Competition · AMC preparation · step 4 of 4
AMC 8 · 2012 · #12
Grade 4 number-theoryPick an answer.
AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Keep only the ones digit
A product's ones digit comes only from its factors' ones digits, so 13²⁰¹² ends the same way as the easier 3²⁰¹².
Multiplying multi-digit numbers and tracking how the ones place comes out of (tens + ones) × (tens + ones) is a Grade 4 place-value insight.
4.NBT.B.5Solve An Easier Related ProblemTable the powers of 3
Compute a few powers of 3 and keep only the ones digit — they come out 3, 9, 7, 1.
Computing 3 × 3, 9 × 3, 27 × 3, … is straight Grade 4 multi-digit multiplication.
4.NBT.B.5Solve An Easier Related ProblemSpot the repeating cycle
The ones digits repeat in a fixed block — the cycle 3, 9, 7, 1 has length 4.
Generating a sequence from a rule and noticing it repeats is exactly the Grade 4 "analyze patterns" standard.
4.OA.C.5Look For A PatternDivide 2012 by the cycle length
Divide the exponent by the cycle length to place 2012 in the cycle: 2012 ÷ 4 leaves remainder 0.
Finding a whole-number quotient with remainder for a 4-digit number divided by a 1-digit divisor is the Grade 4 division standard.
The ones digit of 3²⁰¹² is the same as the ones digit of 3⁴, the final entry of the four-long repeating block of ones digits.
▸ Why?
The ones digits of 3¹, 3², 3³, … settle into a fixed block of four (3, 9, 7, 1) that repeats forever, so the ones digits form a cycle of length 4.
▸ Why?
Each power is the one before it times 3, and the ones digit of a product is fixed by the ones digits of the factors alone, so a given ones digit always yields the same next ones digit; once 3⁵ ends in 3 again — matching 3¹ — the block 3, 9, 7, 1 is forced to repeat.
▸ Why?
When you multiply, the tens place and higher of a factor only ever contribute to the tens place and higher of the product, so nothing above the ones place can change the ones digit of the answer.
▸ Why?
2012 = 4 × 503 is a whole number of complete four-step blocks, so stepping 2012 times around a length-4 cycle ends exactly where a full block ends — the same position as step 4.
Read off the ones digit
Remainder 0 points to the last slot of the block, whose ones digit is 1 — so 13²⁰¹² ends in 1 too.
Mapping a position in a repeating cycle back to its value is the same pattern-analysis skill the table is built on.
4.OA.C.5Look For A PatternThis AMC 8 problem only needs Grade 4 pattern-spotting — find the cycle, divide to find the position, and read off the digit!
- Keep only the ones digit
- Table the powers of 3
- Spot the repeating cycle
- Divide 2012 by the cycle length
- Read off the ones digit
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