AMC 10 · 2025 · #18

Grade 8 number-theory

Archetype Units-Digit Tracking └ Sub-archetype Digitwise Additive Reduction

floor-functionperfect-squaresunits-digit-tracking pattern-recognition ↑ Prerequisites: perfect-squares

📏 Medium solution 💡 3 insights

Problem

Add up

\lfloor\sqrt{k}\rfloor

for every whole number

k

from 1 to 2025, where

\lfloor x\rfloor

means the greatest integer less than or equal to

x

. Report only the ones digit of that total.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Look for a Pattern

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.

1STEP 1

Count how many times each value appears

$\lfloor\sqrt{k}\rfloor=n$ for every $k$ from $n^2$ to $(n+1)^2-1$ , so each value $n$ repeats $2n+1$ times: 1 three times, 2 five times.

⌊√(k)⌋=n for n² ≤ k ≤ (n+1)²-1, #=2n+1

2STEP 2

Write the sum in block form

Blocks for $n=1$ to $44$ cover $k=1$ through $2024$ , and the lone last term $k=2025=45^2$ adds a single $45$ .

S=Σ_n=1⁴⁴ n(2n+1) + 45

3STEP 3

Keep only ones digits

Keep only the ones digit of each block $n(2n+1)$ : for $n=1$ to $10$ they run 3,0,1,6,5,8,5,6,1,0, since $5\cdot11=55$ and $6\cdot13=78$ .

n(2n+1) ones digits, n=1..10: 3,0,1,6,5,8,5,6,1,0

4STEP 4

Use the 10-term cycle

That digit list depends only on the last digit of $n$ , so it repeats every 10: one cycle sums to $35$ , and four cycles up to $n=40$ leave $0$ .

3+0+1+6+5+8+5+6+1+0=35≡5 (mod 10), 4·5=20≡0

5STEP 5

Add the leftover terms and finish

The leftovers $n=41$ to $44$ repeat $3+0+1+6=10$ , so adding the lone $45$ leaves ones digit 5, choice (D).

0+(3+0+1+6)+45≡0+0+5≡5 (mod 10) → (D)

Answer

Track the pieces mod 10: four full cycles give 0, the partial cycle n=41..44 gives 3+0+1+6=10≡0, and the extra 45 gives 5, so 0+0+5=5. This matches an exact computation using the sum formulas: Σ_n=1⁴⁴n(2n+1)=2·(44·45·89)/6+(44·45)/2=58740+990=59730, plus 45 gives 59775, which indeed ends in 5. Both routes agree on 5, ruling out every other choice.

💡Key takeaway

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.

Count how many times each value appears
Write the sum in block form
Keep only ones digits
Use the 10-term cycle
Add the leftover terms and finish

Count how many times each value appears

Write the sum in block form

Keep only ones digits

Use the 10-term cycle

Add the leftover terms and finish

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