freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-64-odd-period-square-roots.md

id, challengeType, title, forumTopicId

id	challengeType	title	forumTopicId
5900f3ac1000cf542c50febf	5	Problem 64: Odd period square roots	302176

Description

All square roots are periodic when written as continued fractions and can be written in the form:

\displaystyle \quad \quad \sqrt{N}=a_0+\frac 1 {a_1+\frac 1 {a_2+ \frac 1 {a3+ \dots}}}

For example, let us consider \sqrt{23}::

\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}

If we continue we would get the following expansion:

\displaystyle \quad \quad \sqrt{23}=4+\frac 1 {1+\frac 1 {3+ \frac 1 {1+\frac 1 {8+ \dots}}}}

The process can be summarized as follows:

\quad \quad a_0=4, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7

\quad \quad a_1=1, \frac 7 {\sqrt{23}-3}=\frac {7(\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2

\quad \quad a_2=3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7

\quad \quad a_3=1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} 7=8+\sqrt{23}-4

\quad \quad a_4=8, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7

\quad \quad a_5=1, \frac 7 {\sqrt{23}-3}=\frac {7 (\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2

\quad \quad a_6=3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7

\quad \quad a_7=1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} {7}=8+\sqrt{23}-4

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.

The first ten continued fraction representations of (irrational) square roots are:

\quad \quad \sqrt{2}=[1;(2)], period = 1

\quad \quad \sqrt{3}=[1;(1,2)], period = 2

\quad \quad \sqrt{5}=[2;(4)], period = 1

\quad \quad \sqrt{6}=[2;(2,4)], period = 2

\quad \quad \sqrt{7}=[2;(1,1,1,4)], period = 4

\quad \quad \sqrt{8}=[2;(1,4)], period = 2

\quad \quad \sqrt{10}=[3;(6)], period = 1

\quad \quad \sqrt{11}=[3;(3,6)], period = 2

\quad \quad \sqrt{12}=[3;(2,6)], period = 2

\quad \quad \sqrt{13}=[3;(1,1,1,1,6)], period = 5

Exactly four continued fractions, for N \le 13, have an odd period.

How many continued fractions for N \le 10\,000 have an odd period?

Instructions

Tests

tests:
  - text: <code>oddPeriodSqrts()</code> should return a number.
    testString: assert(typeof oddPeriodSqrts() === 'number');
  - text: <code>oddPeriodSqrts()</code> should return 1322.
    testString: assert.strictEqual(oddPeriodSqrts(), 1322);

Challenge Seed

function oddPeriodSqrts() {

  return true;
}

oddPeriodSqrts();

Solution

// solution required