Project Euler #34: Digit factorials - review code : Java -

a digital factorial need found out. no. divisible sum of factorials of digits of no. trying on hackerrank : https://www.hackerrank.com/contests/projecteuler/challenges/euler034/

my code passing 1 test case i.e. if n < 20.other test cases not passing.

code below.

import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class solution {     public static void main(string[] args) {         scanner sc = new scanner(system.in);         int n = sc.nextint();         int len,newt=0,pos = 0,ans=0;         long fac,sum;         for(int i=10;i<=n;i++){             newt  = i;             sum =0;             len = string.valueof(newt).length();                  while(len>0){             fac=1;         pos = newt % 10;         newt = newt/10;                       for(int k=1;k<=pos;k++){        fac = fac*k;              }              sum+=fac;                 len--;         }       //      system.out.print(sum+" "+i);       //      system.out.println();                 if(sum%i==0){             ans = i;         }         }        system.out.print(ans);         } } 

your code iterating digits of number n seems overly complex. it's simple this:

if (n <= 0)     throw new illegalargumentexception("not positive number: " + n); (; n > 0; n /= 10) {     int digit = n % 10;     // use digit here } 

your code can improved using methods separate logic independent parts, e.g.

private static int sumcuriousnumbers(int n) {     int sum = 0;     (int = 10; <= n; i++)         if (iscuriousnumber(i))             sum += i;     return sum; } private static boolean iscuriousnumber(int number) {     int sum = 0;     (int = number; > 0; /= 10)         sum += factorial(i % 10);     return (sum % number == 0); } private static int factorial(int n) {     int f = 1;     (int = 1; <= n; i++)         f *= i;     return f; } 

this helps document code , 3 methods can unit-tested too.