269696 is: "; number
PRINT "It's square is "; number*number

### BBC BASIC

Clarity has been preferred over all other considerations. The line LET n = n + 1, for instance, would more naturally be written n% += 1, using an integer variable and a less verbose assignment syntax; but this optimization did not seem to justify the additional explanations Professor Babbage would probably need to understand it.

REM Statements beginning 'REM' are explanatory remarks: the machine will ignore them.

REM We shall test positive integers from 1 upwards until we find one whose square ends in 269,696.

REM A number that ends in 269,696 is one that leaves a remainder of 269,696 when divided by a million.

REM So we are looking for a value of n that satisfies the condition 'n squared modulo 1,000,000 = 269,696', or 'n^2 MOD 1000000 = 269696' in the notation that the machine can accept.

LET n = 0

REPEAT
LET n = n + 1
UNTIL n^2 MOD 1000000 = 269696

PRINT "The smallest number whose square ends in 269696 is" n

PRINT "Its square is" n^2

Output:
The smallest number whose square ends in 269696 is     25264
Its square is 638269696

#### Alternative method

Translation of: PowerShell

The algorithm given in the alternative PowerShell implementation may be substantially more efficient, depending on how long SQR takes, and I think could well be more comprehensible to Babbage.

REM Lines that begin 'REM' are explanatory remarks addressed to the human reader.

REM The machine will ignore them.

LET n = 269696

REPEAT

LET n = n + 1000000

REM Find the next number that ends in 269,696.

REM The function SQR finds the square root.

LET root = SQR n

REM The function INT truncates a real number to an integer.

UNTIL root = INT root

REM If the square root is equal to its integer truncation, then it is an integer: so we have found our answer.

PRINT "The smallest number whose square ends in 269696 is" root

PRINT "Its square is" n

Output:

Identical to the first BBC BASIC version.

### QBasic

Works with: QBasic
Works with: QuickBasic
Translation of: Yabasic
number = 524
DO
NUMBER = number + 2
LOOP UNTIL ((number ^ 2) MOD 1000000) = 269696
PRINT "The smallest number whose square ends in 269696 is: "; number
PRINT "It's square is "; number ^ 2
END


### True BASIC

Translation of: Yabasic
LET number = 2
DO
LET number = number + 2
LOOP UNTIL REMAINDER((number ^ 2), 1000000) = 269696
PRINT "The smallest number whose square ends in 269696 is: "; number
PRINT "It's square is "; number ^ 2
END


### uBasic/4tH

Translation of: FreeBASIC
n = 524                                ' first number to try

Do
' square the number
' look at the last 6 digits, if they match print the number
Until (n*n) % 1000000 = 269696
' increase the number with 2, number end on a 6
n = n + 2
Until n = 99736
' if the number = 99736 then we haved found a smaller number, so stop
' look at the last 6 digits, if they match print the number
Until (n*n) % 1000000 = 269696
' increase the number with 8, number ends on a 4
n = n + 8
Loop

If n = 99736 Then
Print "No smaller number was found"
Else
' we found a smaller number, print the number and its square
Print "The number = ";n ; " and its square = "; n*n
EndIf


### IS-BASIC

100 PROGRAM "Babbage.bas"
110 LET N=2
120 DO
130   LET N=N+2
140 LOOP UNTIL MOD(N*N,1000000)=269696
150 PRINT "The smallest number whose square ends in 269696 is:";N
160 PRINT "It's square is";N^2

Alternative method:

100 PROGRAM "Babbage.bas"
110 LET N=269696
120 DO
130   LET N=N+1000000
140   LET R=SQR(N)
150 LOOP UNTIL R=INT(R)
160 PRINT "The smallest number whose square ends in 269696 is:";R
170 PRINT "It's square is";N

## Batch File

:: This line is only required to increase the readability of the output by hiding the lines of code being executed
@echo off

:: Everything between the lines keeps repeating until the answer is found
:: The code works by, starting at 1, checking to see if the last 6 digits of the current number squared is equal to 269696
::----------------------------------------------------------------------------------
:loop
:: Increment the current number being tested by 1
set /a number+=1

:: Square the current number
set /a numbersquared=%number%*%number%

:: Check if the last 6 digits of the current number squared is equal to 269696, and if so, stop looping and go to the end
if %numbersquared:~-6%==269696 goto end

goto loop
::----------------------------------------------------------------------------------

:end
echo %number% * %number% = %numbersquared%
pause>nul

Output:
25264 * 25264 = 638269696


## Befunge

Befunge is not an easily readable language, but with a basic understanding of the syntax, I think an intelligent person should be able to follow the logic of the code below.

    1+       ::*    "d"::** %       "V8":** -!     #v_ > > > > >
v
increment n  n*n  modulo 1000000  equal to 269696?  v  if false, loop to right
v
v"Smallest number whose square ends in 269696 is "0 <  else output n below
$( "The smallest number that ends with the figures 269696 when squared is " !number ", since the square of " !number " is " !number*!number ) ) ) Output: The smallest number that ends with the figures 269696 when squared is 25264, since the square of 25264 is 638269696 ## C // This code is the implementation of Babbage Problem #include <stdio.h> #include <stdlib.h> #include <limits.h> int main() { int current = 0, //the current number square; //the square of the current number //the strategy of take the rest of division by 1e06 is //to take the a number how 6 last digits are 269696 while (((square=current*current) % 1000000 != 269696) && (square<INT_MAX)) { current++; } //output if (square>+INT_MAX) printf("Condition not satisfied before INT_MAX reached."); else printf ("The smallest number whose square ends in 269696 is %d\n", current); //the end return 0 ; }  Output: The smallest number whose square ends in 269696 is 25264 ## C# namespace Babbage_Problem { class iterateNumbers { public iterateNumbers() { long baseNumberSquared = 0; //the base number multiplied by itself long baseNumber = 0; //the number to be squared, this one will be iterated do //this sets up the loop { baseNumber += 1; //add one to the base number baseNumberSquared = baseNumber * baseNumber; //multiply the base number by itself and store the value as baseNumberSquared } while (Right6Digits(baseNumberSquared) != 269696); //this will continue the loop until the right 6 digits of the base number squared are 269,696 Console.WriteLine("The smallest integer whose square ends in 269,696 is " + baseNumber); Console.WriteLine("The square is " + baseNumberSquared); } private long Right6Digits(long baseNumberSquared) { string numberAsString = baseNumberSquared.ToString(); //this is converts the number to a different type so it can be cut up if (numberAsString.Length < 6) { return baseNumberSquared; }; //if the number doesn't have 6 digits in it, just return it to try again. numberAsString = numberAsString.Substring(numberAsString.Length - 6); //this extracts the last 6 digits from the number return long.Parse(numberAsString); //return the last 6 digits of the number } } }}  Output: The smallest integer whose square ends in 269,696 is 25264 The square is 638269696 ## C++ #include <iostream> int main( ) { int current = 0 ; while ( ( current * current ) % 1000000 != 269696 ) current++ ; std::cout << "The square of " << current << " is " << (current * current) << " !\n" ; return 0 ; }  Output: The square of 25264 is 638269696 !  ## Caché ObjectScript BABBAGE ; start at the integer prior to the square root of 269,696 as it has to be at least that big set i = ($piece($zsqr(269696),".",1,1) - 1) ; piece 1 of . gets the integer portion ; loop forever, incrementing by one, until we find a square ending in 269696 for { set i = i + 1 ; this will start us at the integer value from the piece statement above ; evaluate if the last 6 digits of the square equal 269696 set square = i * i if ($extract(square,$length(square) - 5,$length(square)) = 269696) {
; We match - display the integer and square value to the screen, formatting the numerics with a comma separator as needed.
write !,"Result: "_$fnumber(i,",")_" squared is "_$fnumber(square,",")_".  This ends in 269696."
quit    ; exit for loop
}
}

quit    ; exit routine
Output:
SAMPLES>do ^BABBAGE
Result: 25,264 squared is 638,269,696.  This ends in 269696.

## Clojure

; Defines function named babbage? that returns true if the
; square of the provided number leaves a remainder of 269,696 when divided
; by a million
(defn babbage? [n]
(let [square (* n n)]
(= 269696 (mod square 1000000))))

; Use the above babbage? to find the first positive integer that returns true
; (We're exploiting Clojure's laziness here; (range) with no parameters returns
; an infinite series.)
(first (filter babbage? (range)))

Output:
25264

## COBOL

IDENTIFICATION DIVISION.
PROGRAM-ID. BABBAGE-PROGRAM.
* A line beginning with an asterisk is an explanatory note.
* The machine will disregard any such line.
DATA DIVISION.
WORKING-STORAGE SECTION.
* In this part of the program we reserve the storage space we shall
* be using for our variables, using a 'PICTURE' clause to specify
* how many digits the machine is to keep free.
* The prefixed number 77 indicates that these variables do not form part
* of any larger 'record' that we might want to deal with as a whole.
77  N           PICTURE 99999.
* We know that 99,736 is a valid answer.
77  N-SQUARED   PICTURE 9999999999.
77  LAST-SIX    PICTURE 999999.
PROCEDURE DIVISION.
* Here we specify the calculations that the machine is to carry out.
CONTROL-PARAGRAPH.
PERFORM COMPUTATION-PARAGRAPH VARYING N FROM 1 BY 1
UNTIL LAST-SIX IS EQUAL TO 269696.
STOP RUN.
COMPUTATION-PARAGRAPH.
MULTIPLY N BY N GIVING N-SQUARED.
MOVE N-SQUARED TO LAST-SIX.
* Since the variable LAST-SIX can hold a maximum of six digits,
* only the final six digits of N-SQUARED will be moved into it:
* the rest will not fit and will simply be discarded.
IF LAST-SIX IS EQUAL TO 269696 THEN DISPLAY N.

Output:
25264

## Common Lisp

(defun babbage-test (n)
"A generic function for any ending of a number"
(when (> n 0)
(do* ((i 0 (1+ i))
(d (expt 10 (1+ (truncate (log n) (log 10))))) )
((= (mod (* i i) d) n) i) )))

Output:
(babbage-test 269696)
25264

### Alternate solution

I use Allegro CL 10.1

; Project : Babbage problem

(setq n 1)
(setq bab2 1)
(loop while (/= bab2 269696)
do (setq n (+ n 1))
(setf bab1 (expt n 2))
(setf bab2 (mod bab1 1000000)))
(format t "~a" "The smallest number whose square ends in 269696 is: ")
(write n)
(terpri)
(format t "~a" "Its square is: ")
(write (* n n))


Output:

The smallest number whose square ends in 269696 is: 25264
Its square is: 638269696


### With proper syntax

With some improvements of Common Lisp style

;; * The package definition
(defpackage :babbage
(:use :common-lisp))
(in-package :babbage)

;; * The function
(defun babbage (end)
"Returns the smallest number whose square ends in END."
(loop
:with digits = (ceiling (log end 10))     ; How many digits has end?
:for num :from (isqrt end)                ; The start number
:for square = (expt num 2)                ; The square of num
:for ends = (mod square (expt 10 digits)) ; The last digits
:until (= ends end)
:finally
(format t "The smallest number whose square ends in ~D is: ~D~%" end num)
(format t "Its square is: ~D~%" square)
(return num)))

Output:
ABBAGE> (babbage 269696)
The smallest number whose square ends in 269696 is: 25264
Its square is: 638269696
25264

## Component Pascal

MODULE BabbageProblem;
IMPORT StdLog;

PROCEDURE Do*;
VAR
i: LONGINT;
BEGIN
i := 2;
WHILE (i * i MOD 1000000) # 269696 DO
IF i MOD 10 = 4 THEN INC(i,2) ELSE INC(i,8) END
END;
StdLog.Int(i)
END Do;

END BabbageProblem.

