AMC 10 · 2025 · #2
Grade 5 number-theoryPick an answer.
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 digits repeat every ten.
Only the last digit of a number affects the last digit of its square, so the endings must cycle.
Only the last digit of a number affects the last digit of its square, so the endings must cycle.
▸ Why?
Everything above the ones place is a pile of tens, and a pile of tens never reaches the last digit.
▸ Why?
With only ten possible endings, the pattern returns to its start and repeats forever.
Add one full block
One block sums to 45.
If a chunk repeats, you only need the sum of one chunk.
4.NBT.B.4Identify SubproblemsCount the whole blocks in 2025 terms
2025 is 202 blocks with five terms left.
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
The full blocks give 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
Adding the leftover 25 gives 9115.
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