AMC 10 · 2025 · #18
Grade 8 number-theoryWhat is the ones digit of the sum
⌊1⌋+⌊2⌋+⌊3⌋+⋯+⌊2025⌋?(Recall that ⌊x⌋ represents the greatest integer less than or equal to x.)
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Add up $\lfloor\sqrt{k}\rfloor$ for every whole number $k$ from $1$ to $2025$, where $\lfloor x\rfloor$ is the greatest integer $\le x$, and report only the ones digit of that total.
Givens: The sum $\lfloor\sqrt{1}\rfloor+\lfloor\sqrt{2}\rfloor+\lfloor\sqrt{3}\rfloor+\dots+\lfloor\sqrt{2025}\rfloor$; $\lfloor x\rfloor$ means round down to the nearest whole number; $2025=45^2$, so the last term is $\lfloor\sqrt{2025}\rfloor=45$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $5$, (E) $8$
Unknowns: The ones (units) digit of the whole sum
Understand
Restated: Add up $\lfloor\sqrt{k}\rfloor$ for every whole number $k$ from $1$ to $2025$, where $\lfloor x\rfloor$ is the greatest integer $\le x$, and report only the ones digit of that total.
Givens: The sum $\lfloor\sqrt{1}\rfloor+\lfloor\sqrt{2}\rfloor+\lfloor\sqrt{3}\rfloor+\dots+\lfloor\sqrt{2025}\rfloor$; $\lfloor x\rfloor$ means round down to the nearest whole number; $2025=45^2$, so the last term is $\lfloor\sqrt{2025}\rfloor=45$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $5$, (E) $8$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Adding 2025 floor-roots one at a time is hopeless, but Tool #5 (Look for a Pattern) exposes two patterns that do the heavy lifting: each value $n$ repeats $2n+1$ times, and — because only the ones digit is wanted — the ones digits of the terms cycle every $10$. Tool #9 (Solve an Easier Related Problem) is what lets us throw away the true sizes and keep only ones digits (work mod $10$). Tool #7 (Identify Subproblems) splits the job into 'how many of each value' and 'ones digit of each block', and peels off the lone final term $45$.
Execute — Answer: D
8.EE.A.2 Step 1 Count how many times each value appears
- $\lfloor\sqrt{k}\rfloor$ equals $n$ for every $k$ from $n^2$ up to $(n+1)^2-1$.
- The count of those $k$ is $(n+1)^2-n^2=2n+1$.
- So $1$ appears $3$ times ($k=1,2,3$), $2$ appears $5$ times ($k=4..8$), $3$ appears $7$ times, and in general $n$ appears $2n+1$ times.
💡 Consecutive squares are spaced $2n+1$ apart, so each floor value sits flat for exactly that many steps.
4.OA.C.5 Step 2 Write the sum in block form
- Grouping equal values, the running blocks for $n=1$ to $44$ cover $k=1$ up to $45^2-1=2024$.
- The very last term $k=2025=45^2$ gives a single $45$.
- So the whole sum is $1\cdot3+2\cdot5+3\cdot7+\dots+44\cdot89+45$.
💡 Every value except the lone final $45$ comes in a full block of $2n+1$ copies.
4.NBT.B.5 Step 3 Keep only ones digits
- Because the question asks for the ones digit, replace each block total $n(2n+1)$ by just its ones digit.
- Multiplying out $n(2n+1)$ for $n=1$ to $10$ and keeping the last digit gives $3,0,1,6,5,8,5,6,1,0$.
- For example $5\cdot11=55$ ends in $5$ and $6\cdot13=78$ ends in $8$.
💡 The ones digit of a product depends only on the ones digits of its factors.
4.OA.C.5 Step 4 Use the 10-term cycle
- The ones digit of $n(2n+1)$ depends only on the ones digit of $n$, so the list of ones digits repeats every $10$ terms.
- One full cycle sums to $3+0+1+6+5+8+5+6+1+0=35$, whose ones digit is $5$.
- The terms $n=1$ to $40$ are four complete cycles, contributing $4\cdot5=20$, ones digit $0$.
💡 Once the ones digits repeat, whole cycles can be counted instead of every term.
4.NBT.B.4 Step 5 Add the leftover terms and finish
- Left over are $n=41,42,43,44$, whose ones digits match $n=1,2,3,4$: $3+0+1+6=10$, ones digit $0$.
- Then add the lone final term $45$, ones digit $5$.
- Total ones digit $=0+0+5=5$.
- So the ones digit of the whole sum is $5$, choice (D).
- (A full evaluation confirms $S=59775$, ending in $5$.)
💡 The ones digit of a sum is just the ones digit of the sum of the ones digits.
8.EE.A.2 $\lfloor\sqrt{k}\rfloor$ equals $n$ for every $k$ from $n^2$ up to $(n+1)^2-1$. 4.OA.C.5 Grouping equal values, the running blocks for $n=1$ to $44$ cover $k=1$ up to $4 4.NBT.B.5 Because the question asks for the ones digit, replace each block total $n(2n+1)$ 4.OA.C.5 The ones digit of $n(2n+1)$ depends only on the ones digit of $n$, so the list o 4.NBT.B.4 Left over are $n=41,42,43,44$, whose ones digits match $n=1,2,3,4$: $3+0+1+6=10$ Review
Reasonableness: Track the pieces mod $10$: four full cycles give $0$, the partial cycle $n=41..44$ gives $3+0+1+6=10\equiv0$, and the extra $45$ gives $5$, so $0+0+5=5$. This matches an exact computation using the sum formulas: $\sum_{n=1}^{44}n(2n+1)=2\cdot\frac{44\cdot45\cdot89}{6}+\frac{44\cdot45}{2}=58740+990=59730$, plus $45$ gives $59775$, which indeed ends in $5$. Both routes agree on $5$, ruling out every other choice.
Alternative: Compute the closed form directly: $\sum_{n=1}^{44} n(2n+1)+45 = 2\sum n^2+\sum n+45$. Using $\sum_{1}^{44} n^2=\tfrac{44\cdot45\cdot89}{6}=29370$ and $\sum_{1}^{44} n=990$ gives $2(29370)+990+45=59775$; its ones digit is $5$. This skips the cycle argument but needs the two summation formulas.
CCSS standards used (min grade 8)
8.EE.A.2Use square root and cube root symbols to represent solutions (Reading $\lfloor\sqrt{k}\rfloor$ and knowing $\lfloor\sqrt{k}\rfloor=n$ exactly between the perfect squares $n^2$ and $(n+1)^2$.)4.OA.C.5Generate a number or shape pattern following a given rule (Finding that value $n$ repeats $2n+1$ times and that the ones digits of the terms cycle every $10$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing each product $n(2n+1)$ and reading off its ones digit.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the block ones digits and the leftover terms to get the final ones digit.)
⭐ When only the ones digit matters, keep just the last digit of every piece and watch for a repeating cycle instead of adding thousands of terms.
⭐ When only the ones digit matters, keep just the last digit of every piece and watch for a repeating cycle instead of adding thousands of terms.
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