Steady squares

Steady squares is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Euler Project #284

Task

The 3-digit number 376 in the decimal numbering system is an example of numbers with the special property that its square ends with the same digits: 376*376 = 141376. Let's call a number with this property a steady square. Find steady squares under 10.000

C

<lang c>#include <stdio.h>

1. include <stdbool.h>

bool steady(int n) {

   int mask=1;
   for (int d=n; d; d/=10) mask*=10;
   return (n*n)%mask == n;

}

int main() {

   for (int i=1; i<10000; i++)
       if (steady(i))
           printf("%4d^2 = %8d\n", i, i*i);
   return 0;

}</lang>

Output:

   1^2 =        1
   5^2 =       25
   6^2 =       36
  25^2 =      625
  76^2 =     5776
 376^2 =   141376
 625^2 =   390625
9376^2 = 87909376

CLU

<lang clu>n_digits = proc (n: int) returns (int)

   i: int := 0
   while n>0 do
       i := i+1
       n := n/10
   end
   return(i)

end n_digits

steady = proc (n: int) returns (bool)

   sq: int := n ** 2
   return (sq // 10**n_digits(n) = n)

end steady

start_up = proc ()

   po: stream := stream$primary_output()
   for i: int in int$from_to(1, 10000) do
       if ~steady(i) then continue end
       stream$putright(po, int$unparse(i), 4)
       stream$puts(po, "^2 = ")
       stream$putright(po, int$unparse(i**2), 8)
       stream$putl(po, "")
   end

end start_up</lang>

Output:

   1^2 =        1
   5^2 =       25
   6^2 =       36
  25^2 =      625
  76^2 =     5776
 376^2 =   141376
 625^2 =   390625
9376^2 = 87909376

FreeBASIC

<lang freebasic>function numdig( byval n as uinteger ) as uinteger

   'number of decimal digits in n
   dim as uinteger d=0
   while n
       d+=1
       n\=10
   wend
   return d

end function

function is_steady_square( n as const uinteger ) as boolean

   dim as integer n2 = n^2
   if n2 mod 10^numdig(n) = n then return true else return false

end function

for i as uinteger = 1 to 10000

   if is_steady_square(i) then print using "####^2 = ########";i;i^2

next i</lang>

Output:

  1^2 =        1
  5^2 =       25
  6^2 =       36
 25^2 =      625
 76^2 =     5776
376^2 =   141376
625^2 =   390625

9376^2 = 87909376

Python

<lang python> print("working...") print("Steady squares under 10.000 are:") limit = 10000

for n in range(1,limit):

   nstr = str(n)
   nlen = len(nstr)
   square = str(pow(n,2))
   rn = square[-nlen:]
   if nstr == rn:
      print(str(n) + " " + str(square))

print("done...") </lang>

Output:

working...
Steady squares under 10.000 are:
1 1
5 25
6 36
25 625
76 5776
376 141376
625 390625
9376 87909376
done...

Ring

<lang ring> see "working..." +nl see "Steady squatres under 10.000 are:" + nl limit = 10000

for n = 1 to limit

   nstr = string(n)
   len = len(nstr)
   square = pow(n,2)
   rn = right(string(square),len)
   if nstr = rn
      see "" + n + " " + square + nl
   ok

next

see "done..." +nl </lang>

Output:

working...
Steady squares under 10.000 are:
1 1
5 25
6 36
25 625
76 5776
376 141376
625 390625
9376 87909376
done...

Wren

Library: Wren-fmt

Although it hardly matters for a small range such as this, one can cut down the numbers to be examined by observing that a steady square must end in 0, 1, 5 or 6. <lang ecmascript>import "./fmt" for Fmt

System.print("Steady squares under 10,000:") var finalDigits = [0, 1, 5, 6] for (i in 1..9999) {

   if (!finalDigits.contains(i % 10)) continue
   var sq = i * i
   if (sq.toString.endsWith(i.toString)) Fmt.print("$,5d -> $,10d", i, sq)

}</lang>

Output:

Steady squares under 10,000:
    1 ->          1
    5 ->         25
    6 ->         36
   25 ->        625
   76 ->      5,776
  376 ->    141,376
  625 ->    390,625
9,376 -> 87,909,376