Execute: ^Q BabbageProble.Do
Output:
25264


## D

// It's basically the same as any other version.
// What can be observed is that 269696 is even, so we have to consider only even numbers,
// because only the square of even numbers is even.

import std.math;
import std.stdio;

void main( )
{
// get smallest number <= sqrt(269696)
int k = cast(int)(sqrt(269696.0));

// if root is odd -> make it even
if (k % 2 == 1)
k = k - 1;

// cycle through numbers
while ((k * k) % 1000000 != 269696)
k = k + 2;

// display output
writefln("%d * %d = %d", k, k, k*k);
}

Output:
25264 * 25264 = 638269696


## Dafny

// Helper function for mask: does the actual computation.
decreases v-m
requires 0 <= v && 0 < m
{
if v < m then m else mask_(v,m*10)
}

// Return the smallest power of 10 greater than v.
requires 0 <= v
{
}

// Return true if the last digits of v == suffix.
predicate method EndWith(v:int,suffix:int)
requires 0 <= suffix
{
}

method SmallestSqEndingWith(suffix:int) returns (s:int)
requires 0 < suffix
ensures EndWith(s*s, suffix)
// ensures forall i :: 0 <= i < s ==> !EndWith(i*i,suffix)
decreases *                   // This method may not terminate.
{
s := 0;
// squares is the sequence of s*s. A ghost variable is only used by the
// verification process at compile time.
ghost var squares := [];
while !EndWith(s*s, suffix)
invariant s == |squares|
invariant forall i :: 0 <= i < s ==> squares[i] == i*i && !EndWith(squares[i], suffix)
decreases *
{
squares := squares + [s*s];
s := s + 1;
}
// Leaving the method:
// s*s ends with the suffix.
assert EndWith(s*s, suffix);
// The sequence squares contains i*i for i in [0..s]; none of the elements of
// squares ends with the suffix.
assert s == |squares|;
assert forall i :: 0 <= i < s ==> i*i == squares[i] && !EndWith(squares[i], suffix);
// That last assertion should imply the commented-out post-condition of the
// method, but I'm not sure how to express that.
//
// Conclusion: s is guaranteed to be the smallest number whose square ends
// with the suffix.
}

method Main() decreases *
{
var suffix := 269696;
var smallest := SmallestSqEndingWith(suffix);
print smallest, "\n";
}

## Dart


main() {
var x = 0;
while((x*x)% 1000000 != 269696)
{	x++;}

print('$x'); }  ## Delphi See Pascal. ## Dyalect Simple approach: var i = 0 while i * i % 1000000 != 269696 { i += 1 } print("\(i) is the smallest number that ends with 269696") Using a non-strict range and iteration: for i in 2..Integer.Max { if i * i % 1000000 == 269696 { print("\(i) is the smallest number that ends with 269696") break } } Output: 25264 is the smallest number that ends with 269696 ## EasyLang while n * n mod 1000000 <> 269696 n += 1 . print n  ## EDSAC order code The only short cut in the program below is to note that both 269696 and 1000000 are divisible by 64, so that only numbers n divisible by 8 need be considered. To save time in finding the residue of n^2 modulo 1000000, the program does not compute n^2 but adds n^2 - (n - 8)^2 to the previous residue. This of course is the principle that Babbage used in his Difference Engine. One inconvenience of EDSAC was that a programmer who wanted to define a 35-bit constant by means of pseudo-orders could specify the low and high 17-bit words, but not the middle bit ("sandwich digit"), which had to be 0. So a constant such as 1000000, with middle bit 1, could not be defined by pseudo-orders. In this case the programmer could usually define the negative of the desired constant and work with that instead, as illustrated below. A more convenient way to define 35-bit constants was to use a library subroutine to read them from the input tape in decimal format, with no worries about the sandwich digit. This is demonstrated in the EDSAC solution to the linear congruential generator. [Babbage problem from Rosetta Code website] [EDSAC program, Initial Orders 2] [Library subroutine M3. Pauses the loading, prints header, and gets overwritten when loading resumes. Here, the last character sets the teleprinter to figures.] PFGKIFAFRDLFUFOFE@A6FG@E8FEZPF @&*SOLUTION!TO!BABBAGE!PROBLEM@&# ..PZ [blank tape, needed to mark end of header text] [Library subroutine P6. Prints strictly positive integer. 32 locations; argument at 0, working locations 1, 4, 5] T56K [define load address for subroutine] GKA3FT25@H29@VFT4DA3@TFH30@S6@T1FV4DU4DAFG26@ TFTFO5FA4DF4FS4FL4FT4DA1FS3@G9@EFSFO31@E20@J995FJF!F [Main routine. Load after subroutine P6. Must be at an even address because each double value at the start must be at an even address.] T88K [define absolute load address] GK [set @ (theta) for relative addresses] [Variables] [0] PF PF [trial solution, call it n] [2] PF PF [residue of n^2 modulo 1000000] [4] PF PF [1st difference for n^2] [Constants] [6] P64F PF [2nd difference for n^2, i.e. 128] [8] P4F PF [1st difference for n, i.e. 8] T10#Z PF T10Z [clears sandwich digit between 10 and 11; cf. Wilkes, Wheeler & Gill, 1951, pp 110, 141-2] [10] #1760F V2046F [-1000000] T12#Z PF T12Z [clears sandwich digit between 12 and 13] [12] Q1728F PD [269696] [14] &F [line feed] [15] @F [carriage return] [16] K4096F [teleprinter null] [Enter with acc = 0] [17] T#@ [trial number n := 0] T2#@ [(n^2 mod 1000000) := 0] S6#@ [acc := -128] RD [right shift] T4#@ [(1st difference for n^2) := -64] [Start of loop] [22] TF [clear acc] A#@ [load n] A8#@ [add 8] T#@ [update n] A4#@ [load 1st difference of n^2] A6#@ [add 128] T4#@ [update] A2#@ [load residue of n^2 mod 1000000] A4#@ [add 1st difference] [31] A10#@ [subtract 1000000, by adding -1000000] E31@ [repeat until result < 0] S10#@ [add back 1000000] U2#@ [update residue] S12#@ [subtract target 269696] G22@ [loop back if residue < 269696] [if still here, acc is non-neg mult of 64] S8#@ [test for acc = 0 by subtracting 8] E22@ [loop back if residue > 269696] [Here with the solution] TF [clear acc] A#@ [load solution n] TD [store at absolute address 0 for printing] [42] A42@ [for return from subroutine] G56F [call subroutine to print n] O15@ [print CR] O14@ [print LF] O16@ [print null, to flush printer buffer] ZF [stop] E17Z [define entry point] PF [enter with acc = 0] Output: SOLUTION TO BABBAGE PROBLEM 25264  ## Elena Translation of: Smalltalk ELENA 4.x : import extensions; import system'math; public program() { var n := 1; until(n.sqr().mod:1000000 == 269696) { n += 1 }; console.printLine(n) } Output: 25264  ## Elixir defmodule Babbage do def problem(n) when rem(n*n,1000000)==269696, do: n def problem(n), do: problem(n+2) end IO.puts Babbage.problem(0)  or Stream.iterate(2, &(&1+2)) |> Enum.find(&rem(&1*&1, 1000000) == 269696) |> IO.puts  Output: 25264  ## Erlang -module(solution1). -export([main/0]). babbage(N,E) when N*N rem 1000000 == 269696 -> io:fwrite("~p",[N]); babbage(N,E) -> case E of 4 -> babbage(N+2,6); 6 -> babbage(N+8,4) end. main()-> babbage(4,4).  ## F# let mutable n=1 while(((n*n)%( 1000000 ))<> 269696) do n<-n+1 printf"%i"n  Same as above, sans mutable state. Seq.initInfinite id |> Seq.skipWhile (fun n->(n*n % 1000000) <> 269696) |> Seq.head |> printfn "%d"  ## Factor ! Lines like this one are comments. They are meant for humans to ! read and have no effect on the instructions carried out by the ! computer (aside from Factor's parser ignoring them). ! Comments may appear after program instructions on the same ! line. ! Each word between USING: and ; is a vocabulary. By importing ! a vocabulary in this way, its words are made available for the ! program to use. This is a way to keep the space requirements ! down for deployed programs, and a nice side effect is that it ! gives readers a clue for where to look for documentation. USING: kernel math math.ranges prettyprint sequences ; ! Before the program begins, it's incredibly helpful to have an ! understanding of Factor's dataflow model. Don't worry; it's ! not complicated, but it's confusing to read a Factor program ! without this knowledge. ! Factor is a stack-based language. What this means is that ! there is an implicit data stack in the background, waiting ! to recieve whatever manner of thing we wish to give it. Here ! is a simple arithmetic expression to demonstrate: ! language token | data stack ! ---------------+----------- ! 2 2 ! numbers place themselves on the stack. ! 1 2 1 ! 4 2 1 4 ! + 2 5 ! consume 1 and 4 and leave behind 5. ! * 10 ! consume 2 and 5 and leave behind 10. ! Thus the phrase ! 2 1 4 + * ! in Factor is a way to calculate 2 * (4 + 1). ! We could have also written this as ! 1 4 + 2 * ! with no change in meaning or outcome. ! Because of the way the data stack works, there is no need ! to specify order of operations in the language, because you do ! so inherently by the order you place things on the data stack. ! === BEGIN PROGRAM ============================================ 518 99,736 2 <range> ! Here we place three numbers on the ! stack representing a range of numbers. ! The first, 518, represents the starting ! point of the sequence. 99,736 ! represents the ending point of the ! sequence. 2 represents the "step" of ! the sequence, or a constant distance ! between members. ! <range> takes those three numbers and ! creates an object representing the ! described range of numbers. Computers ! of today are more than capable of ! storing that many numbers, but <range> ! doesn't store them all; it calculates ! the number that is needed at the ! current time. ! The rationale for the sequence is as ! follows. Odd squares are always odd, so ! we don't need to consider them. That's ! why the sequence starts with an even ! number and is incremented by 2. We ! choose 518 to start because it's the ! largest even square less than 269,696. ! We choose 99,736 to end because we ! know it's a solution. [ sq 1,000,000 mod 269,696 = ] ! the [ ... ] form is called a quotation. ! Think of it like a sequence that stores ! code. It's a way to place code on the ! data stack without executing it. This ! is so that it can be used by the find ! word. You could also think of it much ! like a function that hasn't been given ! a name. find ! When we call the find word, there are ! two objects on the stack: a sequence ! and a quotation. find is a word that ! takes a sequence and a quotation and ! applies the quotation to one member of ! the sequence after another. It does ! so until the quotation returns a t ! value (denoting a boolean true) and ! then leaves that number, along with its ! index in the sequence, on the stack. ! Let's take a look at what happens ! for each iteration of find. Let's look ! at what happens with the first number ! in the sequence. ! language token | data stack ! ---------------+----------- ! 518 518 ! 518 is placed on the stack ! from the sequence by find. ! sq 268,324 ! square it ! 1,000,000 268,324 1,000,000 ! place a million on the stack ! mod 268,324 ! take modulus of 268,324 ! and 1,000,000 ! 269,696 268,324 269,696 ! place 269,696 on the stack ! = f ! test 268,324 and 269,696 for ! equality. ! So the square of the first number in ! the sequence, 518, does not end with ! 269,696. We'll try each number in the ! sequence until we get a t. . ! Consume the top member of the data stack and print it out. drop ! find leaves both the found element from the sequence ! and the index at which it was found on the data stack. ! We don't care about the index so we will call drop to ! remove it from the top of the data stack. All programs ! must end with an emtpy data stack. ! Putting the entire program together, it looks like this: ! 518 99,736 2 <range> [ sq 1,000,000 mod 269,696 = ] find . drop  Output: 25264  ## Forth Can a Forth program be made readable to a novice, without getting into what a stack is? We shall see. ( First we set out the steps the computer will use to solve the problem ) : BABBAGE 1 ( start from the number 1 ) BEGIN ( commence a "loop": the computer will return to this point repeatedly ) 1+ ( add 1 to our number ) DUP DUP ( duplicate the result twice, so we now have three copies ) ( We need three because we are about to multiply two of them together to find the square, and the third will be used the next time we go around the loop -- unless we have found our answer, in which case we shall need to print it out ) * ( * means "multiply", so we now have the square ) 1000000 MOD ( find the remainder after dividing it by a million ) 269696 = ( is it equal to 269,696? ) UNTIL ( keep repeating the steps from BEGIN until the condition is satisfied ) . ; ( when it is satisfied, print out the number that allowed us to satisfy it ) ( Now we ask the machine to carry out these instructions ) BABBAGE  Output: 25264 ## Fortran ### First FORTRAN Mister Babbage, I have been working for 2 years in New York on an IBM 704 computer. I have just finished my work on a new language, a language for non-specialists. I called it: FORTRAN (FORmula TRANslator). And with it I solved your problem. Sincerely, John Backus - September 1956  DO 3 N=1,99736 IF(MODF(N*N,1000000)-269696)3,4,3 3 CONTINUE 4 PRINT 5,N 5 FORMAT(I6) STOP  Output: 25264 Mister Babbage, an addendum : I was confident that my program will work because the FORTRAN for IBM 704 rely on a computer with a hardware of 36 bit integers. You already proved than the number N must be less or equal to 99736. But if N were greater than 46340 (√ 231-1 ), half of the programs you see here use 32 bit integers and they would have failed with an overflow exception. Sincerely - J. B. ### Modern Fortran Translation of: F# program babbage implicit none integer :: n n=1 do while (mod(n*n,1000000) .ne. 269696) n = n + 1 end do print*, n end program babbage  ## FreeBASIC ' version 25-10-2016 ' compile with: fbc -s console ' Charles Babbage would have known that only number ending ' on a 4 or 6 could produce a square ending on a 6 ' also any number below 520 would produce a square smaller than 269,696 ' we can stop when we have reached 99,736 ' we know it square and it ends on 269,696 Dim As ULong number = 524 ' first number to try Dim As ULong square, count Do ' square the number square = number * number ' look at the last 6 digits, if they match print the number If Right(Str(square), 6) = "269696" Then Exit Do ' increase the number with 2, number end ons a 6 number = number +2 ' if the number = 99736 then we haved found a smaller number, so stop If number = 99736 Then Exit Do square = number * number ' look at the last 6 digits, if they match print the number If Right(Str(square),6 ) = "269696" Then Exit Do ' increase the number with 8, number ends on a 4 number = number +8 ' go to the first line under "Do" Loop If number = 99736 Then Print "No smaller number was found" Else ' we found a smaller number, print the number and its square Print Using "The number = #####, and its square = ##########,"; number; square End If ' empty keyboard buffer While Inkey <> "" : Wend Print : Print "hit any key to end program" Sleep End  Output: The number = 25,264 and its square = 638,269,696 ## Frink // This is a solver for the Rosetta Code problem "Babbage problem" // https://rosettacode.org/wiki/Babbage_problem i = 1 while true { // mod is the modulus operator. // The == operator is a test for equality. A single = // assigns to a variable. if i² mod million == 269696 { // This prints i and i² println[i + "² = " + i²] exit[] // This terminates the program if a solution is found. } i = i + 1 } Output: 225264² = 638269696 ## FutureBasic window 1 long i for i = 1 to 1000000 if i ^ 2 mod 1000000 == 269696 then exit for next print @"The smallest number whose square ends in 269696 is ";i print @"Its square is ";i ^ 2 HandleEvents Output: The smallest number whose square ends in 269696 is 25264 Its square is 638269696 ## Fōrmulæ Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition. Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used. In this page you can see the program(s) related to this task and their results. ## Gambas Public Sub Main() Dim iNum As Long For iNum = 1 To 100000 If Str(iNum * iNum) Ends "269696" Then Break Next Print "The lowest number squared that ends in '269696' is " & Str(iNum) End  Output: The lowest number squared that ends in '269696' is 25264  ## Go package main import "fmt" func main() { const ( target = 269696 modulus = 1000000 ) for n := 1; ; n++ { // Repeat with n=1, n=2, n=3, ... square := n * n ending := square % modulus if ending == target { fmt.Println("The smallest number whose square ends with", target, "is", n, ) return } } }  Output: The smallest number whose square ends with 269696 is 25264  ## Groovy int n=104; ///starting point while( (n**2)%1000000 != 269696 ) { if (n%10==4) n=n+2; if (n%10==6) n=n+8; } println n+"^2== "+n**2 ;  Output: 25264^2== 638269696  ## Haskell #### head --Calculate squares, testing for the last 6 digits findBabbageNumber :: Integer findBabbageNumber = head (filter ((269696 ==) . flip mod 1000000 . (^ 2)) [1 ..]) main :: IO () main = (putStrLn . unwords) (zipWith (++) (show <$> ([id, (^ 2)] <*> [findBabbageNumber]))
[" ^ 2 equals", " !"])

