The Last Digit of a Monster

Question

What is the last digit of 7^100? No computation of 7^100, obviously, find the structure, use it, and explain why such structure had to exist.

Accepted Answer

Find the cycle. Working last-digits only (i.e. mod 10): 7^1 to 7, quad 7^2 = 49 to 9, quad 7^3 to 9cdot7 = 63 to 3, quad 7^4 to 3cdot7 = 21 to 1, and once you hit 1, the pattern repeats: the last digits cycle through 7, 9, 3, 1 with period 4. Since 100 = 4 cdot 25 is a multiple of 4, 7^100 sits at the end of a cycle, last digit mathbf1. (Slicker: 7^100 = (7^4)^25 = 2401^25, and a number ending in 1 stays ending in 1 under any power.) Why a cycle was guaranteed. The last digit of 7^n+1 is determined by the last digit of 7^n alone, multiplication mod 10 forgets everything else. With only 10 possible states, the sequence must eventually revisit a state and loop (pigeonhole). Because 7 shares no factor with 10, multiplication by 7 is invertible mod 10, so the loop has no lead-in tail and must return to the start, that's why the cycle is clean. The general theorem lurking here: for gcd(a, m) = 1, a^varphi(m) equiv 1 pmod m (Euler), and varphi(10) = 4 explains the period directly. Why this is a quant reflex, not trivia. "Reduce a huge computation to its residue" is the move behind checksum digits, hash-bucket assignment, and day-of-week/settlement-date arithmetic ("what weekday is 90 day

The Last Digit of a Monster

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