# 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:

*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.

For these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer, who has never programmed—in fact, who has never so much as seen a single line of code.

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.

## Contents

- 1 360 Assembly
- 2 Ada
- 3 ALGOL 68
- 4 Aime
- 5 APL
- 6 AppleScript
- 7 AutoHotkey
- 8 AWK
- 9 Batch File
- 10 BASIC
- 11 Befunge
- 12 C
- 13 Common Lisp
- 14 C++
- 15 C#
- 16 Clojure
- 17 COBOL
- 18 Component Pascal
- 19 D
- 20 Dafny
- 21 Dart
- 22 Elena
- 23 Elixir
- 24 Erlang
- 25 F#
- 26 Factor
- 27 Forth
- 28 Fortran
- 29 FreeBASIC
- 30 FutureBasic
- 31 Gambas
- 32 Go
- 33 Groovy
- 34 Haskell
- 35 J
- 36 Java
- 37 JavaScript
- 38 jq
- 39 Julia
- 40 Kotlin
- 41 Liberty BASIC
- 42 Limbo
- 43 Lua
- 44 M2000 Interpreter
- 45 Mathematica
- 46 Microsoft Small Basic
- 47 Modula-2
- 48 NetRexx
- 49 Nim
- 50 OCaml
- 51 PARI/GP
- 52 Pascal
- 53 Perl
- 54 Perl 6
- 55 Phix
- 56 PicoLisp
- 57 PILOT
- 58 PowerShell
- 59 Processing
- 60 Prolog
- 61 PureBasic
- 62 Python
- 63 Racket
- 64 R
- 65 Red
- 66 REXX
- 67 Ring
- 68 Ruby
- 69 Run BASIC
- 70 Rust
- 71 Scala
- 72 Scilab
- 73 SequenceL
- 74 Shen
- 75 Sidef
- 76 Simula
- 77 Smalltalk
- 78 Swift
- 79 Tcl
- 80 TI-83 BASIC
- 81 UNIX Shell
- 82 UTFool
- 83 VAX Assembly
- 84 VBA
- 85 VBScript
- 86 x86 Assembly
- 87 XLISP
- 88 zkl

## 360 Assembly[edit]

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

ST 13,4(15) prolog: link backwards

ST 15,8(13) prolog: link forwards

LR 13,15 prolog: establish addressability

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)

## Ada[edit]

-- 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.

with Ada.Text_IO;

-- 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, ect.

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.

Ada.Text_IO.Put_Line(Number'Image(X));

-- Number'Image(X) converts the value stored at X into a string of

-- printable characters (more specifically, of digits).

-- Ada.Text_IO.Put_Line(...) prints this string, for humans to read.

-- I did already run the program, and it did print out 25264.

end Babbage_Problem;

## ALGOL 68[edit]

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

## Aime[edit]

integer i;

i = sqrt(269696);

while (i * i % 1000000 != 269696) {

i += 1;

}

o_(i, "\n");

## APL[edit]

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↑⍒⍵}

⍝ We can now ask the computer for the answer:

SMALLESTWHERE (SQUAREOF N) ENDSIN 269696

- Output:

25264

## AppleScript[edit]

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

## AutoHotkey[edit]

; 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[edit]

# 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

## Batch File[edit]

:: 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

## BASIC[edit]

### Applesoft BASIC[edit]

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[edit]

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

### BBC BASIC[edit]

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[edit]

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.

### IS-BASIC[edit]

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

## Befunge[edit]

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

>:#,_$ . 55+, @

ouput message then n newline exit

numeric constants explained:

"d" ascii value of 'd', i.e. 100

:: duplicate twice: 100,100,100

** multiply twice: 100*100*100 = 1000000

"V8" ascii values of 'V' and '8', i.e. 86 and 56

: duplicate the '8' (56): 86,56,56

** multiply twice: 86*56*56 = 269696

- Output:

Smallest number whose square ends in 269696 is 25264

## C[edit]

// 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 ((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

## Common Lisp[edit]

(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[edit]

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

## C++[edit]

#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 !

## C#[edit]

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

## Clojure[edit]

; 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 (expt n 2)]

(= 269696 (mod square 1000000))))

; Given a range of positive integers up to 99,736, apply the above babbage?

