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Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:

1634 = 1^4 + 6^4 + 3^4 + 4^4
8208 = 8^4 + 2^4 + 0^4 + 8^4
9474 = 9^4 + 4^4 + 7^4 + 4^4
As 1 = 1^4 is not a sum it is not included.

The sum of these numbers is 1634 + 8208 + 9474 = 19316.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits (largest number is 6 digits long).

Code
Diff
  • function result() {
      let ans = [];
      let fivePow = 1;
      let sumOfPows = 0;
      while(fivePow < 999999) {
        fivePow++;
        sumOfPows = fivePow.toString().split('').reverse().reduce((a,v,i) => a + Math.pow(v,5),0);
        if(sumOfPows == fivePow) ans.push(fivePow);
      }
      return ans.reduce((a,v) => a + v,0);
    }
    • const result = () =>
    • Array.from(new Array(194980),(val,index)=>index)
    • .filter(a => a == a.toString().split("")
    • .map(e => Math.pow(parseInt(e), 5))
    • .reduce((a, b) => a+b))
    • .reduce((a,b) => a+b) - 1;
    • function result() {
    • let ans = [];
    • let fivePow = 1;
    • let sumOfPows = 0;
    • while(fivePow < 999999) {
    • fivePow++;
    • sumOfPows = fivePow.toString().split('').reverse().reduce((a,v,i) => a + Math.pow(v,5),0);
    • if(sumOfPows == fivePow) ans.push(fivePow);
    • }
    • return ans.reduce((a,v) => a + v,0);
    • }

Started with 3, because why waste 2 loops on answers you know aren't correct.

Code
Diff
  • function test(n) {
      let sum=0;
      for (let i=3; i<n; i++){
        if ((i%3==0) || (i%5==0)){
        sum+=i;
        }
      }
       return sum;
    }
    • //if you have any Suggestion or noticed a mistake in my code ...tell me...i would need that...
    • function test(n) {
    • var sum=0;
    • for (i=1; i<n; i++){
    • let sum=0;
    • for (let i=3; i<n; i++){
    • if ((i%3==0) || (i%5==0)){
    • sum+=i;
    • }
    • }
    • return sum;
    • }