2018-09-30 23:01:58 +01:00
										 
									 
								 
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								id: 5900f38b1000cf542c50fe9e
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								challengeType: 5
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								title: 'Problem 31: Coin sums'
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								## Description
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								<section id='description'>
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								In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
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								1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).
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								It is possible to make £2 in the following way:
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								1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
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								How many different ways can £(n) be made using any number of coins?
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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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								tests:
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								  - text: <code>coinSums(50)</code> should return 451.
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								    testString: assert(coinSums(50) == 451, '<code>coinSums(50)</code> should return 451.');
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								  - text: <code>coinSums(100)</code> should return 4563.
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								    testString: assert(coinSums(100) == 4563, '<code>coinSums(100)</code> should return 4563.');
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								  - text: <code>coinSums(150)</code> should return 21873.
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								    testString: assert(coinSums(150) == 21873, '<code>coinSums(150)</code> should return 21873.');
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								  - text: <code>coinSums(200)</code> should return 73682.
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								    testString: assert(coinSums(200) == 73682, '<code>coinSums(200)</code> should return 73682.');
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											2018-09-30 23:01:58 +01:00
										 
									 
								 
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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 coinSums(n) {
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								  // Good luck!
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								  return n;
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								}
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								coinSums(200);
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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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								const coinSums = (n) => {
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								  const getWays = (n, m=8, c=[1, 2, 5, 10, 20, 50, 100, 200]) => {
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								    if (n === 0) return 1;
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								    if (m === 0 || n < 0) return 0;
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								    return getWays(n - c[m - 1], m, c) + getWays(n, m - 1, c);
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								  };
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								  return getWays(n);
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								};
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								```
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								</section>
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