; function, returning only numbers that return true.

(filter babbage? (range 99736))

- Output:

(25264)

## COBOL[edit]

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

## Component Pascal[edit]

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[edit]

// 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[edit]

// Helper function for mask: does the actual computation.

function method mask_(v:int,m:int):int

decreases v-m

requires 0 <= v && 0 < m

ensures v < mask_(v,m)

{

if v < m then m else mask_(v,m*10)

}

// Return the smallest power of 10 greater than v.

function method mask(v:int):int

requires 0 <= v

ensures v < mask(v)

{

mask_(v,10)

}

// Return true if the last digits of v == suffix.

predicate method EndWith(v:int,suffix:int)

requires 0 <= suffix

{

v % mask(suffix) == 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[edit]

main() {

var x = 0;

while((x*x)% 1000000 != 269696)

{ x++;}

print('$x');

}

## Elena[edit]

ELENA 3.4 :

import extensions.

import system'math.

public program

[

var n := 1.

until(n sqr; mod:1000000 == 269696)

[

n += 1.

].

console printLine(n)

]

- Output:

25264

## Elixir[edit]

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[edit]

-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#[edit]

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[edit]

! 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[edit]

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[edit]

### First FORTRAN[edit]

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 (√ 2^{31}-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[edit]

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[edit]

' 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

## FutureBasic[edit]

include "ConsoleWindow"

DIM AS 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

- Output:

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

## Gambas[edit]

**Click this link to run this code**

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[edit]

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[edit]

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[edit]

#### head[edit]

--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 !

#### Safe.headMay[edit]

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 Safe (headMay)

maybeBabbage :: Integer -> Maybe Integer

maybeBabbage upperLimit =

headMay

(filter ((269696 ==) . flip rem 1000000) ((^ 2) <$> [1 .. upperLimit]))

main :: IO ()

main = do

let upperLimit = 100000

putStrLn

(case maybeBabbage upperLimit of

Nothing ->

intercalate (show upperLimit) ["No such number found below ", " ..."]

Just x ->

intercalate

" ^ 2 -> "

(fmap show ([floor . sqrt . fromInteger, id] <*> pure x)))

- Output:

25264 ^ 2 -> 638269696

#### Suffixes and integer roots[edit]

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)

babbagePairs :: [[Integer]]

babbagePairs =

[0,1000000 ..] >>= -- Drawing from a succession of N * 10^6

\x ->

let y = (x + 269696) -- The next number ending in 269696,

r = (sqrt . fromIntegral) 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.

main :: IO ()

main = do

let arrowed = intercalate " ^ 2 -> " . fmap show

mapM_ putStrLn (arrowed <$> take 10 babbagePairs)

- 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 inspect 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:

import Data.List (intercalate)

babbagePairs :: [[Integer]]

babbagePairs =

[0,10000 ..] >>=

\x ->

([(:) <*> return . (^ 2)] <*> ((x +) <$> [0264, 5264, 9736, 4736])) >>=

\[a, b] ->

[ [a, b]

| ((269696 ==) . flip rem 1000000) b ]

main :: IO ()

main = do

let arrowed = intercalate " ^ 2 -> " . fmap show

mapM_ putStrLn (arrowed <$> take 4000 babbagePairs)

## J[edit]

The key to understandability is a mix of hopefully adequate notation and some level of verifiability.

So let's break the implementation into some named pieces and present enough detail that a mathematician can verify that the result is both consistent and reasonable:

square=: ^&2

modulo1e6=: 1000000&|

trythese=: i. 1000000 NB. first million nonnegative integers

which=: I. NB. position of true values

which 269696=modulo1e6 square trythese NB. right to left <-

25264 99736 150264 224736 275264 349736 400264 474736 525264 599736 650264 724736 775264 849736 900264 974736

The smallest of these values is 25264.

#### Alternatively, inspired by the APL example that makes the sentence sound natural[edit]

NB. In the interactive environment.

NB. First here, Mr Babbage, we'll make the computer's words more meaningful to an english speaker.

NB. The first is the "head" of a list, written with these inviting open arms that embrace one small dot :

first=: {.

