From charlesreid1

Problem

Problem 52 on Project Euler: https://projecteuler.net/problem=52

Solution

Link to solution on git.charlesreid1.com: https://charlesreid1.com:3000/cs/euler/src/master/scratch/Round2_050-070/052

Explanation of Solution

The key here was structuring the program so that it uses chained method calls, so that you stop multiplying and move on to the next candidate as soon as you have a failure. That, combined with the concept that we wanted the same SET of digits, meant we needed a method that would turn an arbitrary integer into a set of digits, and we could compare as we go.

Here's how the final product looks:

public class PermutedMultiples { 
	public static void main(String[] args) { 
		boolean found = false;
		Set<Integer> digits = new HashSet<Integer>();

		int n = 100;
		while(!found) { 
			// 1x 
			Set<Integer> s = getSet(n);
			if(s.equals(getSet(2*n))) {
				if(s.equals(getSet(3*n))) {
					if(s.equals(getSet(4*n))) {
						if(s.equals(getSet(5*n))) {
							if(s.equals(getSet(6*n))) {
								System.out.println("x = "+n);
								found = true;
							}
						}
					}
				}
			}
			
			n++;
		}
	}

	public static Set<Integer> getSet(int n) { 
		Set<Integer> result = new HashSet<Integer>();
		while(n!=0) { 
			int r = n%10;
			n = n/10;
			result.add(r);
		}
		return result;
	}

}

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