Project Euler – Problem 71 Solution

Yan Cui

I help clients go faster for less using serverless technologies.

Problem

Consider the fraction, n/d, where n and d are positive integers. If n < d and HCF(n,d)=1, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for d <= 8 in ascending order of size, we get:

1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8

It can be seen that 2/5 is the fraction immediately to the left of 3/7.

By listing the set of reduced proper fractions for d <= 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.

Solution

	#time
	let rec gcd a b =
	if b = 0
	then abs a
	else gcd b (a % b)
	let ``3/7`` = 3.0 / 7.0
	let answer =
	[8..1000000]
	|> List.map (fun d -> int (floor (``3/7`` * float d)), d)
	|> List.filter (fun (n, d) -> gcd n d = 1)
	|> List.maxBy (fun (n, d) -> float n / float d)
	|> fst // gets just the numerator

view raw gistfile1.fs hosted with by GitHub

This problem is fairly easy, given that the answer we’re looking for much be very close to 3 / 7 (0.4285714286) I simply iterate through the denominators, d, and find the closest numerator, n, which will produce a value less than 3 / 7. Then filter the set so we end up with only the n, d pairs that have a GCD of 1 and pick the numerator from the n, d pair whose n / d fraction is the biggest.