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* fix: clean-up Project Euler 121-140 * fix: corrections from review Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com> * fix: missing backticks Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com> * fix: corrections from review Co-authored-by: Tom <20648924+moT01@users.noreply.github.com> * fix: missing delimiter Co-authored-by: Sem Bauke <46919888+Sembauke@users.noreply.github.com> Co-authored-by: Kristofer Koishigawa <scissorsneedfoodtoo@gmail.com> Co-authored-by: Tom <20648924+moT01@users.noreply.github.com>
66 lines
1.2 KiB
Markdown
66 lines
1.2 KiB
Markdown
---
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id: 5900f3e61000cf542c50fef9
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title: 'Problem 122: Efficient exponentiation'
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challengeType: 5
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forumTopicId: 301749
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dashedName: problem-122-efficient-exponentiation
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---
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# --description--
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The most naive way of computing $n^{15}$ requires fourteen multiplications:
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$$n × n × \ldots × n = n^{15}$$
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But using a "binary" method you can compute it in six multiplications:
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$$\begin{align}
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& n × n = n^2\\\\
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& n^2 × n^2 = n^4\\\\
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& n^4 × n^4 = n^8\\\\
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& n^8 × n^4 = n^{12}\\\\
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& n^{12} × n^2 = n^{14}\\\\
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& n^{14} × n = n^{15}
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\end{align}$$
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However it is yet possible to compute it in only five multiplications:
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$$\begin{align}
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& n × n = n^2\\\\
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& n^2 × n = n^3\\\\
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& n^3 × n^3 = n^6\\\\
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& n^6 × n^6 = n^{12}\\\\
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& n^{12} × n^3 = n^{15}
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\end{align}$$
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We shall define $m(k)$ to be the minimum number of multiplications to compute $n^k$; for example $m(15) = 5$.
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For $1 ≤ k ≤ 200$, find $\sum{m(k)}$.
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# --hints--
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`efficientExponentation()` should return `1582`.
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```js
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assert.strictEqual(efficientExponentation(), 1582);
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```
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# --seed--
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## --seed-contents--
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```js
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function efficientExponentation() {
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return true;
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}
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efficientExponentation();
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```
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# --solutions--
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```js
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// solution required
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```
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