AMC 8 · 2012 · #12

Grade 4 number-theory

Archetype Modular Cycle Frequency Count └ Sub-archetype Power-Cycle Residue

units-digit-trackingexponentsmodular-arithmeticpattern-recognition pattern-recognitionmodular-arithmetic-mod-10 ↑ Prerequisites: multi-digit-arithmeticexponents

📏 Short solution 💡 3 insights

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Problem

What single digit appears in the ones place of 13²⁰¹²? We are not asked for the whole number, only its last digit.

Pick an answer.

(A)

$hspace{.05in}1$

(B)

$hspace{.05in}3$

(C)

$hspace{.05in}5$

(D)

$hspace{.05in}7$

(E)

$hspace{.05in}9$

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

The exponent 2012 is way too big to compute, so we use Tool #9 (Easier Problem): compute the first few powers 3¹, 3², 3³, … and look only at their ones digits. Tool #5 (Look for a Pattern) then takes over — powers of 3 cycle through ones digits {3, 9, 7, 1} every 4 steps. Once the cycle length is known, finding the 2012th term becomes a single division-with-remainder problem. We avoid Tool #13 (Algebra) because a clean cycle pattern makes algebra unnecessary.

1STEP 1

A product's ones digit comes only from its factors' ones digits, so 13²⁰¹² ends the same way as the easier 3²⁰¹².

ones digit of 13²⁰¹² = ones digit of 3²⁰¹²

2STEP 2

Compute a few powers of 3 and keep only the ones digit — they come out 3, 9, 7, 1.

3¹ = 3 → 3; 3² = 9 → 9; 3³ = 27 → 7; 3⁴ = 81 → 1; 3⁵ = 243 → 3

3STEP 3

The ones digits repeat in a fixed block — the cycle 3, 9, 7, 1 has length 4.

cycle = (3, 9, 7, 1), length = 4

4STEP 4

Divide the exponent by the cycle length to place 2012 in the cycle: 2012 ÷ 4 leaves remainder 0.

2012 ÷ 4 = 503 remainder 0

5STEP 5

Remainder 0 points to the last slot of the block, whose ones digit is 1 — so 13²⁰¹² ends in 1 too.

ones digit of 13²⁰¹² = 1 → (A)

Answer

Sanity-check the cycle on an exponent we can verify: 3⁸ = 6561, which ends in 1. Our rule says 8 ÷ 4 has remainder 0, so position "last in cycle" = 1. Matches. Try 3¹⁰ = 59049, ending in 9: 10 ÷ 4 = 2 remainder 2, so position 2 in the cycle = 9. Matches again. The cycle is reliable, so 3²⁰¹² ending in 1 is trustworthy.

💡Key takeaway

This AMC 8 problem only needs Grade 4 pattern-spotting — find the cycle, divide to find the position, and read off the digit!

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