149 lines
3.7 KiB
Markdown
149 lines
3.7 KiB
Markdown
---
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id: 5900f3ac1000cf542c50febf
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title: 'Problem 64: Odd period square roots'
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challengeType: 5
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forumTopicId: 302176
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dashedName: problem-64-odd-period-square-roots
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---
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# --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 $N \\le n$ have an odd period?
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# --hints--
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`oddPeriodSqrts(13)` should return a number.
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```js
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assert(typeof oddPeriodSqrts(13) === 'number');
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```
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`oddPeriodSqrts(500)` should return `83`.
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```js
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assert.strictEqual(oddPeriodSqrts(500), 83);
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```
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`oddPeriodSqrts(1000)` should return `152`.
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```js
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assert.strictEqual(oddPeriodSqrts(1000), 152);
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```
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`oddPeriodSqrts(5000)` should return `690`.
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```js
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assert.strictEqual(oddPeriodSqrts(5000), 690);
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```
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`oddPeriodSqrts(10000)` should return `1322`.
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```js
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assert.strictEqual(oddPeriodSqrts(10000), 1322);
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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 oddPeriodSqrts(n) {
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return true;
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}
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oddPeriodSqrts(13);
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```
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# --solutions--
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```js
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function oddPeriodSqrts(n) {
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// Based on https://www.mathblog.dk/project-euler-continued-fractions-odd-period/
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function getPeriod(num) {
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let period = 0;
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let m = 0;
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let d = 1;
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let a = Math.floor(Math.sqrt(num));
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const a0 = a;
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while (2 * a0 !== a) {
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m = d * a - m;
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d = Math.floor((num - m ** 2) / d);
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a = Math.floor((Math.sqrt(num) + m) / d);
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period++;
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}
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return period;
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}
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function isPerfectSquare(num) {
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return Number.isInteger(Math.sqrt(num));
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}
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let counter = 0;
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for (let i = 2; i <= n; i++) {
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if (!isPerfectSquare(i)) {
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if (getPeriod(i) % 2 !== 0) {
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counter++;
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}
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}
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}
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return counter;
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}
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```
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