Output:
25264 ^ 2 equals 638269696 !

Or, if we incline to the nullius in verba approach, are not yet convinced that there really are any such numbers below 100,000, and look uncertainly at head – a partial function which simply fails on empty lists, we could import the Safe module, and use the headMay alternative, which, more cautiously and experimentally, returns a Maybe value:

import Data.List (intercalate)
import Data.Maybe (maybe)

maybeBabbage :: Integer -> Maybe Integer
maybeBabbage upperLimit =
(filter ((269696 ==) . flip rem 1000000) ((^ 2) <$> [1 .. upperLimit])) main :: IO () main = do let upperLimit = 100000 putStrLn$
maybe
(intercalate (show upperLimit) ["No such number found below ", " ..."])
(intercalate " ^ 2  ->  " .
fmap show . (<*>) [floor . sqrt . fromInteger, id] . pure)
(maybeBabbage upperLimit)

Output:
25264 ^ 2  ->  638269696

#### Suffixes and integer roots

The inverse approach, which gets us to the first number in just 638 tests, is to append a 269696 suffix to each successive integer, filtering for results with integer square roots.

We can then harvest as many as we need from an infinite stream of babbages, Mr Babbage.

import Data.List (intercalate)

--------------------- BABBAGE PROBLEM --------------------

babbagePairs :: [[Integer]]
babbagePairs =
[0, 1000000 ..]
>>= \x ->                     -- Drawing from a succession of N * 10^6
let y = (x + 269696)        -- The next number ending in 269696,
r = root y              -- its square root,
i = floor r             -- and the integer part of that root.
in [ [i, y]                -- Root and square harvested together,
| r == fromIntegral i -- only if that root is an integer.
]

root :: Integer -> Double
root = sqrt. fromIntegral

--------------------------- TEST -------------------------
main :: IO ()
main = mapM_ (putStrLn . arrowed) $take 10 babbagePairs arrowed :: [Integer] -> String arrowed = intercalate " ^ 2 -> " . fmap show  Output: 25264 ^ 2 -> 638269696 99736 ^ 2 -> 9947269696 150264 ^ 2 -> 22579269696 224736 ^ 2 -> 50506269696 275264 ^ 2 -> 75770269696 349736 ^ 2 -> 122315269696 400264 ^ 2 -> 160211269696 474736 ^ 2 -> 225374269696 525264 ^ 2 -> 275902269696 599736 ^ 2 -> 359683269696 A quick glance at these results suggests that Mr Babbage would have done well to consider more closely the way in which the final digits of the square constrain the final digits of the root. We can get to the solution almost immediately, after only a handful of tests, well within the reach of pencil and paper, if we discern that the root itself, to produce the 269692 suffix in its square, must have one of only four different final digit sequences: (0264, 5264, 4736, or 9736). With a machine, this approach can industrialise the babbage harvest, yielding thousands of pairs in less than a second: ---------------------- BABBAGE PAIRS --------------------- babbagePairs :: [(Integer, Integer)] babbagePairs = [0, 10000 ..] >>= \x -> ( ((,) <*> (^ 2)) . (x +) <$> [264, 5264, 9736, 4736]
)
>>= \(a, b) ->
[ (a, b)
| ((269696 ==) . flip rem 1000000) b
]

--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_
putStrLn
( (\(a, b) -> show a <> " ^2 -> " <> show b)
<$> take 2000 babbagePairs )  ## Hoon :: This is Hoon, a language for writing human-legible :: instructions to a machine called Urbit. :: :: A pair of non-alphanumeric symbols is called a rune. :: Each rune begins a unique expression. :: :: An expression can be an instruction to the machine, :: or a description of essential information. :: Each rune specifies a different expression. :: Each expression can contain other expressions. :: (In practice, every expression contains :: at least one of the expressions that follow it.) :: :: :: tells the machine to ignore the rest of a line :: these lines allow commentary for a human reader :: like the question this program will answer: :: :: What is the smallest positive integer whose :: square ends in the digits 269,696? :: :: The program of instructions for solving this, :: uninterrupted by commentary, is: :: :: :- %say :: |= [*] :: :- %noun :: ^- @ud :: =/ n 0 :: |- :: ?: =(269.696 (mod (pow n 2) 1.000.000)) :: n :: %=$
::    n  +(n)
::  ==
::
::  The first three significant lines describe
::  two things: our program's input, and its structure.
::  They specify the program requires no input.
::  Otherwise, we can ignore them for our purposes.
::
::  ^-  describes the desired output of our program.
::  @ud signifies an unknown positive integer.
::      The output will be a positive integer.
::      The output will be our answer.
:-  %say
|=  [*]
:-  %noun
^-  @ud
::
::  =/ assigns a value to a name, for future reference.
::     Here we assign value 0 to "n", as in algebra.
=/  n  0
::
::  |- begins a recursive, iterative process.
::     It might not end until a condition is met.
|-
::
::  ?: does two things.
::     First, it returns a Boolean true-or-false answer
::     to the question that follows.
::     Second, it evaluates one of two expressions
::     branching on whether the answer is true or false.
::
::  Here we rephrase our question for the machine:
::  "Is 269,696 equal to the remainder of
::  n^2 divided by 1,000,000?"
?:  =(269.696 (mod (pow n 2) 1.000.000))
::
::  If so, we get our answer: the current value of n.
n
::
::  If not, we run the whole program again,
::  but this time n is replaced by (n+1).
::  n will be incremented by 1 until our
::  ?: question has the answer "true",
::  upon which it will return n.
::  If n is 0, it will now be 1.
::  If n is 25,263, it will now be 25,264 which,
::  as this program shows, is the smallest positive
::  integer whose square ends in the digits 269,696.
::
::  %= runs the whole program again, but
::     with the listed names, on the left, assigned
::     new values on the right. It could take an
::     arbitrary number of children, so unlike the
::     other runes here which take a fixed number,
::     this expression must end in a == rune, which
::     just brings the expression to an end.