NB. The small i. notation denotes "all integers up to 100000". You've already found a solution in that range.

n=: i. 100000

NB. This is how we write squaring.

squareof=: *:

NB. In our notation, a dyad is a word that takes an x value on the left and an y value on the right.

ends=: dyad : ' x = 1000000 | y '

NB. This dyad selects values from the list x, as marked by the list y

where=: dyad : ' y # x '

NB. Now that we defined our words, we can ask our question with them :

first n where 269696 ends squareof n

25264

NB. With a bit of habit, you won't need to define words in english anymore.

NB. The following easily relates word for word to the sentence we've written :

{. (i.100000) #~ 269696 = 1000000 | *: i.100000

25264

NB. Like all mathematical notations, in J you see patterns that suggest simplification :

{. I. 269696 = 1000000 | *: i.100000

25264

## Java[edit]

public class Test {

public static void main(String[] args) {

// let n be zero

int n = 0;

// repeat the following action

do {

// increase n by 1

n++;

// while the modulo of n times n is not equal to 269696

} while (n * n % 1000_000 != 269696);

// show the result

System.out.println(n);

}

}

25264

## JavaScript[edit]

#### Iteration[edit]

// Every line starting with a double slash will be ignored by the processing machine,

// just like these two.

//

// Since the square root of 269,696 is approximately 519, we create a variable named "n"

// and give it this value.

n = 519

// The while-condition is in parentheses

// * is for multiplication

// % is for modulo operation

// != is for "not equal"

while ( ((n * n) % 1000000) != 269696 )

n = n + 1

// n is incremented until the while-condition is met, so n should finally be the

// smallest positive integer whose square ends in the digits 269,696. To see n, we

// need to send it to the monitoring device (named console).

console.log(n)

#### Array constructor[edit]

JavaScript's Array constructor lets us filter an array automatically populated with a function of the element index. This proves faster than setting up and running a while loop test, and we can make it particularly efficient by testing the potential squares rather than the potential roots.

Starting with numbers which end in 269696, and filtering for those which have integer roots, so that we reach **25264 ^2 -> 638269696** after only 638 tests.

(() => {

// BABBAGE ---------------------------------------------------------------

// babbageNumbers :: Int -> [(Int, Int)]

const babbageNumbers = intTests =>

map(x => [Math.sqrt(x), x], filter(

hasIntegerRoot,

takeNfromSeries(intTests, x => (x * 1000000) + 269696)

));

// hasIntegerRoot :: Int -> Bool

const hasIntegerRoot = n => {

const r = Math.sqrt(n);

return Math.floor(r) === r;

};

// takeNFromSeries :: Int -> (Int -> a) -> [a]

const takeNfromSeries = (n, f) =>

Array.from({

length: n

}, (_, i) => f(i));

// GENERIC FUNCTIONS -----------------------------------------------------

// curry :: ((a, b) -> c) -> a -> b -> c

const curry = f => a => b => f(a, b);

// filter :: (a -> Bool) -> [a] -> [a]

const filter = (f, xs) => xs.filter(f);

// intercalate :: String -> [a] -> String

const intercalate = (s, xs) => xs.join(s);

// map :: (a -> b) -> [a] -> [b]

const map = (f, xs) => xs.map(f);

// unlines :: [String] -> String

const unlines = xs => xs.join('\n');

// TEST ------------------------------------------------------------------

return unlines(

map(

curry(intercalate)(' ^2 -> '),

babbageNumbers(1000000) // Testing 10^6 numbers ending in 269696

)

);

})();

- 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 650264 ^2 -> 422843269696 724736 ^2 -> 525242269696 775264 ^2 -> 601034269696 849736 ^2 -> 722051269696 900264 ^2 -> 810475269696 974736 ^2 -> 950110269696

## jq[edit]

$ jq -n '1 | until( .*. | tostring | test("269696$"); .+1)' 25264

In words: start with n=1; if the decimal representation of n*n ends with 269696 then print n, otherwise increment n and restart.

**Note for Mr Babbage.**

Dear Sir. The answer to your most excellent problem is 25,264. For a demonstration, come join us in the 21st century and see for yourself: Ceci est une |."

