AMC 10 · 2023 · #8

Grade 4 arithmetic
units-digit-trackingmodular-arithmetic-mod-10exponentspattern-recognition pattern-recognitionmodular-arithmetic-mod-10easier-related-problem ↑ Prerequisites: units-digit-trackingexponents
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Problem
Find the units digit (the ones digit) of the very large number 2022²⁰²³ + 2023²⁰²².

Pick an answer.

(A)
7
(B)
1
(C)
9
(D)
5
(E)
3

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Computing 2022²⁰²³ exactly is impossible by hand, so Tool #9 (Easier Problem) says: shrink the question. The units digit of 2022²⁰²³ is the same as the units digit of 2²⁰²³, and the units digit of 2023²⁰²² is the same as the units digit of 3²⁰²² — both follow because the units digit of a product depends only on the units digits of the factors. Tool #5 (Pattern) then takes over: list the first few units digits of powers of 2 and powers of 3, notice each cycles with period 4, and use 2023 mod 4 and 2022 mod 4 to land in the right slot. Tool #7 (Subproblems) keeps the two pieces clean — solve each units digit separately, then add and read the ones place. No algebra, no big arithmetic.

1STEP 1

A power's last digit depends only on its base's last digit, so 2022²⁰²³ matches 2²⁰²³ and 2023²⁰²² matches 3²⁰²².

2022²⁰²³ ≡ 2²⁰²³ (mod 10), 2023²⁰²² ≡ 3²⁰²² (mod 10)
2STEP 2

List the units digit of 2^k for small k to spot a pattern.

k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 ; 2^k & 2 & 4 & 8 & 16 & 32 & 64 & 128 & 256 ; units & 2 & 4 & 8 & 6 & 2 & 4 & 8 & 6
3STEP 3

The block 2, 4, 8, 6 repeats every 4, so 2023 mod 4 = 3 lands on the 3rd entry, 8.

2023 ÷ 4 = 505 remainder 3 → block entry 3 = 8
4STEP 4

Repeat for powers of 3. List the units digits of 3^k for small k.

k & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 ; 3^k & 3 & 9 & 27 & 81 & 243 & 729 & 2187 & 6561 ; units & 3 & 9 & 7 & 1 & 3 & 9 & 7 & 1
5STEP 5

The block 3, 9, 7, 1 also repeats every 4, so 2022 mod 4 = 2 lands on the 2nd entry, 9.

2022 ÷ 4 = 505 remainder 2 → block entry 2 = 9
6STEP 6

Add the two units digits: 8 + 9 = 17, whose ones digit is 7.

8 + 9 = 17 → units digit = 7 → (A)
Answer
7
Quick checks. (1) The cycle-length claim survives small spot-checks: 2⁴ = 16 ends in 6 (slot 4 ✓), 2⁵ = 32 ends in 2 (slot 1, since 5 mod 4 = 1 ✓), 3⁴ = 81 ends in 1 (slot 4 ✓). (2) Endpoint: replacing 2023 with 3 (so the exponent is 3, remainder 3) gives 2³ = 8, matching the formula. (3) Eliminate distractors: (B) 1 would need both exponents to land in the same column (6 + ? ending in 1 requires 5, not happening); (D) 5 never appears as a units digit of 2^k or 3^k; (E) 3 is just the units digit of 3¹ — a bait answer.
💡Key takeaway

This AMC 10 problem only needs Grade 4 division-with-remainder and pattern-spotting you already know — the units digit of 2^k cycles through 2, 4, 8, 6 and the units digit of 3^k cycles through 3, 9, 7, 1, so 2023 mod 4 = 3 picks 8 and 2022 mod 4 = 2 picks 9, and 8 + 9 = 17 ends in 7.