A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:
1/2 = 0.5
1/3 = 0.(3)
1/4 = 0.25
1/5 = 0.2
1/6 = 0.1(6)
1/7 = 0.(142857)
1/8 = 0.125
1/9 = 0.(1)
1/10 = 0.1'
Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.
Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
Interesting problem. Brute force is obvious. But, with a little big of math, this problem is rather simple. In order to check if we have created recurring cycle of a fraction, for example, how do we know to stop when 1/7 = 0.1428571? How do we know when we should stop at 7? Why not at 1? Sure, 1 does show up at the beginning, but it may lead to 142857123?
Let k be the length of the recurring cycle, and let n be the denomiator, then, $$ 10^k * \frac{1}{n} $$ This represents a rational number with integer portion of the number as the repeating part. Then, if we subtract 1/n, we get the repeating portion! $$ 10^k * \frac{1}{n} - \frac{1}{n} = g $$ For some g, where g is an integer. Let's multiply by n: $$ 10^k - 1 = gn $$ Let's (mod n): $$ 10^k - 1 \equiv0\pmod{n}$$ So, all we have to check is when is 10^k - 1 divisibl by n!
How about cycles that have non-repeating portion? For example, 0.166666? We can take care of the non-repeats by getting rid of any factors of two's or five's in the number. So, we ahve two function dedicated to this: Two(), Five()
Answer: 983
Runtime: 144 ms