## Mathematica/Wolfram Language

Solving up to his guess would show that there is indeed a smaller integer with that property.

 Solve[Mod[x^2, 10^6] == 269696 && 0 <= x <= 99736, x, Integers]

Output:
{{x->25264},{x->99736}}

## MAXScript

-- MAXScript : Babbage problem : N.H.
posInt = 1
while posInt < 1000000 do
(
if (matchPattern((posInt * posInt) as string) pattern: "*269696") then exit
posInt += 1
)
Print "The smallest number whose square ends in 269696 is " + ((posInt) as string)
Print "Its square is " + (((pow posInt 2) as integer) as string)
Output:

Output to MAXScript Listener:

"The smallest number whose square ends in 269696 is 25264"
"Its square is 638269696"


## Microsoft Small Basic

Translation of: VBscript
' Babbage problem
' The quote (') means a comment
' The equals sign (=) means assign
n = 500
' 500 is stored in variable n*n
' 500 because 500*500=250000 less than 269696

' The nitty-gritty is in the 3 lines between "While" and "EndWhile".
' So, we start with 500, n is being incremented by 1 at each round
' while its square (n*n) (* means multiplication) does not have
' a remainder (function Math.Remainder) of 269696 when divided by one million.
' This means that the loop will stop when the smallest positive integer
' whose square ends in 269696
' is found and stored in n.
' (<>)  means "not equal to"
While Math.Remainder( n*n , 1000000 ) <> 269696
n = n + 1
EndWhile

' (TextWindow.WriteLine) displays the string to the monitor
' (+) concatenates strings or variables to be displayed
TextWindow.WriteLine("The smallest positive integer whose square ends in 269696 is " + (n) + ".")
TextWindow.WriteLine("Its square is " + (n*n) + ".")

' End of Program.
Output:
The smallest positive integer whose square ends in 269696 is 25264.
Its square is 638269696.


## MiniScript

// Lines that start with "//" are "comments" that are ignored
// by the computer.  We use them to explain the code.

// Start by finding the smallest number that could possibly
// square to 269696.  sqrt() returns the square root of a
// number, and floor() truncates any fractional part.
k = floor(sqrt(269696))

// Since 269696 is even, we are only going to consider even
// roots.  We use the % (modulo) operator, which returns the
// remainder after division, to tell if k is odd; if so, we
// add 1 to make it even.
if k % 2 == 1 then k = k + 1

// Now we count up by 2 from k, until we find a number that,
// when squared, ends in 269696 (using % again).
while k^2 % 1000000 != 269696
k = k + 2
end while

// The first such number we find is our answer.
print k + "^2 = " + k^2

Output:
25264^2 = 638269696

## Modula-2

MODULE BabbageProblem;
FROM FormatString IMPORT FormatString;
FROM RealMath IMPORT sqrt;

VAR
buf : ARRAY[0..63] OF CHAR;
k : INTEGER;
BEGIN
(* Find the greatest integer less than the square root *)
k := TRUNC(sqrt(269696.0));

(* Odd numbers cannot be solutions, so decrement *)
IF k MOD 2 = 1 THEN
DEC(k);
END;

(* Find a number that meets the criteria *)
WHILE (k*k) MOD 1000000 # 269696 DO
INC(k,2)
END;

FormatString("%i * %i = %i", buf, k, k, k*k);
WriteString(buf);

END BabbageProblem.


## Nanoquery

n = 0

while not (n ^ 2 % 1000000) = 269696
n += 1
end

println n
Output:
25264

## NetRexx

Translation of: Groovy
/* NetRexx */
options replace format comments java crossref symbols nobinary utf8
numeric digits 5000 -- set up numeric precision

babbageNr = babbage() -- call a function to perform the analysis and capture the result
babbageSq = babbageNr ** 2  -- calculate the square of the result
-- display results using a library function
System.out.printf("%,10d\u00b2 == %,12d%n", [Integer(babbageNr), Integer(babbageSq)])
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- A function method to answer Babbage's question:
--   "What is the smallest positive integer whose square ends in the digits 269,696?"
--     — Babbage, letter to Lord Bowden, 1837;
--       see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.
--   (He thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.)

method babbage() public static binary
n = int 104 -- (integer arithmatic)
-- begin a processing loop to determine the value
-- starting point: 104
loop while ((n * n) // 1000000) \= 269696
-- loop continues while the remainder of n squared divided by 1,000,000 is not equal to 269,696
if n // 10 == 4 then do
-- increment n by 2 if the remainder of n divided by 10 equals 4
n = n + 2
end
if n // 10 == 6 then do
-- increment n by 8 if the remainder of n divided by 10 equals 6
n = n + 8
end
end

return n -- end the function and return the result

Output:
    25,264² ==  638,269,696

## NewLisp

;;;	Start by assigning n the integer square root of 269696
;;;	minus 1 to be even
(setq n 518)
;;;	Increment n by 2 till the last 6 digits of its square are 269696
(while (!= (% (* n n) 1000000) 269696)
(++ n 2))
;;;	Show the result and its square
(println n "^2 = " (* n n))

Output:
25264^2 = 638269696


## Nim

var n : int = 0
while n*n mod 1_000_000 != 269_696:
inc(n)
echo n

Output:
25264

## Objeck

class Babbage  {
function : Main(args : String[]) ~ Nil {
cur := 0;
do {
cur++;
}
while(cur * cur % 1000000 <> 269696);

cur_sqr := cur * cur;
"The square of {$cur} is {$cur_sqr}!"->PrintLine();
}
}
Output:
The square of 25264 is 638269696!


## OCaml

let rec f a=
if (a*a) mod 1000000 != 269696
then f(a+1)
else a
in
let a= f 1 in
Printf.printf "smallest positive integer whose square ends in the digits 269696 is %d\n" a


## Ol

(print
(let loop ((i 2))
(if (eq? (mod (* i i) 1000000) 269696)
i
(loop (+ i 1)))))

Output:
25264

## PARI/GP

m=269696;
k=1000000;
{for(n=1,99736,
\\ Try each number in this range, from 1 to 99736
if(denominator((n^2-m)/k)==1, \\ Check if n squared, minus m, is divisible by k
return(n) \\ If so, return this number and STOP.
)
)}

## Pascal

program BabbageProblem;
(* Anything bracketed off like this is an explanatory comment. *)
var n : longint; (* The VARiable n can hold a 'long', ie large, INTeger. *)
begin
repeat
n := n + 2 (* Increase n by 2. *)
until (n * n) mod 1000000 = 269696;
(* 'n * n' means 'n times n'; 'mod' means 'modulo'. *)
write(n)
end.

Output:
25264

## Perl

### Naive

#!/usr/bin/perl
use strict ;
use warnings ;

my $current = 0 ; while ( ($current ** 2 ) % 1000000  != 269696 ) {
$current++ ; } print "The square of$current is " . ($current *$current) . " !\n" ;

Output:
The square of 25264 is 638269696 !

### More efficient and well documented

I am pretty sure that comments and efficient code would be appreciated by Mr. Babbage.

# Dear Mr. Babbage,
# in the following code, $number is the variable to contain the number # we're looking for. '%' is the modulo division operator and '!=' # means "not equal to". # We start to search with 518, because no smaller number can be even # squared to 269696. We also increment the number always by 2, as no # odd number can be part of the solution. We do this, because # computational power - even in 2022 - does not grow on trees. # And even if it did, we have less of them than you did. my$number = 518;
while ( ($number ** 2) % 1000000 != 269696 ) {$number += 2 ;
}
print "The square of $number is " . ($number ** 2) . " !\n";

Output:
The square of 25264 is 638269696 !

## Phix

We can omit anything odd, as any odd number squared is obviously always odd.
Mr Babbage might need the whole "i is a variable" thing explained, and that "?i" prints the value of i, nowt else springs to mind.

with javascript_semantics
for i=2 to 99736 by 2 do
if remainder(i*i,1000000)=269696 then ?i exit end if
end for

Output:
25264


### Proper method

This might confuse poor old Mr Babbage and require much more explanation, but I like to think he'd be quietly impressed with the approach.
Create a list of one-digit numbers whose square ends in 6, then use that list to create/filter all two-digit numbers whose square ends in 96, and so on.
For simplicity and with some poetic licence we start with {0} as the list of all zero-digit numbers that end in "".

with javascript_semantics
procedure solve_babbage_problem() // (so that return quits 3 loops)
sequence s = {0}, t
integer p10 = 1, r10 = 10, cc = 0
for digits = 1 to 6 do
t = {}
for prefix=0 to 9 do
for i=1 to length(s) do
integer cand = prefix*p10+s[i]
atom cand2 = cand*cand
cc += 1
if remainder(cand2,r10) = remainder(269696,r10) then
if digits=6 then
printf(1,"Solution: %d (%d calcs)\n",{cand,cc})
return
end if
t &= cand
end if
end for
end for
s = t
printf(1,"%d-digit candidates: %v\n",{digits,s})
p10 *= 10
r10 *= 10
end for
end procedure
solve_babbage_problem()

Output:

As you can see this would be significantly quicker (by hand) than adding 2 about 12,500 times, or even starting from 514 and only checking numbers that end in 4 or 6, still nearly 5,000 of 'em, whereas this is a mere 193 calculations/squares. Of course the answer we are looking for is the first/lowest 6-digit candidate found. Note that 400264 (and others) being a potential answer precludes the use of a depth-first search, forcing a full breadth-first approach.

