179 lines
2.1 KiB
Markdown
179 lines
2.1 KiB
Markdown
---
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id: 5900f3ac1000cf542c50febf
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challengeType: 5
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title: 'Problem 64: Odd period square roots'
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---
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## Description
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<section id='description'>
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All square roots are periodic when written as continued fractions and can be written in the form:
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√N = a0 +
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1
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a1 +
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1
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a2 +
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1
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a3 + ...
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For example, let us consider √23:
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√23 = 4 + √23 — 4 = 4 +
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1
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= 4 +
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1
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1√23—4
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1 +
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√23 – 37
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If we continue we would get the following expansion:
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√23 = 4 +
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1
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1 +
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1
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3 +
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1
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1 +
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1
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8 + ...
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The process can be summarised as follows:
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a0 = 4,
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1√23—4
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=
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√23+47
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= 1 +
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√23—37
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a1 = 1,
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7√23—3
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=
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7(√23+3)14
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= 3 +
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√23—32
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a2 = 3,
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2√23—3
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=
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2(√23+3)14
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= 1 +
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√23—47
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a3 = 1,
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7√23—4
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=
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7(√23+4)7
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= 8 +
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√23—4
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a4 = 8,
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1√23—4
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=
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√23+47
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= 1 +
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√23—37
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a5 = 1,
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7√23—3
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=
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7(√23+3)14
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= 3 +
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√23—32
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a6 = 3,
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2√23—3
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=
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2(√23+3)14
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= 1 +
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√23—47
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a7 = 1,
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7√23—4
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=
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7(√23+4)7
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= 8 +
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√23—4
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It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
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The first ten continued fraction representations of (irrational) square roots are:
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√2=[1;(2)], period=1
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√3=[1;(1,2)], period=2
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√5=[2;(4)], period=1
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√6=[2;(2,4)], period=2
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√7=[2;(1,1,1,4)], period=4
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√8=[2;(1,4)], period=2
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√10=[3;(6)], period=1
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√11=[3;(3,6)], period=2
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√12= [3;(2,6)], period=2
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√13=[3;(1,1,1,1,6)], period=5
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Exactly four continued fractions, for N ≤ 13, have an odd period.
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How many continued fractions for N ≤ 10000 have an odd period?
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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>euler64()</code> should return 1322.
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testString: assert.strictEqual(euler64(), 1322, '<code>euler64()</code> should return 1322.');
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```
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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 euler64() {
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// Good luck!
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return true;
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}
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euler64();
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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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</section>
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