Babbage problem: Difference between revisions
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+25264 when squared is: +638269696
</pre>
=={{header|Amazing Hopper}}==
<syntaxhighlight lang="c">
#include <basico.h>
algoritmo
decimales '0'
número = 0, i=10
ciclo:
iterar grupo( número+=2, \
#( (número^2) % 1000000 != 269696 ), ; )
imprimir ("The smallest number whose square ends in 269696 is: ",\
número,\
"\nIt's square is: ", #(número ^ 2), NL )
i--, jnz(ciclo)
terminar
</syntaxhighlight>
{{out}}
<pre>
The smallest number whose square ends in 269696 is: 25264
It's square is: 638269696
The smallest number whose square ends in 269696 is: 99736
It's square is: 9947269696
The smallest number whose square ends in 269696 is: 150264
It's square is: 22579269696
The smallest number whose square ends in 269696 is: 224736
It's square is: 50506269696
The smallest number whose square ends in 269696 is: 275264
It's square is: 75770269696
The smallest number whose square ends in 269696 is: 349736
It's square is: 122315269696
The smallest number whose square ends in 269696 is: 400264
It's square is: 160211269696
The smallest number whose square ends in 269696 is: 474736
It's square is: 225374269696
The smallest number whose square ends in 269696 is: 525264
It's square is: 275902269696
The smallest number whose square ends in 269696 is: 599736
It's square is: 359683269696
The smallest number whose square ends in 269696 is: 650264
It's square is: 422843269696
</pre>
<p>Comentarios "in spanish":</p>
<p>
Documentación:</p>
<p>Establece que se trabajará con números sin decimales:</p>
"decimales(0)" (A)
La variable que tendrá el número cuyo cuadrado finaliza
con los dígitos 269696. Se inicializa con valor 0:
"número" (B)
La variable que controlará 11 iteraciones del proceso
principal. Se inicializa con 10, y se decrementará hasta
llegar a 0, de acuerdo a lo descrito en 1.4, 1.5 y 1.6:
"i" (C)
Esto permitirá encontrar 11 números que cumplan con lo
buscado.
-------------------------------------------------------------
Establece una etiqueta de salto, donde la ejecución del
programa saltará si se cumple una condición, que será
explicada en 1.4 y 1.5:
"CICLO:" (1.1)
Una macro que iterará el proceso principal donde por cada
incremento de "número" evaluará si los últimos dígitos del
cuadrado de "número" son los dígitos buscados por el señor
Babbage:
"iterar grupo" (1.2)
Explicación del proceso descrito en 1.2:
número+=2 (1.2.1) : incrementará el valor de la
variable "número" de 2 en 2.
número^2 (1.2.2) : cuadrado de "número".
% : resto módulo, resto de la división.
1000000 : esto permite obtener el resto de la
división de tamaño 6 dígitos.
Si el resultado de "(número^2)%1000000" es igual a
269696, entonces, "número" es el número buscado.
Una macro que dispondrá los mensajes y los resultados de
1.2.1 y 1.2.2, los que serán mostrados en pantalla:
"IMPRIMIR" (1.3)
Dispone el valor de la variable "i" para ser evaluada por
"JNZ", y luego la decrementará en 1:
"i--" (1.4)
Evalúa: si la variable no es cero, entonces saltará a la
posición del programa señalada en 1.1;
"JNZ(CICLO)" (1.5)
de lo contrario, si "i" es cero, entonces continuará con
el resto del programa, que será "terminar" la ejecución del
algoritmo:
"TERMINAR" (1.6)
Comentarios:
JNZ : significa Jump if Non-Zero.
#() : macro que permite resolver expresiones infijas en
tiempo de compilación.
</p>
=={{header|APL}}==
Line 1,668 ⟶ 1,790:
=={{header|Elena}}==
{{trans|Smalltalk}}
ELENA
<syntaxhighlight lang="elena">import extensions;
import system'math;
Line 1,676 ⟶ 1,798:
var n := 1;
until(n.sqr().mod
{
n += 1
Line 2,543 ⟶ 2,665:
<syntaxhighlight lang="julia">
"""
babbage(x::Integer)
Returns the smallest positive integer whose square ends in `x`
"""
function babbage(x::Integer)
i = big(0) # start with 0 and increase by 1 until target reaached
d = 10^ndigits(x) # smallest power of 10 greater than x
while (i
i += 1 # try next squre of numbers 0, 1, 2, ...
end
return i
end
</syntaxhighlight>
Line 2,670 ⟶ 2,798:
{{out}}
<pre>{{x->25264},{x->99736}}</pre>
=={{header|MATLAB}}==
<syntaxhighlight lang="MATLAB">
clear all;close all;clc;
BabbageProblem();
function BabbageProblem
% Initialize x to 524, as the square root of 269696 is approximately 519.something
x = 524;
% Loop until the square of x modulo 1000000 equals 269696
while mod(x^2, 1000000) ~= 269696
% If the last digit of x is 4, increment x by 2
% Otherwise, increment x by 8
if mod(x, 10) == 4
x = x + 2;
else
x = x + 8;
end
end
% Display the result
fprintf('The smallest positive integer whose square ends in 269696 = %d\n', x);
end
</syntaxhighlight>
{{out}}
<pre>
The smallest positive integer whose square ends in 269696 = 25264
</pre>
=={{header|Maxima}}==
Line 3,639 ⟶ 3,797:
This could certainly be written more concisely. Extra verbiage is included to make the process more clear.
<syntaxhighlight lang="raku" line># For all
for 1 .. Inf -> $integer {
Line 3,925 ⟶ 4,083:
Which outputs the same thing as above.
=={{header|S-BASIC}}==
Because we have to use double-precision floating point to represent the required number of digits, and because the approach to calculating a double-precision n mod 1000000 to isolate the right-most six digits of the square is particularly inefficient, the program will take a long time to find the solution (but it will, eventually!)
<syntaxhighlight lang="BASIC">
$lines
$constant true = FFFFH
$constant false = 0
var n, sq, r = real.double
var done = integer
print "Finding smallest number whose square ends in 269696"
n = 520 rem - no smaller number has a square that large
done = false
rem - no need to search beyond the number Babbage already knew
while not done and n <= 99736.0 do
begin
sq = n * n
rem - compute sq mod 1000000 by repeated subtraction
r = sq
while r >= 1000000.0 do
r = r - 1000000.0
if r = 269696.0 then
begin
print using "The smallest number is ######"; n
print using "and its square is ##,###,###,###"; sq
done = true
end
rem - only even numbers can have a square ending in 6
n = n + 2
end
end
</syntaxhighlight>
{{out}}
<pre>
Finding smallest number whose square ends in 269696
The smallest number is 25264
and its square is 638,269,696
</pre>
=={{header|Scala}}==
Line 4,435 ⟶ 4,636:
638269696
</pre>
=={{header|Transd}}==
Line 4,793 ⟶ 4,980:
=={{header|Wren}}==
{{libheader|Wren-fmt}}
<syntaxhighlight lang="
The answer must be an even number and it can't be less than the square root of 269,696.
So, if we start from that, keep on adding 2 and squaring it we'll eventually find the answer.
Line 4,799 ⟶ 4,986:
*/
import "./fmt" for Fmt // this enables us to format numbers with thousand separators
var start = 269696.sqrt.ceil // get the next integer higher than (or equal to) the square root
start = (start/2).ceil * 2 // if it's odd, use the next even integer
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