2018-09-30 23:01:58 +01:00
										 
									 
								 
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								id: 5900f3e61000cf542c50fef9
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								challengeType: 5
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								title: 'Problem 122: Efficient exponentiation'
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											2019-08-05 09:17:33 -07:00
										 
									 
								 
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								forumTopicId: 301749
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											2018-09-30 23:01:58 +01:00
										 
									 
								 
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								---
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								## Description
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								<section id='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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								</section>
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								## Instructions
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								<section id='instructions'>
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								</section>
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								## Tests
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								<section id='tests'>
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								```yml
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											2018-10-04 14:37:37 +01:00
										 
									 
								 
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								tests:
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								  - text: <code>euler122()</code> should return 1582.
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											2019-07-25 08:44:15 -07:00
										 
									 
								 
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								    testString: assert.strictEqual(euler122(), 1582);
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								</section>
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								## Challenge Seed
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								<section id='challengeSeed'>
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								<div id='js-seed'>
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								```js
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								function euler122() {
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								  // Good luck!
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								  return true;
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								}
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								euler122();
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								```
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								</div>
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								</section>
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								## Solution
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								<section id='solution'>
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								```js
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								// solution required
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								```
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											2019-07-18 08:24:12 -07:00
										 
									 
								 
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											2018-09-30 23:01:58 +01:00
										 
									 
								 
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								</section>
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