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1999 AMC 8

Problem 24

Problem 24 · 1999 AMC 8 Stretch
Number Theory units-digitcyclicity

When 19992000 is divided by 5, the remainder is

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Answer: D — 1.
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Hint 1 of 2
You'll never compute 1999²⁰⁰⁰ — and you don't have to. The remainder when dividing by 5 depends only on the units digit, and the units digit of a power depends only on the units digit of the base. So really you're asking: what does 9²⁰⁰⁰ end in?
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Hint 2 of 2
Units digits of powers always fall into a short repeating cycle. List a few powers of 9 and the pattern jumps out; then use whether the exponent is odd or even.
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Approach: the units digit cycles — find where the even exponent lands
  1. Dividing by 5 only cares about the last digit (10, 20, 30, … are all multiples of 5), and the last digit of a power only depends on the base's last digit, 9.
  2. Powers of 9 cycle in their units digit: 9, 81, 729, … → 9, 1, 9, 1, … Odd exponent → ends in 9, even exponent → ends in 1. Since 2000 is even, 1999²⁰⁰⁰ ends in 1.
  3. A number ending in 1 is one more than a multiple of 10 (hence of 5), so the remainder is 1. The transferable idea: for last-digit or mod-5/mod-10 questions, ignore the giant number and ride the short repeating cycle of units digits.
Another way — reduce the base mod 5 first:
  1. Modulo 5, 1999 leaves remainder 4 (since 2000 is a multiple of 5), so 1999²⁰⁰⁰ ≡ 4²⁰⁰⁰ (mod 5).
  2. And 4 ≡ −1 (mod 5), so 4²⁰⁰⁰ ≡ (−1)²⁰⁰⁰ = 1 (mod 5).
  3. The remainder is 1. Spotting that the base is −1 away from a multiple of 5 makes an even power collapse to 1 instantly — a clean trick once you've met modular arithmetic.
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