Leftovers - Futility Closet

Leftovers

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

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

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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.