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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:

√N = a0 + 1

a1 + 1

a2 + 1

a3 + ...

For example, let us consider √23:

√23 = 4 + √23 — 4 = 4 + 1 = 4 + 1

1√23—4

1 + √23 – 37

If we continue we would get the following expansion:

√23 = 4 + 1

1 + 1

3 + 1

1 + 1

8 + ...

The process can be summarised as follows:

a0 = 4,

1√23—4 = √23+47 = 1 + √23—37 a1 = 1,

7√23—3 = 7(√23+3)14 = 3 + √23—32 a2 = 3,

2√23—3 = 2(√23+3)14 = 1 + √23—47 a3 = 1,

7√23—4 = 7(√23+4)7 = 8 + √23—4 a4 = 8,

1√23—4 = √23+47 = 1 + √23—37 a5 = 1,

7√23—3 = 7(√23+3)14 = 3 + √23—32 a6 = 3,

2√23—3 = 2(√23+3)14 = 1 + √23—47 a7 = 1,

7√23—4 = 7(√23+4)7 = 8 + √23—4

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.

The first ten continued fraction representations of (irrational) square roots are: √2=[1;(2)], period=1 √3=[1;(1,2)], period=2 √5=[2;(4)], period=1 √6=[2;(2,4)], period=2 √7=[2;(1,1,1,4)], period=4 √8=[2;(1,4)], period=2 √10=[3;(6)], period=1 √11=[3;(3,6)], period=2 √12= [3;(2,6)], period=2 √13=[3;(1,1,1,1,6)], period=5 Exactly four continued fractions, for N ≤ 13, have an odd period. How many continued fractions for N ≤ 10000 have an odd period?

Instructions

Tests

tests:
  - text: <code>euler64()</code> should return 1322.
    testString: assert.strictEqual(euler64(), 1322);

Challenge Seed

function euler64() {
  // Good luck!
  return true;
}

euler64();

Solution

// solution required

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