1-digit candidates: {4,6}
2-digit candidates: {14,36,64,86}
3-digit candidates: {236,264,736,764}
4-digit candidates: {264,2236,2764,4736,5264,7236,7764,9736}
5-digit candidates: {264,24736,25264,49736,50264,74736,75264,99736}
Solution: 25264 (193 calcs)


For reference and from a previous version, the full list of 6-digit candidates would be:

6-digit candidates: {25264,99736,150264,224736,275264,349736,400264,474736,525264,599736,650264,724736,775264,849736,900264,974736}


## PHP

<?php

for (
$i = 1 ; // Initial positive integer to check ($i * $i) % 1000000 !== 269696 ; // While i*i does not end with the digits 269,696$i++                                        // ... go to next integer
);

echo $i, ' * ',$i, ' = ', ($i *$i), PHP_EOL;  // Print the result

Output:
25264 * 25264 = 638269696

## Picat

### Iterative

main =>
N = 1,
while (N**2 mod 1000000 != 269696)
N := N + 1
end,
println(N).
Output:
25264

### Constraint modelling

import cp.

main =>
N*N mod 1000000 #= 269696,   % ends with "269696"
N #> 0,                      % positive integer
solve($[min(N)],[N]), % minimize N println(n=N). Output: 25264 ## PicoLisp : (for N 99736 # Iterate N from 1 to 99736 (T (= 269696 (% (* N N) 1000000)) N) ) # Stop if remainder is 269696 -> 25264 ## PILOT Remark:Lines identified as "remarks" are intended for the human reader, and will be ignored by the machine. Remark:A "compute" instruction gives a value to a variable. Remark:We begin by making the variable n equal to 2. Compute:n = 2 Remark:Lines beginning with asterisks are labels. We can instruct the machine to "jump" to them, rather than carrying on to the next instruction as it normally would. *CheckNextNumber Remark:In "compute" instructions, "x * y" should be read as "x times y" and "x % y" as "x modulo y". Compute:square = n * n Compute:lastSix = square % 1000000 Remark:A "jump" instruction that includes an equation or an inequality in parentheses jumps to the designated label if and only if the equation or inequality is true. Jump( lastSix = 269696 ):*FoundIt Remark:If the last six digits are not equal to 269696, add 2 to n and jump back to "CheckNextNumber". Compute:n = n + 2 Jump:*CheckNextNumber *FoundIt Remark:Type, i.e. print, the result. The symbol "#" means that what follows is one of our variables and the machine should type its value. Type:The smallest number whose square ends in 269696 is #n. Its square is #square. Remark:The end. End: ## Plain English To run: Start up. Put 1 into a number. Loop. Divide the number times the number by 1000000 giving a quotient and a remainder. If the remainder is 269696, break. Bump the number. Repeat. Write "The answer is " then the number on the console. Wait for the escape key. Shut down. Output: The answer is 25264  ## PL/I /* Babbage might have used a difference engine to compute squares. */ /* The algorithm used here uses only additions to form successive squares. */ /* Since there is no guarantee that the final square will not exceed a */ /* 32-bit integer word, modulus is formed to limit the magnitude of the */ /* squares, since we are really interested only in the last six digits of */ /* the square. */ Babbage_problem: procedure options (main); /* R. Vowels, 19 Dec. 2021 */ declare n fixed decimal (5); declare (odd, sq) fixed binary (31); odd = 3; sq = 4; /* the initial square is 4 */ do n = 3 to 99736; odd = odd + 2; sq = sq + odd; /* form the next square */ if sq >= 1000000 then sq = sq - 1000000; /* keep the remainder */ if sq = 269696 then leave; end; put ('The smallest number whose square ends in 269696 is ' || trim(n) ); put skip list ('The corresponding square is ' || trim (n*n) ); /* Even if the number had been 99736, n*n would not have overflowed */ /* because decimal arithmetic allows up to 15 decimal digits. */ end Babbage_problem; Output: The smallest number whose square ends in 269696 is 25264 The corresponding square is 638269696  ## PowerShell ########################################################################################### # # Definitions: # # Lines that begin with the "#" symbol are comments: they will be ignored by the machine. # # ----------------------------------------------------------------------------------------- # # While # # Run a command block based on the results of a conditional test. # # Syntax # while (condition) {command_block} # # Key # # condition If this evaluates to TRUE the loop {command_block} runs. # when the loop has run once the condition is evaluated again. # # command_block Commands to run each time the loop repeats. # # As long as the condition remains true, PowerShell reruns the {command_block} section. # # ----------------------------------------------------------------------------------------- # # * means 'multiplied by' # % means 'modulo', or remainder after division # -ne means 'is not equal to' # ++ means 'increment variable by one' # ########################################################################################### # Declare a variable,$integer, with a starting value of 0.

$integer = 0 while (($integer * $integer) % 1000000 -ne 269696) {$integer++
}

# Show the result.

$integer  Output: 25264  Alternative method Works with: PowerShell version 2 By looping through potential squares instead of potential square roots, we reduce the number of loops by a factor of 40. # Start with the smallest potential square number$TestSquare = 269696

#  Test if our potential square is a square
#  by testing if the square root of it is an integer
#  Test if the square root is an integer by testing if the remainder
#  of the square root divided by 1 is greater than zero
#  % is the remainder operator
#  -gt is the "greater than" operator

#  While the remainder of the square root divided by one is greater than zero
While ( [Math]::Sqrt( $TestSquare ) % 1 -gt 0 ) { # Add 100,000 to get the next potential square number$TestSquare = $TestSquare + 1000000 } # This will loop until we get a value for$TestSquare that is a square number

#  Caclulate the root
$Root = [Math]::Sqrt($TestSquare )

#  Display the result and its square
$Root$TestSquare

Output:
25264
638269696

## Processing

// Lines that begin with two slashes, thus, are comments: they
// will be ignored by the machine.

// First we must declare a variable, n, suitable to store an integer:

int n;

// Each statement we address to the machine must end with a semicolon.

// To begin with, the value of n will be zero:

n = 0;

// Now we must repeatedly increase it by one, checking each time to see
// whether its square ends in 269,696.

// We shall do this by seeing whether the remainder, when n squared
// is divided by one million, is equal to 269,696.

do {
n = n + 1;
} while (n * n % 1000000 != 269696);

// To read this formula, it is necessary to know the following
// elements of the notation:
//     * means 'multiplied by'
//     % means 'modulo', or remainder after division
//     != means 'is not equal to'

// Now that we have our result, we need to display it.

// println is short for 'print line'

println(n);

Output:
25264

## Prolog

Works with Swi-Prolog version 7+

:- use_module(library(clpfd)).

babbage_(B, B, Sq) :-
B * B #= Sq,
number_chars(Sq, R),
append(_, ['2','6','9','6','9','6'], R).
babbage_(B, R, Sq) :-
N #= B + 1,
babbage_(N, R, Sq).

babbage :-
once(babbage_(1, Num, Square)),
format('lowest number is ~p which squared becomes ~p~n', [Num, Square]).

Output:
1 ?- babbage.
lowest number is 25264 which squared becomes 638269696
true.


## PureBasic

EnableExplicit
Macro putresult(n)
If OpenConsole("Babbage_problem")
PrintN("The smallest number whose square ends in 269696 is " + Str(n))
Input()
EndIf
EndMacro

CompilerIf #PB_Processor_x64
#MAXINT = 1 << 63 - 1
CompilerElseIf #PB_Processor_x86
#MAXINT = 1 << 31 - 1
CompilerEndIf

#GOAL = 269696
#DIV  = 1000000
Define n.i, q.i = Int(Sqr(#MAXINT))

For n = 2 To q Step 2
If (n*n) % #DIV = #GOAL : putresult(n) : Break : EndIf
Next

Output:
The smallest number whose square ends in 269696 is 25264

## Python

# Lines that start by # are a comments:
# they will be ignored by the machine

n=0 # n is a variable and its value is 0

# we will increase its value by one until
# its square ends in 269,696

while n**2 % 1000000 != 269696:

# n**2 -> n squared
# %    -> 'modulo' or remainer after division
# !=   -> not equal to

n += 1 # += -> increase by a certain number

print(n) # prints n

# short version
>>> [x for x in range(30000) if (x*x) % 1000000 == 269696] [0]
25264

Output:
25264

Or, generating a non-finite stream of numbers which are the sum of 269696 and some integer multiple of one million, and also have an integer square root:

Works with: Python version 3.7
'''Babbage problem'''

from math import (floor, sqrt)
from itertools import (islice)

# squaresWithSuffix :: Int -> Gen [Int]
def squaresWithSuffix(n):
'''Non finite stream of squares with a given suffix.'''
stem = 10 ** len(str(n))
i = 0
while True:
i = until(lambda x: isPerfectSquare(n + (stem * x)))(
succ
)(i)
yield n + (stem * i)
i = succ(i)

# isPerfectSquare :: Int -> Bool
def isPerfectSquare(n):
'''True if n is a perfect square.'''
r = sqrt(n)
return r == floor(r)

# TEST ----------------------------------------------------

# main :: IO ()
def main():
'''Smallest positive integers whose squares end in the digits 269,696'''
print(
fTable(main.__doc__ + ':\n')(
lambda n: str(int(sqrt(n))) + '^2'
)(repr)(identity)(
take(10)(squaresWithSuffix(269696))
)
)

# GENERIC -------------------------------------------------

# identity :: a -> a
def identity(x):
'''The identity function.'''
return x

# succ :: Enum a => a -> a
def succ(x):
'''The successor of a value.
For numeric types, (1 +).
'''
return 1 + x if isinstance(x, int) else (
chr(1 + ord(x))
)

# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''
return lambda xs: (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)

# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.
'''
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)

# FORMATTING ----------------------------------------------
# fTable :: String -> (a -> String) ->
#                     (b -> String) -> (a -> b) -> [a] -> String
def fTable(s):
'''Heading -> x display function -> fx display function ->
f -> xs -> tabular string.
'''
def go(xShow, fxShow, f, xs):
ys = [xShow(x) for x in xs]
w = max(map(len, ys))
return s + '\n' + '\n'.join(map(
lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),
xs, ys
))
return lambda xShow: lambda fxShow: lambda f: lambda xs: go(
xShow, fxShow, f, xs
)

# MAIN ---
if __name__ == '__main__':
main()

Output:
Smallest positive integers whose squares end in the digits 269,696:

25264^2 -> 638269696
99736^2 -> 9947269696
150264^2 -> 22579269696
224736^2 -> 50506269696
275264^2 -> 75770269696
349736^2 -> 122315269696
400264^2 -> 160211269696
474736^2 -> 225374269696
525264^2 -> 275902269696
599736^2 -> 359683269696

[Finished in 0.285s]

As a footnote on what Babbage might have managed with pencil and paper – applying the squaresWithSuffix(n) function to shorter suffixes (6, 96, 696 ...) enables us to explore the way in which the final digits of the integer root constrain and determine those of the perfect square. It quickly becomes apparent that Mr Babbage need only have considered roots ending in the digit sequences 264 or 736, a constraint which, had he deduced it with pencil and paper, would have allowed him to reach 25264 after testing the squares of only 24 other numbers.

## Quackery

(   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   )

( The computer will ignore anything between parentheses, providing there is
a space or carriage return on either side of each parenthesis.            )

(   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   )

( A number squared is that number, multiplied by itself.

Here, "dup" means "make a duplicate" and "*" means "multiply two numbers",
So "dup *" means "make a duplicate of a number and multiply it by itself".
We will tell the computer that this action is called "squared".           )

[ dup * ] is squared

( "squared" takes a number and returns that number, squared.                )

(   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   )

( A number n ends with the digits 269696 if n mod 1000000 = 269696,
When expressed in postfix notation this is: 1000000 mod 269696 =
We will tell the computer that this test is called "ends.with.269696".    )

[ 1000000 mod 269696 = ] is ends.with.269696

( "ends.with.269696" takes a number and returns true if it ends with 269696,
and false if it does not.                                                 )

(   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   )

( The square root of 269696 is approximately equal to 518.35, so we need not
consider numbers less than 519.

