9 Digit Problem

CJam - 26

1{;9,:)[email protected]{_3$<i\%+}/}g

It's randomized but works pretty fast with the java interpreter. It can take a couple of minutes with the online interpreter.


1 pushes 1 (to be explained later)
{…}g is a do-while loop
; removes a value from the stack (initially the 1 we started with)
9, makes the array [0 ... 8]
:) increments the array elements, resulting in [1 ... 9]
_ duplicates the array
mr shuffles the array
s converts to string
[email protected] pushes 0 then brings the other copy of the array on top
{…}/ is a for-each loop (over the numbers 1 ... 9)
_ duplicates the current number (let's call it "k")
3$ copies the numeric string from the stack
<i gets the substring with the first k characters then converts to integer
\% swaps with the other copy of k then gets the remainder (%k)
+ adds the remainder to the previous value on the stack (initially 0 from above)
At this point, we have the numeric string on the stack, followed by a 0 if the number matches all the requirements (i.e. all the remainders were 0) or a non-0 value otherwise.
The top of the stack becomes the do-while loop condition. It is popped and the loop continues if the condition was true.
If we found the solution, the condition is 0 (false), the loop ends and the rest of the stack (the numeric string) is printed.
If it's not the solution, the condition is the non-0 value (true) and the loop continues with the string on the stack. The string gets popped at the start of the next iteration (so the loop expects a value on the stack, and that's the reason for the initial 1).

Thanks Dennis for making the code shorter and more convoluted :p

Javascript (E6) 105 125 134

Recursive building of the number, each step check the divisibility.
Runtime near 0 sec
No I/O this time, as the OP asked for a program to find the number, and the number is found and automatically logged to console

Golfed more Courtesy of MT0

(Q=(n,d,b)=>([(m=n+(s=[...b]).splice(i,1))%d||Q(m,d+1,s)for(i in b)],d>9&&(Q.z=n),Q.z))('',1,'123456789')


for(i=0;s=[...b],m=n+s.splice(i,1),b[i];i++)m%d||Q(m,d+1,s);d>9&&(Q.z=n);return Q.z;


(Q=(n='', d=1, b=[...'123456789'], i) => {
   for(i=0; s=[...b], m=n+s.splice(i,1), b[i]; i++)
     m % d || Q(m,d+1,s);
   d > 9 && (Q.z=n);
   return Q.z;


With 3 little changes, you can use the same function to find longer numbers using base > 10. For instance in base 14 ...

  for(i=0; s=[...b], m = n+s.splice(i,1), b[i]; i++)
    parseInt(m,14)%d || Q(m,d+1,s);
  d>13 && (Q.z=n);
  return Q.z;



Q=(n,d,b,i,c,z)=>{ // i,c,z fake parameters instead of vars.
  for (i=0; b[i]; i++)
    m = n + s.splice(i,1);
    if (m % d == 0)
      if (z = d<9 ? Q(m, d+1, s) : m) return z;

Python3, 214, 199, 184, 176, 174, 171, 165, 150, 146

from itertools import*
g=lambda i,d:d==1!=print(i)or int(i[9:])%d==0!=g(i[:-1],d-1)
for x in permutations("123456789"):g("".join(map(str,x))*2,9)



This is my first golf script. Hope you like it :)


Code Golf