118 lines
2.9 KiB
Markdown
118 lines
2.9 KiB
Markdown
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---
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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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forumTopicId: 302176
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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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$\displaystyle \quad \quad \sqrt{N}=a_0+\frac 1 {a_1+\frac 1 {a_2+ \frac 1 {a3+ \dots}}}$
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For example, let us consider $\sqrt{23}:$:
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$\quad \quad \sqrt{23}=4+\sqrt{23}-4=4+\frac 1 {\frac 1 {\sqrt{23}-4}}=4+\frac 1 {1+\frac{\sqrt{23}-3}7}$
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If we continue we would get the following expansion:
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$\displaystyle \quad \quad \sqrt{23}=4+\frac 1 {1+\frac 1 {3+ \frac 1 {1+\frac 1 {8+ \dots}}}}$
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The process can be summarized as follows:
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$\quad \quad a_0=4, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7$
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$\quad \quad a_1=1, \frac 7 {\sqrt{23}-3}=\frac {7(\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2$
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$\quad \quad a_2=3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7$
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$\quad \quad a_3=1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} 7=8+\sqrt{23}-4$
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$\quad \quad a_4=8, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7$
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$\quad \quad a_5=1, \frac 7 {\sqrt{23}-3}=\frac {7 (\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2$
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$\quad \quad a_6=3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7$
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$\quad \quad a_7=1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} {7}=8+\sqrt{23}-4$
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It can be seen that the sequence is repeating. For conciseness, we use the notation $\sqrt{23}=[4;(1,3,1,8)]$, 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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$\quad \quad \sqrt{2}=[1;(2)]$, period = 1
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$\quad \quad \sqrt{3}=[1;(1,2)]$, period = 2
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$\quad \quad \sqrt{5}=[2;(4)]$, period = 1
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$\quad \quad \sqrt{6}=[2;(2,4)]$, period = 2
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$\quad \quad \sqrt{7}=[2;(1,1,1,4)]$, period = 4
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$\quad \quad \sqrt{8}=[2;(1,4)]$, period = 2
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$\quad \quad \sqrt{10}=[3;(6)]$, period = 1
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$\quad \quad \sqrt{11}=[3;(3,6)]$, period = 2
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$\quad \quad \sqrt{12}=[3;(2,6)]$, period = 2
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$\quad \quad \sqrt{13}=[3;(1,1,1,1,6)]$, period = 5
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Exactly four continued fractions, for $N \le 13$, have an odd period.
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How many continued fractions for $N \le 10\,000$ 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>oddPeriodSqrts()</code> should return a number.
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testString: assert(typeof oddPeriodSqrts() === 'number');
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- text: <code>oddPeriodSqrts()</code> should return 1322.
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testString: assert.strictEqual(oddPeriodSqrts(), 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 oddPeriodSqrts() {
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return true;
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}
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oddPeriodSqrts();
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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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