## Julia[edit]

function babbage(x::Integer)

i = big(0)

d = floor(log10(x)) + 1

while i ^ 2 % 10 ^ d != x

i += 1

end

return i

end

- Output:

julia> babbage(269696) 25264

## Kotlin[edit]

fun main(args: Array<String>) {

var number = 520L

var square = 520 * 520L

while (true) {

val last6 = square.toString().takeLast(6)

if (last6 == "269696") {

println("The smallest number is $number whose square is $square")

return

}

number += 2

square = number * number

}

}

- Output:

The smallest number is 25264 whose square is 638269696

## Liberty BASIC[edit]

Now Mr. Babbage -- May I call you Charlie? No. OK -- we'll first start with 'n' equal to zero, then multiply it by itself to square it. If the last six digits of the result are not 269696, we'll add one to 'n' then go back and square it again. On our modern computer it should only take a moment to find the answer...

[start]

if right$(str$(n*n),6)="269696" then

print "n = "; using("###,###", n);

print " n*n = "; using("###,###,###,###", n*n)

end if

if n<100000 then n=n+1: goto [start]

print "Program complete."

Eureka! We found it! --

n = 25,264 n*n = 638,269,696

n = 99,736 n*n = 9,947,269,696

Program complete.

Now my question for you, Sir, is how did you know that the square of ANY number would end in 269696?? Oh, and by the way, 99,736 is an answer too.

## Limbo[edit]

implement Babbage;

include "sys.m";

sys: Sys;

print: import sys;

include "draw.m";

draw: Draw;

Babbage : module

{

init : fn(ctxt : ref Draw->Context, args : list of string);

};

init (ctxt: ref Draw->Context, args: list of string)

{

sys = load Sys Sys->PATH;

current := 0;

while ((current * current) % 1000000 != 269696)

current++;

print("%d", current);

}

## Lua[edit]

-- get smallest number <= sqrt(269696)

k = math.floor(math.sqrt(269696))

-- if root is odd -> make it even

if k % 2 == 1 then

k = k - 1

end

-- cycle through numbers

while not ((k * k) % 1000000 == 269696) do

k = k + 2

end

io.write(string.format("%d * %d = %d\n", k, k, k * k))

## M2000 Interpreter[edit]

Def Long k=1000000, T=269696, n

n=Sqrt(269696)

For n=n to k {

If n^2 mod k = T Then Exit

}

Report format$("The smallest number whose square ends in {0} is {1}, Its square is {2}", T, n, n**2)

## Mathematica[edit]

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}}

## Microsoft Small Basic[edit]

' 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.

## Modula-2[edit]

MODULE BabbageProblem;

FROM FormatString IMPORT FormatString;

FROM RealMath IMPORT sqrt;

FROM Terminal IMPORT WriteString,ReadChar;

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);

ReadChar

END BabbageProblem.

## NetRexx[edit]

/* 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

## Nim[edit]

var n : int = 0

while n*n mod 1_000_000 != 269_696:

inc(n)

echo n

## OCaml[edit]

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

## PARI/GP[edit]

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[edit]

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

n := 2; (* Start with n equal to 2. *)

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[edit]

#!/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 !

## Perl 6[edit]

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

Alternatively, the following just may be declarative enough to allow Babbage to understand what's going on:

say $_ if ($_² % 1000000 == 269696) for 1..99736;

- Output:

25264 99736

## Phix[edit]

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.

for i=2 to 99736 by 2 do

if remainder(i*i,1000000)=269696 then ?i exit end if

end for

- Output:

25264

## PicoLisp[edit]

: (for N 99736 # Iterate N from 1 to 99736

(T (= 269696 (% (* N N) 1000000)) N) ) # Stop if remainder is 269696

-> 25264

## PILOT[edit]

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:

## PowerShell[edit]

###########################################################################################

#

# 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**

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[edit]

// 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[edit]

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[edit]

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[edit]

# 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:

from math import (floor, sqrt)

from itertools import (islice)

# main :: IO ()

def main():

for b in take(10)(squaresWithSuffix(269696)):

print (

f'{int(sqrt(b))}^2 -> {b}'

)

# squaresWithSuffix :: Int -> Gen [Int]

def squaresWithSuffix(n):

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):

r = sqrt(n)

return r == floor(r)

# GENERIC ABSTRACTIONS ------------------------------------

# succ :: Enum a => a -> a

# succ :: Int -> Int

def succ(x):

return 1 + x

# take :: Int -> [a] -> [a]

# take :: Int -> String -> String

def take(n):

return lambda xs: (

xs[0:n]

if isinstance(xs, list)

else list(islice(xs, n))

)

# until :: (a -> Bool) -> (a -> a) -> a -> a

def until(p):

def go(f, x):

v = x

while not p(v):

v = f(v)

return v

return lambda f: lambda x: go(f, x)

# MAIN ---

main()

- 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 [Finished in 0.259s]

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.

