A problem from Litton Mathematical Recreations, which attributes it to Fermat circa 1635:

What is the remainder upon dividing 5^999,999 by 7?

When successive powers of 5 are divided by 7, the remainders form a repeating series:
5¹ / 7 = 0 remainder 5
5² / 7 = 3 remainder 4
5³ / 7 = 17 remainder 6
5⁴ / 7 = 89 remainder 2
5⁵ / 7 = 446 remainder 3
5⁶ / 7 = 2232 remainder 1
5⁷ / 7 = 11160 remainder 5
5⁸ / 7 = 55803 remainder 4
5⁹ / 7 = 279017 remainder 6
5¹⁰ / 7 = 1395089 remainder 2
…
The 999,999th term of the series is 6.