Starting withe 518, we will intruct the computer to repeatedly add 2 to
the number using the command "2 +" (because the number we are looking for
must be an even number as the square root of any even perfect square
must be an even number), then square a duplicate of the number until a
value is found of which its square ends with 269696.

"until" will take the truth value returned by "ends.with.269696" and, if
that value is "false", will cause the previous commands from the nearest
"[" preceding "until" to be repeated.

When the value is "true" the computer will proceed to the word "echo",
which will take the calculated solution and display it on the screen.     )

518 [ 2 + dup squared ends.with.269696 until ] echo

(   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   .   )
Output:
25264

## R

babbage_function=function(){
n=0
while (n**2%%1000000!=269696) {
n=n+1
}
return(n)
}
babbage_function()[length(babbage_function())]

Output:
25264

## Racket

;; Text from a semicolon to the end of a line is ignored
;; This lets the racket engine know it is running racket
#lang racket

;; “define” defines a function in the engine
;; we can use an English name for the function
;; a number ends in 269696 when its remainder when
;; divided by 1000000 is 269696 (we omit commas in
;; numbers... they are used for another reason).
(define (ends-in-269696? x)
(= (remainder x 1000000) 269696))

;; we now define another function square-ends-in-269696?
;; actually this is the composition of ends-in-269696? and
;; the squaring function (which is called “sqr” in racket)
(define square-ends-in-269696? (compose ends-in-269696? sqr))

;; a for loop lets us iterate (it’s a long Latin word which
;; Victorians are good at using) over a number range.
;;
;; for/first go through the range and break when it gets to
;; the first true value
;;
;; (in-range a b) produces all of the integers from a (inclusive)
;; to b (exclusive). Because we know that 99736² ends in 269696,
;; we will stop there. The add1 is to make in-range include 99736
;;
;; we define a new variable, so that we can test the verity of
;; our result
(define first-number-that-when-squared-ends-in-269696
(for/first ((i ; “i” will become the ubiquetous looping variable of the future!
; when returns when only the first one that matches
#:when (square-ends-in-269696? i))
i))

;; display prints values out; newline writes a new line (otherwise everything
;; gets stuck together)
(display first-number-that-when-squared-ends-in-269696)
(newline)
(display (sqr first-number-that-when-squared-ends-in-269696))
(newline)
(newline)
(display (ends-in-269696? (sqr first-number-that-when-squared-ends-in-269696)))
(newline)
(display (square-ends-in-269696? first-number-that-when-squared-ends-in-269696))
(newline)
;; that all seems satisfactory

Output:
25264
638269696

#t
#t

## Raku

(formerly Perl 6)

This could certainly be written more concisely. Extra verbiage is included to make the process more clear.

# For all positives integers from 1 to Infinity
for 1 .. Inf -> $integer { # calculate the square of the integer my$square = $integer²; # print the integer and square and exit if the square modulo 1000000 is equal to 269696 print "{$integer}² equals $square" and exit if$square mod 1000000 == 269696;
}

Output:
25264² equals 638269696

More concisely (but clear enough to allow Mr. Babbage to understand what's going on?):

.say and exit if $_² % 1e6 == 269696 for ^∞;  Output: 25264 ## Red Red [] number: 510 ;; starting number ;; repeat, until the last condition in the block is true until [ number: number + 2 ;; only even numbers can have even squares ;; The word modulo computes the non-negative remainder of the ;; first argument divided by the second argument. ;; ** => Returns a number raised to a given power (exponent) 269696 = modulo (number ** 2) 1000000 ] ?? number  Output: number: 25264 ## REXX Each of the first three REXX programs were constructed to be as simple as possible so as to be easily understood by a mathematican (in Mr. Charles Babbage's time). For instance, in Charles Babbage's era, multiplication of j and j would be jj (implied multiplication), so it would be necessary to explain that an asterisk (*) means multiplication. Another form of multiplication in his day would be: j x j or j ∙ j. Most mathematical or algebraic texts used simple letters for values (we would now call them variables when dealing with computer programs). Fortunately, the REXX language uses decimal numbers, so binary values don't need to be explained. So, with that in mind, the use of (REXX) arithermetic operators were explained within a comment, as well as trying to explain some statements in the REXX language. And, further comments probably should've been added to explain what a comment is in the REXX language (and for that matter, comments should've been included for each and every REXX statement explaining what the instruction (statement) does and what the nomenclature means. Fortunately, most REXX statements are easy understood (one of REXX's design goals) and (most likely) are intuitively understood (at least on a fundamental or basic level), although a do or for loop would be confusing without a detailed explanation of what a loop is. A computer program (in his day) would've been undstood to be list of instructions to a (human) computer to be performed, quite literally. For instance, Mr. Babbage would know of a typewriter (machine), and, to the human computers in his day and age, they would've known (or intuited) what a type or print instruction would do, and he would recognize that the (human) computer would use a typewriter. An idle monitor on the other hand (I suspect), he would be baffled and he wouldn't probably know of it's function. When not displaying anything, it looks like an ineffective mirror, possibly having a blinking cursor, whatever that is. (In those days of yore, a cursor was the movable transparent slide on a slide rule.) If this were a computer program to be shown to a computer programming novice (albeit a very intelligent polymath novice), the computer program would also have a lot more comments, notes, and accompanying verbiage which would/could/should explain: • what a (computer program) comment looks like • what a computer is • what a computer program is • how a computer stores numbers and such • what are variables and how to store stuff in them • how the do loop works (initial value, incrementation, etc) • how an assignment = operator works • how a comparison == operator works • how an if statement works • what a (computer program) statement is • what the * operator is and how it does multiplication • what the + operator is and how it does addition • what the // operator is and how it does division remainder • what the right BIF does • who what a BIF is and how it returns a value • how/when the then cause gets executed (after an if) • explain how/why an end statement is needed for a do loop • explain how a leave statement works • ··· the say is probably the only statement that is self─explanatory ### examine the right-most six digits of square /*REXX program finds the lowest (positive) integer whose square ends in 269,696. */ do j=2 by 2 until right(j * j, 6) == 269696 /*start J at two, increment by two. */ end /*◄── signifies the end of the DO loop.*/ /* [↑] * means multiplication. */ say "The smallest integer whose square ends in 269,696 is: " j  output: The smallest integer whose square ends in 269,696 is: 25264  ### examine remainder after dividing by one million /*REXX program finds the lowest (positive) integer whose square ends in 269,696. */ do j=2 by 2 /*start J at two, increment by two. */ if ((j * j) // 1000000) == 269696 then leave /*is square mod one million our target?*/ end /*◄── signifies the end of the DO loop.*/ /* [↑] // is division remainder.*/ say "The smallest integer whose square ends in 269,696 is: " j  output is identical to the 1st REXX version. ### examine only numbers ending in 4 or 6 /*REXX program finds the lowest (positive) integer whose square ends in 269,696. */ /*─────────────────── we will only examine integers that are ending in four or six. */ do j=4 by 10 /*start J at four, increment by ten.*/ k = j /*set K to J's value. */ if right(k * k, 6) == 269696 then leave /*examine right-most 6 decimal digits. */ /* == means exactly equal to. */ k = j+2 /*set K to J+2 value. */ if right(k * k, 6) == 269696 then leave /*examine right-most 6 decimal digits. */ end /*◄── signifies the end of the DO loop.*/ /* [↑] * means multiplication. */ say "The smallest integer whose square ends in 269,696 is: " k  output is identical to the 1st REXX version. ### start with smallest possible number /*REXX ---------------------------------------------------------------- * The solution must actually be larger than sqrt(269696)=519.585 *--------------------------------------------------------------------*/ z=0 Do i=524 By 10 Until z>0 If right(i*i,6)==269696 then z=i Else Do j=i+2 if right(j*j,6)==269696 then z=j End End Say "The smallest integer whose square ends in 269696 is:" z Say ' 'z'**2 =' z**2  Output: The smallest integer whose square ends in 269696 is: 25264 25264**2 = 638269696 ## Ring n = 0 while pow(n,2) % 1000000 != 269696 n = n + 1 end see "The smallest number whose square ends in 269696 is : " + n + nl see "Its square is : " + pow(n,2) Output: The smallest number whose square ends in 269696 is : 25264 Its square is : 638269696  ## Ruby n = 0 n = n + 2 until (n*n).modulo(1000000) == 269696 print n  ## Run BASIC for n = 1 to 1000000 if n^2 MOD 1000000 = 269696 then exit for next PRINT "The smallest number whose square ends in 269696 is "; n PRINT "Its square is "; n^2 The smallest number whose square ends in 269696 is 25264 Its square is 638269696  ## Rust ### Imperative version fn main() { let mut current = 0; while (current * current) % 1_000_000 != 269_696 { current += 1; } println!( "The smallest number whose square ends in 269696 is {}", current ); }  Output: The smallest number whose square ends in 269696 is 25264  ### Finding in infinite range fn main() { if let Some(n) = (1..).find(|x| x * x % 1_000_000 == 269_696) { println!("The smallest number whose square ends in 269696 is {}", n) } }  Which outputs the same thing as above. ## Scala object BabbageProblem { def main( args:Array[String] ): Unit = { var x : Int = 524 // Sqrt of 269696 = 519.something while( (x * x) % 1000000 != 269696 ){ if( x % 10 == 4 ) x = x + 2 else x = x + 8 } println("The smallest positive integer whose square ends in 269696 = " + x ) } }  Output: The smallest positive integer whose square ends in 269696 = 25264  or same solution with a functional implementation import scala.annotation.tailrec object BabbageProblem { @tailrec def findItFrom(x: Int): Int = if ((x * x) % 1000000 == 269696) x else findItFrom( if (x % 10 == 4) x + 2 else x + 8 ) def findIt: Int = findItFrom(524) // Sqrt of 269696 = 519.something def main(args: Array[String]): Unit = println("The smallest positive integer whose square ends in 269696 = " + findIt) }  Output: The smallest positive integer whose square ends in 269696 = 25264  ## Scheme (define (digits n) (string->list (number->string n))) (define (ends-with list tail) ;; does list end with tail? (starts-with (reverse list) (reverse tail))) (define (starts-with list head) (cond ((null? head) #t) ((null? list) #f) ((equal? (car list) (car head)) (starts-with (cdr list) (cdr head))) (else #f))) (let loop ((i 1)) (if (ends-with (digits (* i i)) (digits 269696)) i (loop (+ i 1)))) ;; 25264  ## Scilab Translation of: Pascal n=2; flag=%F while ~flag n = n+2; if pmodulo(n*n,1000000)==269696 then flag=%T; end end disp(n);  Output:  25264. ## Seed7 $ include "seed7_05.s7i";

const proc: main is func
local
var integer: current is 0;
begin
while current ** 2 rem 1000000 <> 269696 do
incr(current);
end while;
writeln("The square of " <& current <& " is " <& current ** 2);
end func;
Output:
The square of 25264 is 638269696


## SenseTalk

(*
What is the smallest positive integer whose square ends in the digits 269,696?
*)

put 1 into int

repeat forever
if int squared ends with "269696" then
put "The smallest positive integer whose square ends in the digits 269696 is" && int
exit all -- don't keep repeating forever!
end if

end repeat

Note that because SenseTalk sees things as an ordinary person does, there is no need to make a distinction between a number and its text representation.

