9. Finding the Last Two Digits of Powers

This is a high-value exam trick for questions asking “find the last two digits of aⁿ.”

Case A — Base ends in 1: If the last two digits of the base are “x1” (tens digit x, units digit 1), then for (x1)ⁿ: - Last two digits = units digit is always 1; tens digit = (x × n) mod 10

Case B — Base ends in 3, 7, or 9: First convert to a power that ends in 1: - If base ends in 3 or 7: a⁴ always ends in 1 (e.g., 3⁴=81, 7⁴=2401) - If base ends in 9: a² always ends in 1 (e.g., 9²=81) Then split the exponent accordingly and apply Case A to the “ends in 1” part, separately computing the small leftover power.

Why it works: This follows from the fact that for any number coprime to 10, its powers’ last two digits repeat with a cycle that divides 20 (a consequence of Euler’s theorem, since φ(100) relates to 40 but the actual multiplicative order divides 20 for numbers coprime to 10). The “ends in 1” shortcut is simply the easiest special case of this cycle to compute by hand, so we manipulate every problem to land there.