965 B
965 B
id | title | challengeType | forumTopicId | dashedName |
---|---|---|---|---|
5900f3e61000cf542c50fef9 | Problem 122: Efficient exponentiation | 5 | 301749 | problem-122-efficient-exponentiation |
--description--
The most naive way of computing n15 requires fourteen multiplications:
n × n × ... × n = n15
But using a "binary" method you can compute it in six multiplications:
n × n = n2n2 × n2 = n4n4 × n4 = n8n8 × n4 = n12n12 × n2 = n14n14 × n = n15
However it is yet possible to compute it in only five multiplications:
n × n = n2n2 × n = n3n3 × n3 = n6n6 × n6 = n12n12 × n3 = n15
We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
For 1 ≤ k ≤ 200, find ∑ m(k).
--hints--
euler122()
should return 1582.
assert.strictEqual(euler122(), 1582);
--seed--
--seed-contents--
function euler122() {
return true;
}
euler122();
--solutions--
// solution required