Project Euler – Problem 64 Solution


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


For example, let us consider ?23:


If we continue we would get the following expansion:


The process can be summarised as follows:


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?


(see full solution here).


Based on the algorithm on continued fractions from Wikipedia, we can implement the expansion algorithm as:


From the Wikipedia page:

The algorithm can also terminate on ai when ai = 2 a0, which is easier to implement.

which corresponds to the termination condition we have in the Repeat active pattern (which also checks if the accumulator is empty):


Also, this algorithm doesn’t work on numbers that are perfect squares, i.e. 4, 9, 16, … hence we need to exclude them when searching for our answer.

Here’s the solution in full:



This solution took 92ms to execute on my machine.

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Yan Cui

I’m an AWS Serverless Hero and the author of Production-Ready Serverless. I have run production workload at scale in AWS for nearly 10 years and I have been an architect or principal engineer with a variety of industries ranging from banking, e-commerce, sports streaming to mobile gaming. I currently work as an independent consultant focused on AWS and serverless.

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