AMC 10 · 2023 · #8
Grade 4 arithmeticProblem
What is the units digit of ?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Find the units digit (the ones digit) of the very large number $2022^{2023} + 2023^{2022}$.
Givens: Two huge powers added together: $2022^{2023}$ and $2023^{2022}$; $2022$ ends in $2$ and $2023$ ends in $3$; Only the units digit of the sum is asked — not the full number; Answer choices: (A) $7$, (B) $1$, (C) $9$, (D) $5$, (E) $3$
Unknowns: The units digit of $2022^{2023} + 2023^{2022}$
Understand
Restated: Find the units digit (the ones digit) of the very large number $2022^{2023} + 2023^{2022}$.
Givens: Two huge powers added together: $2022^{2023}$ and $2023^{2022}$; $2022$ ends in $2$ and $2023$ ends in $3$; Only the units digit of the sum is asked — not the full number; Answer choices: (A) $7$, (B) $1$, (C) $9$, (D) $5$, (E) $3$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
Computing $2022^{2023}$ exactly is impossible by hand, so Tool #9 (Easier Problem) says: shrink the question. The units digit of $2022^{2023}$ is the same as the units digit of $2^{2023}$, and the units digit of $2023^{2022}$ is the same as the units digit of $3^{2022}$ — 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 $2023 \bmod 4$ and $2022 \bmod 4$ 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.
Execute — Answer: A
3.OA.B.5 Step 1 - Replace $2022$ and $2023$ by their units digits.
- Because multiplying long numbers only mixes the ones place with the ones place, the units digit of $2022^{2023}$ equals the units digit of $2^{2023}$, and the units digit of $2023^{2022}$ equals the units digit of $3^{2022}$.
💡 Stripping each base to its units digit is the easier-problem move — Grade 3 properties of multiplication on the ones place.
3.OA.D.9 Step 2 List the units digit of $2^k$ for small $k$ to spot a pattern.
💡 Writing out a small table to see a repeating block — Grade 3 arithmetic pattern hunting.
4.NBT.B.6 Step 3 - The block $(2, 4, 8, 6)$ has length $4$ and repeats forever.
- To find the units digit of $2^{2023}$, divide $2023$ by $4$ and read off the remainder.
- Remainder $1, 2, 3, 0$ matches the $1$st, $2$nd, $3$rd, $4$th entries of the block.
💡 One division-with-remainder picks the cycle slot — Grade 4 multi-digit division.
3.OA.D.9 Step 4 - Repeat for powers of $3$.
- List the units digits of $3^k$ for small $k$.
💡 Same small-table technique catches the $(3, 9, 7, 1)$ block — Grade 3 arithmetic pattern hunting.
4.NBT.B.6 Step 5 - The block $(3, 9, 7, 1)$ also has length $4$.
- Find the slot for exponent $2022$ via remainder of $2022 \div 4$.
- Remainder $2$ picks the second entry.
💡 Same Grade 4 division-with-remainder trick to read the right slot.
1.NBT.C.4 Step 6 - Add the two units digits and take the units digit of the sum.
- $8 + 9 = 17$, whose ones digit is $7$.
💡 Adding two single-digit numbers and reading the ones place — Grade 1 single-digit sum.
3.OA.B.5 Replace $2022$ and $2023$ by their units digits. Because multiplying long number 3.OA.D.9 List the units digit of $2^k$ for small $k$ to spot a pattern. 4.NBT.B.6 The block $(2, 4, 8, 6)$ has length $4$ and repeats forever. To find the units d 3.OA.D.9 Repeat for powers of $3$. List the units digits of $3^k$ for small $k$. 4.NBT.B.6 The block $(3, 9, 7, 1)$ also has length $4$. Find the slot for exponent $2022$ 1.NBT.C.4 Add the two units digits and take the units digit of the sum. $8 + 9 = 17$, whos Review
Reasonableness: Quick checks. (1) The cycle-length claim survives small spot-checks: $2^4 = 16$ ends in $6$ (slot $4$ ✓), $2^5 = 32$ ends in $2$ (slot $1$, since $5 \bmod 4 = 1$ ✓), $3^4 = 81$ ends in $1$ (slot $4$ ✓). (2) Endpoint: replacing $2023$ with $3$ (so the exponent is $3$, remainder $3$) gives $2^3 = 8$, 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 $3^1$ — a bait answer.
Alternative: Tool #13 (Convert to Algebra) — modular arithmetic notation: $2^{2023} \equiv 2^{2023 \bmod 4} \equiv 2^3 \equiv 8 \pmod{10}$ and $3^{2022} \equiv 3^{2022 \bmod 4} \equiv 3^2 \equiv 9 \pmod{10}$, so the sum $\equiv 17 \equiv 7 \pmod{10}$. Same arithmetic, with the $\bmod 10$ formalism instead of the picture of repeating blocks. The block picture is friendlier; the $\pmod{10}$ notation is more compact.
CCSS standards used (min grade 4)
3.OA.B.5Apply properties of operations as strategies to multiply and divide (Reasoning that multiplying long numbers only mixes their ones places, so $2022^{2023}$ has the same units digit as $2^{2023}$.)3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Spotting the repeating blocks $(2, 4, 8, 6)$ and $(3, 9, 7, 1)$ in the units digits of powers of $2$ and $3$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Using $2023 \div 4 = 505$ remainder $3$ and $2022 \div 4 = 505$ remainder $2$ to pick the correct slot in each four-long block.)1.NBT.C.4Add within 100 including a two-digit number and a one-digit number (Computing $8 + 9 = 17$ and reading off the ones digit $7$.)
⭐ 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 $2023 \bmod 4 = 3$ picks $8$ and $2022 \bmod 4 = 2$ picks $9$, and $8 + 9 = 17$ ends in $7$.
⭐ 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 $2023 \bmod 4 = 3$ picks $8$ and $2022 \bmod 4 = 2$ picks $9$, and $8 + 9 = 17$ ends in $7$.