## SequenceL

main() := babbage(0);

babbage(current) :=
current when current * current mod 1000000 = 269696
else
babbage(current + 1);
Output:
cmd:> babbage.exe
25264


## Shen

(define babbage
N -> N where (= 269696 (shen.mod (* N N) 1000000)))
N -> (babbage (+ N 1))

(babbage 1)

Output:
25264

## Sidef

var n = 0
while (n*n % 1000000 != 269696) {
n += 2
}
say n

Output:
25264


## Simula

BEGIN
INTEGER PROBE, SQUARE;
BOOLEAN DONE;

WHILE NOT DONE DO BEGIN
PROBE := PROBE + 1;
SQUARE := PROBE * PROBE;
IF MOD(SQUARE, 1000000) = 269696 THEN BEGIN

OUTTEXT("THE SMALLEST NUMBER: ");
OUTINT(PROBE,0);
OUTIMAGE;

OUTTEXT("THE SQUARE : ");
OUTINT(SQUARE,0);
OUTIMAGE;

DONE := TRUE;
END;
END;

END
Output:
THE SMALLEST NUMBER: 25264
THE SQUARE : 638269696


## Smalltalk

"We use one variable, called n. Let it initially be equal to 1. Then keep increasing it by 1 for only as long as the remainder after dividing by a million is not equal to 269,696; finally, show the value of n."
| n |
n := 1.
[ n squared \\ 1000000 = 269696 ] whileFalse: [ n := n + 1 ].
n

Output:
25264

## SQL

Works with: ORACLE 19c
/*
This code is an implementation of Babbage Problem in SQL ORACLE 19c
p_max   -- upper bound of the cycle
v_max   -- safe determination of the upper bound of the cycle
v_start -- safe starting point
*/
with
function babbage(p_ziel in varchar2, p_max integer) return varchar2 is
v_max    integer := greatest(p_max,to_number('1E+'||length(to_char(p_ziel))));
v_start  number := case when substr(p_ziel,1,1)='0' then ceil(sqrt('1'||p_ziel)) else ceil(sqrt(p_ziel)) end;
v_length number := to_number('1E+'||length(to_char(p_ziel)));
begin
-- first check
if substr(p_ziel,-1) in (2,3,7,8) then
return 'The exact square of an integer cannot end with '''||substr(p_ziel,-1)||''', so there is no such smallest number whose square ends in '''||p_ziel||'''';
end if;
-- second check
if regexp_count(p_ziel,'([^0]0{1,}$)')=1 and mod(regexp_count(regexp_substr(p_ziel,'(0{1,}$)'),'0'),2)=1 then
return 'An exact square of an integer cannot end with an odd number of zeros, so there is no such smallest number whose square ends in '''||p_ziel||'''';
end if;
-- main cycle
while v_start < v_max loop
exit when mod(v_start**2,v_length) = p_ziel;
v_start := v_start + 1;
end loop;
-- output
if v_start = v_max then
return 'There is no such smallest number before '||v_max||' whose square ends in '''||p_ziel||'''';
else
return ''||v_start||' is the smallest number, whose square '||regexp_replace(to_char(v_start**2),'(\d{1,})('||p_ziel||')','\1''\2''')||' ends in '''||p_ziel||'''';
end if;
--
end;

--Test
select babbage('222',100000) as res from dual
union all
select babbage('33',100000) as res from dual
union all
select babbage('17',100000) as res from dual
union all
select babbage('888',100000) as res from dual
union all
select babbage('1000',100000) as res from dual
union all
select babbage('000',100000) as res from dual
union all
select babbage('269696',100000) as res from dual -- strict Babbage Problem
union all
select babbage('269696',10) as res from dual
union all
select babbage('169696',10) as res from dual
union all
select babbage('19696',100000) as res from dual
union all
select babbage('04',100000) as res from dual;

Output:
The exact square of an integer cannot end with '2', so there is no such smallest number whose square ends in '222'
The exact square of an integer cannot end with '3', so there is no such smallest number whose square ends in '33'
The exact square of an integer cannot end with '7', so there is no such smallest number whose square ends in '17'
The exact square of an integer cannot end with '8', so there is no such smallest number whose square ends in '888'
An exact square of an integer cannot end with an odd number of zeros, so there is no such smallest number whose square ends in '1000'
100 is the smallest number, whose square 10'000' ends in '000'
25264 is the smallest number, whose square 638'269696' ends in '269696'
25264 is the smallest number, whose square 638'269696' ends in '269696'
There is no such smallest number before 1000000 whose square ends in '169696'
12236 is the smallest number, whose square 1497'19696' ends in '19696'
48 is the smallest number, whose square 23'04' ends in '04'


## Swift

import Swift

for i in 2...Int.max {
if i * i % 1000000 == 269696 {
print(i, "is the smallest number that ends with 269696")
break
}
}

Output:
25264 is the smallest number that ends with 269696

## Tcl

Hope Mr Babbage can understand this one-liner...

for {set i 1} {![string match *269696 [expr $i*$i]]} {incr i} {}
puts "$i squared is [expr$i*$i]"  25264 squared is 638269696  ### Scalable version I have added some comments to the code, but left out a Tcl tutorial. # The last k digits of a square x**2 are determined by the last k digits of ## the number x, and independent of all higher digits. ## We search for possible endings (suffixes) of x from the right, and extend ## by one digit in each step. That way we think to have an algorithm, ## which scales very well to really long suffixes. ## Also, when there is no solution, we detect that and terminate orderly. namespace path {::tcl::mathop} ;# commands like: + - * interp alias {} LEN {} string length ## Normalize a string (of digits) as (decimal) number. ## Tcl still thinks, numbers starting with "0" are meant octal (base 8). ## We do not even have a qualifying prefix like "0d". ## Here we just handle such leading zeroes. proc normNumber {str} { set str [string trimleft$str 0]    ;# drop all leading zeroes
if {$str eq ""} { ;# if string is completely empty set str "0" ;# we leave a single zero digit } return$str
}

##      If possible, cuts off some left part of $str to yield a suffix ## of at most length$wantlen.
proc cutSuff {str wantlen} {
set havelen [LEN $str] set toskip [-$havelen $wantlen] return [string range$str $toskip end] ;# treats negative$toskip as 0
}

##      We have some numeric suffix string $str, and we need it for ## the specified length$wantlen, either by taking a shorter suffix,
##      or by filling up with zeroes at the left.
proc numToExactSuffLen {str wantlen} {
set havelen [LEN $str] if {$havelen > $wantlen} { ;# cut down in length set toskip [expr {$havelen - $wantlen}] return [string range$str $toskip end] } elseif {$havelen < $wantlen} { ;# pad zeroes at left end set topad [expr {$wantlen - $havelen}] return "[string repeat "0"$topad]$str" } return$str
}

##      Compute the square of a number, given as a string
proc stringSquare {str} {
set numstr [normNumber $str] ;# make proper number from string return [*$numstr $numstr] ;# square the number } ## We search for a square that has suffix$totend.
##      So far we have constructed suffixes of some (small) length k,
##      which produce squares with a suitable suffix to match $totend ## in the last k digits. These suffixes are collected in list$sofar.
##      We compute the list of suffix candidates with length 1 greater.
proc nextList {totsuff sofar} {
## Determine the length $olen of the members in list$sofar.
if {[llength $sofar]} { ;# has elements set olen [LEN [lindex$sofar 0]]        ;# check out first element
} else {
set olen 0                              ;# empty list
}

## Determine $nlen, the new length we want to contruct suffixes for. set nlen [+ 1$olen]                ;# we prepend 1 digit

## Determine the suffix we have to construct here, i.e.
## $totsuff reduced to the length we construct, here. if {$nlen <= [LEN $totsuff]} { set wantsuff [cutSuff$totsuff $nlen] } else { ## We do not have enough input (from$totsuff) to further limit
## the squares of constructed numbers.  All possible left
## extensions will do.  This can happen e.g. for $totsuff = "" set wantsuff$totsuff           ;# that all we need to match
}
set wantlen [LEN $wantsuff] ## We are going to construct all suffixes one longer as those ## in list$sofar, by prepending all decimal digits.
set res {}                  ;# resulting list of new suffixes
foreach d {0 1 2 3 4 5 6 7 8 9} {
foreach e $sofar { set cand$d$e ## Now we need to know the ending of the square of$cand,
## We take care that $cand may be not noprmalized. set sq [stringSquare$cand]         ;# square it
incr ::didSqs                       ;# count this squaring

## Check for a solution for our new suffix list
if {$wantsuff eq [numToExactSuffLen$sq $wantlen]} { lappend res$cand
}

## Check for a solution for the final job: the suffix of $sq ## must match, and$cand must be a positive number.
if {[string match *$totsuff$sq] && ($d > 0)} { lappend ::sols$cand
puts "(sol after $::didSqs squarings:$cand)"
}
}
}
return $res } set ::didSqs 0 ;# count squarings proc searchSquareSuff {totsuff} { set ::sols {} ;# not yet collected any solution set sufflist [list ""] ;# just the empty suffix: 1-elem list set maxsufflen [+ 1 [LEN$totsuff]]
for {set sufflen 1} {$sufflen <=$maxsufflen} {incr sufflen} {
set sufflist [ nextList $totsuff$sufflist ]
set elems [llength $sufflist] if {0 ==$elems} {
break
}
if {[llength $::sols]} { set sol [normNumber [lindex$::sols 0]]
puts ""
puts "Smallest number with suffix $totsuff is$sol"
puts " since its square is [* $sol$sol]."
if {1 < [llength $::sols]} { puts "More solutions: [lrange$::sols 1 end]"
}
break
}
## Without any solution so far, we show the suffix list.
## It is the basis for further computations, and could be checked.
puts "  List of suffixes of length $sufflen has$elems elements:"
puts "    {$sufflist}" } if {![llength$::sols]} {
puts ""
puts "No solution for $totsuff" } } if {[llength$::argv]} {
foreach a $::argv { searchSquareSuff$a
}
} else {
searchSquareSuff 269696
}
puts "(did $::didSqs squarings upto now)" ## You may want to try 08315917380318501319044 for solution 9999999156746824862  Output:  List of suffixes of length 1 has 2 elements: {4 6} List of suffixes of length 2 has 4 elements: {14 36 64 86} List of suffixes of length 3 has 4 elements: {236 264 736 764} List of suffixes of length 4 has 8 elements: {0264 2236 2764 4736 5264 7236 7764 9736} (sol after 131 squarings: 25264) (sol after 190 squarings: 99736) Smallest number with suffix 269696 is 25264 since its square is 638269696. More solutions: 99736 (did 190 squarings upto now) ## TI-83 BASIC Works with: TI-83 BASIC version TI-84Plus 2.55MP In 1996 was manufactured the TI-83, a handheld graphing calculators with a basic language called TI-83 Basic. The language is small and neat. For example to store 500 into a variable, it is done without twisting the meaning in mathematics of the equal sign (=). 536→N Do not be attracted by brute force, let's do some basic maths: As  N²=1000000·A+269696 269696=27×72×43 1000000=26×56 269696 mod 64 = 0 & 1000000 mod 64 = 0 ⇒ N² mod 64 = 0 ⇒ N mod 8 = 0 √ 269696 =519.32 => N≥520 520=8×5×13 528=16×3×11 536=8×67 N must ends by 4 or 6 ⇒ N≥536  So, Lord Babbage here is your program: 536→N While remainder(N*N,1000000)≠269696 i+8→N End Disp N And within a minute you have the answer: Output:  25264 Done  And you can check the square: Input: 25264²  Output: 638269696  ## Transd #lang transd MainModule : { rem: 269696, _start: (lambda (with n (to-Int (sqrt rem)) (while (neq (mod (pow n 2) 1000000) rem) (+= n 1)) (lout n) ) ) }  Output: 25264  ## UNIX Shell Works with: Bourne Again Shell Works with: Korn Shell Works with: Z Shell # Program to determine the smallest positive integer whose square # has a decimal representation ending in the digits 269,696. # Start with the smallest positive integer of them all let trial_value=1 # Compute the remainder when the square of the current trial value is divided # by 1,000,000.␣ while (( trial_value * trial_value % 1000000 != 269696 )); do # As long as this value is not yet 269,696, increment # our trial integer and try again. let trial_value=trial_value+1 done # To get here we must have found an integer whose square meets the # condition; display that final result echo$trial_value

