AMC 10 · 2003 · #16

Grade 4 number-theory
Archetype Units-Digit Tracking └ Sub-archetype Power Cycle Lookup
units-digit-trackingmodular-arithmetic-mod-10pattern-recognition pattern-recognition ↑ Prerequisites: units-digit-tracking
📏 Medium solution 💡 2 insights
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Problem
Find the units digit (the ones place) of the number 13²⁰⁰³ — that is, the last digit you would see if you actually multiplied thirteen by itself 2003 times.

Pick an answer.

(A)
1
(B)
3
(C)
7
(D)
8
(E)
9

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Computing 13²⁰⁰³ is impossible by hand, so the plan is to track only the one piece that matters — the units digit. Tool #9 (Solve an Easier Related Problem) makes the first cut: when you multiply, the units digit of the product depends only on the units digits of the factors, so 13²⁰⁰³ has the same units digit as 3²⁰⁰³. That replaces 13 with the much smaller 3. Tool #5 (Look for a Pattern) then handles the giant exponent: the units digits of 3¹, 3², 3³, … repeat in a short cycle, so instead of 2003 multiplications we just find where 2003 lands in that cycle. Tool #3 (Eliminate Possibilities) confirms the result is one of the listed choices.

1STEP 1

Only the base's units digit matters

A product's ones place comes only from the factors' ones places, so 13²⁰⁰³ ends in the same digit as 3²⁰⁰³.

units digit of 13²⁰⁰³ = units digit of 3²⁰⁰³
2STEP 2

List units digits of powers of 3

3¹=3, 3²=9, 3³=27, 3⁴=81, 3⁵=243: the units digits run 3, 9, 7, 1 in a cycle of length 4.

3, 9, 7, 1, 3, 9, 7, 1_repeats, …
3STEP 3

Locate 2003 in the length-4 cycle

Divide the exponent by 4: 2003 = 4 × 500 + 3, so the remainder is 3 — the third slot of the block.

2003 = 4 × 500 + 3 → remainder 3
4STEP 4

Read off the answer

The third entry of the block 3, 9, 7, 1 is 7, so 13²⁰⁰³ ends in 7 — choice (C).

3rd term of (3, 9, 7, 1) = 7 → (C)
Answer
7
The check is that the four possible units digits of a power of 3 are exactly 3, 9, 7, 1, so the answer must be one of those — this immediately rules out (D) 8, which can never be the units digit of a power of 3. Among the remaining choices, the exponent 2003 leaves remainder 3 when divided by 4, and every exponent congruent to 3 (like 3, 7, 11, …) gives units digit 7; you can confirm on the small case 3³ = 27, which indeed ends in 7. So (C) is consistent, not a fluke of the large exponent.
💡Key takeaway

Only the last digit of the base matters, and last digits of powers of 3 cycle 3, 9, 7, 1; since 2003 leaves remainder 3 after dividing by 4, the last digit is the third one, 7.

  • Only the base's units digit matters
  • List units digits of powers of 3
  • Locate 2003 in the length-4 cycle
  • Read off the answer

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