Problem 12 · 2012 AMC 8
Medium
Number Theory
units-digit-cycle
What is the units digit of 132012?
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Answer: A — 1.
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Hint 1 of 2
You'll never compute 132012 — and you don't have to. When you multiply, only the units digit of each factor affects the units digit of the answer, so 132012 ends in the same digit as 32012.
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Hint 2 of 2
Now list a few powers of 3 and watch the last digit: 3, 9, 7, 1, then 3 again. It cycles with period 4 — so the answer depends only on where 2012 lands in that cycle of 4.
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Approach: units digits repeat in a short cycle — find the position
- The units digit of a product depends only on the units digits being multiplied, so 132012 ends in the same digit as 32012. The big base is irrelevant.
- List the last digits of powers of 3: 31→3, 32→9, 33→7, 34→1, 35→3… They loop every 4 powers: (3, 9, 7, 1).
- So divide the exponent by 4 and look at the leftover: 2012 = 4 × 503 with leftover 0, meaning we land exactly on the end of a cycle — the 4th spot, which is 1.
- This transfers to every units-digit problem: last digits always cycle (period 1, 2, or 4 for single digits); find the cycle length, then reduce the exponent by that length. A leftover of 0 lands on the last entry, not the first.
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