freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-122-efficient-exponentiation.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3e61000cf542c50fef9	Problem 122: Efficient exponentiation	5	301749	problem-122-efficient-exponentiation

--description--

The most naive way of computing n^{15} requires fourteen multiplications:

n × n × \ldots × n = n^{15}

But using a "binary" method you can compute it in six multiplications:

$$\begin{align} & n × n = n^2\\ & n^2 × n^2 = n^4\\ & n^4 × n^4 = n^8\\ & n^8 × n^4 = n^{12}\\ & n^{12} × n^2 = n^{14}\\ & n^{14} × n = n^{15} \end{align}$$

However it is yet possible to compute it in only five multiplications:

$$\begin{align} & n × n = n^2\\ & n^2 × n = n^3\\ & n^3 × n^3 = n^6\\ & n^6 × n^6 = n^{12}\\ & n^{12} × n^3 = n^{15} \end{align}$$

We shall define m(k) to be the minimum number of multiplications to compute n^k; for example m(15) = 5.

For 1 ≤ k ≤ 200, find \sum{m(k)}.

--hints--

efficientExponentation() should return 1582.

assert.strictEqual(efficientExponentation(), 1582);

--seed--

--seed-contents--

function efficientExponentation() {

  return true;
}

efficientExponentation();

--solutions--

// solution required