##
**Problem:**

145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: as 1! = 1 and 2! = 2 are not sums they are not included.

##
**Solution:**

40730

##
**Code:**

The solution may include methods that will be found here: Library.java .

The solution may include methods that will be found here: Library.java .

public interface EulerSolution{ public String run(); }

/* * Solution to Project Euler problem 34 * By Nayuki Minase * * http://nayuki.eigenstate.org/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions */ public final class p034 implements EulerSolution { public static void main(String[] args) { System.out.println(new p034().run()); } public String run() { // As stated in the problem, 1 = 1! and 2 = 2! are excluded. // If a number has at least n >= 8 digits, then even if every digit is 9, // n * 9! is still less than the number (which is at least 10^n). int sum = 0; for (int i = 3; i < 10000000; i++) { if (i == factorialDigitSum(i)) sum += i; } return Integer.toString(sum); } // Hard-coded values for factorial(0), factorial(1), ..., factorial(9) private static int[] FACTORIAL = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880}; private static int factorialDigitSum(int x) { int sum = 0; while (x != 0) { sum += FACTORIAL[x % 10]; x /= 10; } return sum; } }

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