53 lines
965 B
Markdown
53 lines
965 B
Markdown
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---
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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 n15 requires fourteen multiplications:
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n × n × ... × n = n15
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But using a "binary" method you can compute it in six multiplications:
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n × n = n2n2 × n2 = n4n4 × n4 = n8n8 × n4 = n12n12 × n2 = n14n14 × n = n15
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However it is yet possible to compute it in only five multiplications:
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n × n = n2n2 × n = n3n3 × n3 = n6n6 × n6 = n12n12 × n3 = n15
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We shall define m(k) to be the minimum number of multiplications to compute nk; for example m(15) = 5.
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For 1 ≤ k ≤ 200, find ∑ m(k).
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# --hints--
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`euler122()` should return 1582.
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```js
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assert.strictEqual(euler122(), 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 euler122() {
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return true;
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}
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euler122();
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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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