Output:
25264

### Efficient version, pen-and-pencil method

Works with: Bourne Again Shell

The simple method above requires more than 20000 multiplications and would run days on Babbage's Analytical Engine (AE), if he would have managed to build such a machine.

As he had found a solution with pen and paper, and would have programmed the AE correspondingly, the following solution could have been used on the AE.

It should run on any shell, including the original Bourne Shell, if the arithmetic expressions are replaced by "expr".

#!/bin/dash

#  Babbage problem:
#  	What is the smallest (positive) integer whose square ends in the digits 269,696?
#
#  He found the second to smallest number (99736 instead of 25264) using pencil and paper,
#  and would not have wasted hours of computing time on his (planned) Analytical Engine (AE).
#
#  As most human computers know, a square must end in 0, 1, 4, 5, 6 or 9.
#  because the squares of 0 to 9 end in 0, 1, 4, 9, 6, 5, 6, 9, 4, 1.
#  Thus, the result must have the last digits 14 or 16 in the above case.
#
#  So the algorithm starts with the set {0} and an increment of 1,
#  squaring all numbers of 0+i, and keeping only those that have
#  the correct end digit.
#  Then, the new i is 10*i, and a new set of two digit endings
#  created from the old set of one digit endings, and so on.
#
#  As the AE did not have arrays or the like, the sets must be punched
#  on cards and read in for the next round.
#  The classical (original) Bourne Shell did not have arrays,
#  so you may use this script on very old machines, if 'expr' is used
#  And so his script works with 'dash', the standard command interpreter
#  for non-interactive use.
#
#  To prove the speed, try 1234554321 instead of 269696,
#  the practicall immedidate answer should be 1250061111,
#  while the simple method will take hours.
#
#  Note that this method will stop if there is no solution,
#  while the simple method continues endlessly.

# filename for workfile(s)
wrk=$(basename$0 .sh).data

# set $e to desired ending. Leading zeroes are ignored. e=${1:-269696}

# set the modulus $m to the power of 10 above$e
m=1
while test $m -le$e
do m=$((m*10)) done #$a is number to add in each round (power of 10)
a=1

# first workfile contains just the number 0
echo 0 >$wrk # test all workfile numbers with another digit in front while test$a -lt $m # until the increment excees the modulus do mm=$((a*10))			# modulus in this round
ee=$((e % mm)) # ending in this round cat$wrk |			# numbers from current workfile
do y=$x # first number to test is the number read while test$y -le $((x+mm-1)) do z=$(($y *$y))	# calculate the square
z=$(($z % $mm)) # ending in this round if test$z -eq $ee then echo$y	# candidate for next round
fi
y=$(($y + $a)) # advance leftmost digit done done >$wrk.new		# create new workfile
# next round
a=$((a*10)) # another leftmost digit mv$wrk.new $wrk # cycle workfiles done # find each number in the last workfile if x*x mod m = e # ending in$e and modulus in $m cat$wrk |			# numbers from last workfile
do y=$(($x * $x)) # check y=$(($y %$m))
if test $y -eq$e
then echo $x # solution found fi done | sort -n | # numbers in ascending order head -n 1 # show only smallest  Output: $ sh babbage.sh
25264
$sh babbage.sh 1234554321 1250061111  ## UTFool ··· http://rosettacode.org/wiki/Babbage_problem ··· ■ BabbageProblem § static ▶ main • args⦂ String[] for each number from √269696 up to √Integer.MAX_VALUE if ("⸨number × number⸩").endsWith "269696" System.exit number ## VAX Assembly 36 35 34 33 32 31 00000008'010E0000' 0000 1 result: .ascid "123456" ;output buffer 0000 000E 2 retlen: .word 0 ;$fao_s bytes written
4C 55 21 00000018'010E0000' 0010     3 format: .ascid  "!UL"                   ;unsigned decimal
001B     4
0000  001B     5 .entry  bab,0
55   D4  001D     6         clrl    r5                      ;result
001F     7 10$: 55 D6 001F 8 incl r5 56 00 55 55 7A 0021 9 emul r5,r5,#0,r6 ;mulr.rl, muld.rl, add.rl, prod.wq 51 50 56 000F4240 8F 7B 0026 10 ediv #1000000,r6,r0,r1 ;divr.rl, divd.rq, quo.wl, rem.wl 51 00041D80 8F D1 002F 11 cmpl #269696,r1 E7 12 0036 12 bneq 10$                     ;not equal - try next
0038    13
0038    14         $fao_s - ;convert integer to text 0038 15 ctrstr = format, - 0038 16 outlen = retlen, - 0038 17 outbuf = result, - 0038 18 p1 = r5 B1 AF C1 AF B0 004A 19 movw retlen, result ;adjust length AE AF 7F 004F 20 pushaq result 00000000'GF 01 FB 0052 21 calls #1, g^lib$put_output
04  0059    22         ret
005A    23 .end    bab
Fmt.print("Its square is $,d.", sq) break // break from the loop and end the program } i = i + 2 // increase 'i' by 2 }  Output: The lowest number whose square ends in 269,696 is 25,264. Its square is 638,269,696.  ## x86 Assembly AT&T syntax Works with: gas # What is the lowest number whose square ends in 269,696? # At the very end, when we have a result and we need to print it, we shall use for the purpose a program called PRINTF, which forms part of a library of similar utility programs that are provided for us. The codes given here will be needed at that point to tell PRINTF that we are asking it to print a decimal integer (as opposed to, for instance, text): .data decin: .string "%d\n\0" # This marks the beginning of our program proper: .text .global main main: # We shall test numbers from 1 upwards to see whether their squares leave a remainder of 269,696 when divided by a million. # We shall be making use of four machine 'registers', called EAX, EBX, ECX, and EDX. Each can hold one integer. # Move the number 1,000,000 into EBX: mov$1000000, %ebx

# The numbers we are testing will be stored in ECX. We start by moving a 1 there:

mov    $1, %ecx # Now we need to test whether the number satisfies our requirements. We shall want the computer to come back and repeat this sequence of instructions for each successive integer until we have found the answer, so we put a label ('next') to which we can refer. next: # We move (in fact copy) the number stored in ECX into EAX, where we shall be able to perform some calculations upon it without disturbing the original: mov %ecx, %eax # Multiply the number in EAX by itself: mul %eax # Divide the number in EAX (now the square of the number in ECX) by the number in EBX (one million). The quotient -- for which we have no use -- will be placed in EAX, and the remainder in EDX: idiv %ebx # Compare the number in EDX with 269,696. If they are equal, jump ahead to the label 'done': cmp$269696,  %edx
je     done

# Otherwise, increment the number in ECX and jump back to the label 'next':

inc    %ecx
jmp    next

# If we get to the label 'done', it means the answer is in ECX.

done:

# Put a reference to the codes for PRINTF into EAX:

lea    decin,    %eax

# Now copy the number in ECX, which is our answer, into an area of temporary storage where PRINTF will expect to find it:

push   %ecx

# Do the same with EAX -- giving the code for 'decimal integer' -- and then call PRINTF to print the answer:

push   %eax
call   printf

# The pieces of information we provided to PRINTF are still taking up some temporary storage. They are no longer needed, so make that space available again:

add    $8, %esp # Place the number 0 in EAX -- a conventional way of indicating that the program has finished correctly -- and return control to whichever program called this one: mov$0,       %eax
ret

# The end.
Output:
25264

## XLISP

; The computer will evaluate expressions written in -- possibly nested -- parentheses, where the first symbol gives the operation and any subsequent symbols or numbers give the operands.

; For instance, (+ (+ 2 2) (- 7 5)) evaluates to 6.

; We define our problem as a function:

(define (try n)

; We are looking for a value of n that leaves 269,696 as the remainder when its square is divided by a million.

; The symbol * stands for multiplication.

(if (= (remainder (* n n) 1000000) 269696)

; If this condition is met, the function should give us the value of n:

n

; If not, it should try n+1:

(try (+ n 1))))

; We supply our function with 1 as an initial value to test, and ask the computer to print the final result.

(print (try 1))

Output:
25264

## XPL0

Some preliminary tests show that the number to be squared (N) must end with 4 or 6 to get N^2 ending with 6. Additional tests show N must end 236, 264, 736 or 764. This leaves only 100 tests to determine the final number, which could be done by hand.

int N, C, R;
[C:= 0;
for N:= sqrt(269696) to -1>>1 do                \to infinity (2^31-1)
if rem(N/10)=4 or rem(N/10)=6 then            \must end 6: 4^2=16; 6^2=36
[R:= rem(N/100);
if R=14 or R=36 or R=64 or R=86 then        \14^2=196, etc.
[R:= rem(N/1000);
if R=236 or R=264 or R=736 or R=764 then \236^2=55696, etc.
[C:= C+1;                            \count remaining tests
if rem(N*N/1_000_000) = 269_696 then
[IntOut(0, N);
Text(0, "^^2 = ");
IntOut(0, N*N);
CrLf(0);
IntOut(0, C);
CrLf(0);
exit;
];
];
];
];
]
Output:
25264^2 = 638269696
100


## Yabasic

// Charles Babbage habría sabido que solo un número que termina en 4 o 6
// podría producir un cuadrado que termina en 6, y cualquier número por
// debajo de 520 produciría un cuadrado menor que 269696. Podemos detenernos

number = 524 // primer numero a probar
repeat
number = number + 2
until mod((number ^ 2), 1000000) = 269696
print "El menor numero cuyo cuadrado termina en 269696 es: ", number
print "Y su cuadrado es: ", number*number
Output:
El menor numero cuyo cuadrado termina en 269696 es: 25264

// The magic number is 269696, so, starting about its square root,
const N=269696; [500..].filter1(fcn(n){ n*n%0d1_000_000 == N })
25264