AMC 10 · 2025 · #2
Easy mode Grade 5Square each whole number from 1 up to 2025. For every square, write down only its last digit (the ones digit). For example, 12=1 gives 1, 22=4 gives 4, 32=9 gives 9, and 42=16 gives 6. Add up all 2025 of these last digits. What is the total?
Pick an answer.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: For each of the first $2025$ positive perfect squares $1^2,2^2,3^2,\dots,2025^2$, take only the ones digit. Add all $2025$ of those ones digits together and report the total.
Givens: Jerry lists the ones digit of $n^2$ for every $n$ from $1$ to $2025$; The listed digits begin $1,4,9,6,5,6,\dots$ (the ones digits of $1^2,2^2,3^2,4^2,5^2,6^2,\dots$); Answer choices: (A) $9025$, (B) $9070$, (C) $9090$, (D) $9115$, (E) $9160$
Unknowns: The sum of all $2025$ ones digits that Jerry wrote down
Understand
Restated: For each of the first $2025$ positive perfect squares $1^2,2^2,3^2,\dots,2025^2$, take only the ones digit. Add all $2025$ of those ones digits together and report the total.
Givens: Jerry lists the ones digit of $n^2$ for every $n$ from $1$ to $2025$; The listed digits begin $1,4,9,6,5,6,\dots$ (the ones digits of $1^2,2^2,3^2,4^2,5^2,6^2,\dots$); Answer choices: (A) $9025$, (B) $9070$, (C) $9090$, (D) $9115$, (E) $9160$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems
Adding $2025$ separate ones digits by hand is hopeless, so Tool #5 (Look for a Pattern) is the key move: the ones digit of $n^2$ 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.
Execute — Answer: D
4.NBT.B.5 Step 1 Find the repeating block of ten
- The ones digit of a product depends only on the ones digits of the factors, so the ones digit of $n^2$ is set entirely by the ones digit of $n$.
- That means the ones digits of squares repeat every $10$ terms.
- Square the digits $1$ through $10$ and keep each ones digit: $1,4,9,6,5,6,9,4,1,0$.
- The next square, $11^2=121$, ends in $1$ again, so this block of ten repeats forever.
💡 Only the last digit of a number affects the last digit of its square, so the endings must cycle.
4.NBT.B.4 Step 2 Add one full block
- Add the ten ones digits in a single block to see what each complete cycle contributes: $1+4+9+6+5+6+9+4+1+0$.
- Pairing $1+9$, $4+6$, $9+1$, $6+4$ gives four tens, plus the leftover $5$, for $45$.
- Every full block of ten squares adds $45$ to the running total.
💡 If a chunk repeats, you only need the sum of one chunk.
5.NBT.B.6 Step 3 Count the whole blocks in 2025 terms
- Split the $2025$ terms into groups of $10$.
- Dividing, $2025\div10=202$ with a remainder of $5$, so $2025=202\times10+5$.
- That is $202$ complete blocks of ten, with $5$ terms left over at the end.
💡 The digits left of the ones place, $202$, count how many complete tens fit inside $2025$.
5.NBT.B.5 Step 4 Total from the full blocks
- Each of the $202$ complete blocks contributes $45$, so the full blocks together give $202\times45$.
- Compute it as $200\times45+2\times45=9000+90=9090$.
💡 Multiplying the block sum by the number of blocks handles all the repeats at once.
4.NBT.B.4 Step 5 Add the leftover five terms
- The $5$ leftover terms correspond to $n=2021,2022,2023,2024,2025$, whose ones digits match the start of the block: $1,4,9,6,5$.
- These add to $1+4+9+6+5=25$.
- Add this to the block total: $9090+25=9115$.
- That is choice (D).
- The trap value $9090$ in (C) is exactly what you get if you forget the $5$ leftover terms.
💡 The leftover terms just restart the same block, so their digits are the block's first few.
4.NBT.B.5 The ones digit of a product depends only on the ones digits of the factors, so t 4.NBT.B.4 Add the ten ones digits in a single block to see what each complete cycle contri 5.NBT.B.6 Split the $2025$ terms into groups of $10$. Dividing, $2025\div10=202$ with a re 5.NBT.B.5 Each of the $202$ complete blocks contributes $45$, so the full blocks together 4.NBT.B.4 The $5$ leftover terms correspond to $n=2021,2022,2023,2024,2025$, whose ones di Review
Reasonableness: A rough estimate confirms the size: the average ones digit is $\frac{45}{10}=4.5$, and $4.5\times2025\approx9112$, right beside $9115$. The total must also end in a $5$: $202\times45$ ends in $0$ and the leftover $25$ ends in $5$, so the sum ends in $5$ — matching (D) $9115$ and ruling out (B), (C), (E). Choice (C) $9090$ is the answer only if you drop the last $5$ terms, and (A) $9025$ comes from miscounting the blocks, so (D) is the consistent result.
Alternative: Instead of splitting off the leftover, note that the digits for $n=1$ to $5$ ($1,4,9,6,5$ summing to $25$) and for $n=6$ to $10$ ($6,9,4,1,0$ summing to $20$) make each half-block. There are $202$ full blocks ($202\times45=9090$) and then one extra first-half ($25$), giving $9090+25=9115$ again.
CCSS standards used (min grade 5)
4.NBT.B.5Multiply whole numbers using strategies based on place value (Reasoning that the ones digit of $n^2$ depends only on the ones digit of $n$, so the ones digits repeat every ten squares.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Summing one block to $45$, summing the leftover digits to $25$, and combining $9090+25=9115$.)5.NBT.B.6Find whole-number quotients and remainders (Dividing $2025$ by $10$ to get $202$ full blocks with a remainder of $5$ terms.)5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Computing $202\times45=9090$, the total contributed by all the full blocks.)
⭐ Ones digits of squares repeat every ten numbers, so count the full blocks, multiply by one block's sum, then add the few leftovers.
⭐ Ones digits of squares repeat every ten numbers, so count the full blocks, multiply by one block's sum, then add the few leftovers.
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