## Racket[edit]

;; 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!

(in-range 1 (add1 99736)))

; 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

## R[edit]

babbage_function=function(){

n=0

while (n**2%%1000000!=269696) {

n=n+1

}

return(n)

}

babbage_function()[length(babbage_function())]

- Output:

25264

## Red[edit]

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[edit]

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[edit]

/*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[edit]

/*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 1
^{st}REXX version.

### examine only numbers ending in 4 or 6[edit]

/*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 1
^{st}REXX version.

### start with smallest possible number[edit]

/*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[edit]

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[edit]

n = 0

n = n + 2 until (n*n).modulo(1000000) == 269696

print n

## Run BASIC[edit]

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[edit]

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

## Scala[edit]

//Babbage Problem

object babbage{

def main( args:Array[String] ){

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 )

}

}

## Scilab[edit]

n=2;

flag=%F

while ~flag

n = n+2;

if pmodulo(n*n,1000000)==269696 then

flag=%T;

end

end

disp(n);

- Output:

25264.

## SequenceL[edit]

main() := babbage(0);

babbage(current) :=

current when current * current mod 1000000 = 269696

else

babbage(current + 1);

- Output:

cmd:> babbage.exe 25264

## Shen[edit]

(define babbage

N -> N where (= 269696 (shen.mod (* N N) 1000000)))

N -> (babbage (+ N 1))

(babbage 1)

- Output:

25264

## Sidef[edit]

var n = 0

while (n*n % 1000000 != 269696) {

n += 2

}

say n

- Output:

25264

## Simula[edit]

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[edit]

"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

## Swift[edit]

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[edit]

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

## TI-83 BASIC[edit]

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=2^{7}×7^{2}×43 1000000=2^{6}×5^{6}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

## UNIX Shell[edit]

# 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

## UTFool[edit]

···

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[edit]

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

$ run bab

25264

## VBA[edit]

Sub Baggage_Problem()

Dim i As Long

'We can start at the square root of 269696

i = 520

'269696 is a multiple of 4, 520 too

'so we can increment i by 4

Do While ((i * i) Mod 1000000) <> 269696

i = i + 4 'Increment by 4

Loop

Debug.Print "The smallest positive integer whose square ends in the digits 269 696 is : " & i & vbCrLf & _

"Its square is : " & i * i

End Sub

The smallest positive integer whose square ends in the digits 269 696 is : 25264 Its square is : 638269696

## VBScript[edit]

'Sir, this is a script that could solve your problem.

'Lines that begin with the apostrophe are comments. The machine ignores them.

'The next line declares a variable n and sets it to 0. Note that the

'equals sign "assigns", not just "relates". So in here, this is more

'of a command, rather than just a mere proposition.

n = 0

'Starting from the initial value, which is 0, n is being incremented

'by 1 while its square, n * n (* means multiplication) does not have

'a modulo 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. Before I forget, "<>" basically

'means "not equal to".

Do While ((n * n) Mod 1000000) <> 269696

n = n + 1 'Increment by 1.

Loop

'The function "WScript.Echo" displays the string to the monitor. The

'ampersand concatenates strings or variables to be displayed.

WScript.Echo("The smallest positive integer whose square ends in 269696 is " & n & ".")

WScript.Echo("Its square is " & n*n & ".")

'End of Program.

- Output:

The smallest positive integer whose square ends in 269696 is 25264. Its square is 638269696.

## x86 Assembly[edit]

**AT&T syntax**

# 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[edit]

; 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

## zkl[edit]

// The magic number is 269696, so, starting about its square root,

// find the first integer that, when squared, its last six digits are the magic number.

// The last digits are found with modulo, represented here by the % symbol

const N=269696; [500..].filter1(fcn(n){ n*n%0d1_000_000 == N })

- Output:

25264

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