AMC 10 · 2025 · #2
Grade 5 number-theoryPick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Adding 2025 separate ones digits by hand is hopeless, so Tool #5 (Look for a Pattern) is the key move: the ones digit of n² is fixed by the ones digit of n, so the list of ones digits repeats in blocks of 10. Once the repeating block is known, Tool #7 (Identify Subproblems) splits the count 2025 into whole blocks plus a short leftover, turning one giant sum into a small multiplication plus a tiny add.
Find the repeating block of ten
The ones digit of n² is fixed by the ones digit of n, so squaring 1 through 10 gives a block that repeats: 1,4,9,6,5,6,9,4,1,0.
Only the last digit of a number affects the last digit of its square, so the endings must cycle.
Only a number's last digit affects the last digit of its square, so the endings must cycle.
▸ Why?
A number is its digits weighted by their places, so higher places never touch the lowest one.
▸ Why?
With only ten possible last digits, the endings return to where they began and repeat.
Add one full block
Add one block: 1+4+9+6+5+6+9+4+1+0 pairs into four tens plus a 5, so every full cycle contributes 45.
If a chunk repeats, you only need the sum of one chunk.
4.NBT.B.4Identify SubproblemsCount the whole blocks in 2025 terms
Group the 2025 terms by ten: 2025÷10 leaves 202 full blocks with 5 terms left over.
The digits left of the ones place, 202, count how many complete tens fit inside 2025.
5.NBT.B.6Identify SubproblemsTotal from the full blocks
Each full block adds 45, so the 202 blocks give 202×45 = 9000+90 = 9090.
Multiplying the block sum by the number of blocks handles all the repeats at once.
5.NBT.B.5Identify SubproblemsAdd the leftover five terms
The leftovers n=2021 to 2025 end in 1,4,9,6,5, summing to 25, so 9090+25 = 9115 — choice (D).
The leftover terms just restart the same block, so their digits are the block's first few.
4.NBT.B.4Identify SubproblemsOnes digits of squares repeat every ten numbers, so count the full blocks, multiply by one block's sum, then add the few leftovers.
- Find the repeating block of ten
- Add one full block
- Count the whole blocks in 2025 terms
- Total from the full blocks
- Add the leftover five terms