Euler Problem 71

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: We seek the fraction a/b < 3/7 with denominator at most 1,000,000 that minimizes $\frac37 - \frac{a}{b} = \frac{3b-7a}{7b}$.

The minimum value occurs when the numerator is 1 and the denominator is as large as possible.

The integer solutions to $3b - 7a = 1$ are given by $b = 5 + 7k$ and $a = 2 + 3k$. The largest value of $b$ occurs when $k = 142856$, so $a = 2 + 3(142856) = 428570$.