Maybe like this:
function sumMul(n, m) { if (m <= n) return "INVALID"; const l = m - (m % n || n); return (n + l) / 2 * l / n; }
This is based on the fact that the average of [2, 8] is the same as the average of [2, 4, 6, 8], which means that (2 + 8) / 2 === (2 + 4 + 6 + 8) / 4, where 2 + 4 + 6 + 8 is the sum you're looking for.
[2, 8]
[2, 4, 6, 8]
(2 + 8) / 2 === (2 + 4 + 6 + 8) / 4
2 + 4 + 6 + 8
Good luck!
Great solution bro, I just tested it and it works 🔥
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Maybe like this:
This is based on the fact that the average of
[2, 8]is the same as the average of[2, 4, 6, 8], which means that(2 + 8) / 2 === (2 + 4 + 6 + 8) / 4, where2 + 4 + 6 + 8is the sum you're looking for.Good luck!
Great solution bro, I just tested it and it works 🔥