2025 AMC 10B Problem 18

P16-20~5 minPrint

Sequences & Series Number Properties Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

What is the ones digit of the sum $\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \dots + \lfloor \sqrt{2025} \rfloor?$ (Recall that $\lfloor x \rfloor$ represents the greatest integer less than or equal to $x$ .)

Show solution without answering

Key insight

The value n appears 2n + 1 times, from n² to just before (n+1)². So add n(2n + 1) for n = 1 to 44, plus 45.

Solution

Look at the first few terms.

$k$	$1$ to $3$	$4$ to $8$	$9$ to $15$	$16$ to $24$
$\lfloor \sqrt{k} \rfloor$	$1$	$2$	$3$	$4$
how many terms	$3$	$5$	$7$	$9$

The value $n$ starts at $k = n^2$ and lasts until just before $(n+1)^2$ . That is $(n+1)^2 - n^2 = 2n + 1$ terms.

Where does it stop? $2025 = 45^2$ . So the values $1, 2, \ldots, 44$ each get their full stretch, ending at $k = 2024$ . Then $k = 2025$ alone gives $45$ .

3. Write the sum.
$S = \bigl(1\cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \cdots + 44 \cdot 89\bigr) + 45.$
Each piece is $n(2n+1) = 2n^2 + n$ , so
$S = 2\,(1^2 + 2^2 + \cdots + 44^2) + (1 + 2 + \cdots + 44) + 45.$

4. Use the standard sum formulas.
$1 + 2 + \cdots + 44 = \frac{44 \cdot 45}{2} = 990, \qquad 1^2 + 2^2 + \cdots + 44^2 = \frac{44 \cdot 45 \cdot 89}{6} = 29370.$

5. Add.
$S = 2 \cdot 29370 + 990 + 45 = 58740 + 990 + 45 = 59775.$
The ones digit is $5$ . (You can see this even sooner: $58740$ and $990$ both end in $0$ , so the ones digit comes from the $45$ .)

The answer is $\boxed{\textbf{(D)}\ 5}$ .

Why this works

A floor function stays the same for long stretches, so instead of working out $2025$ terms, ask how long each stretch is. The gaps between squares are the odd numbers $3, 5, 7, \ldots$ , which turns the sum into a sum of squares. Always check the top end, where the last stretch is usually cut short; here it is just one term.

Alternative approach

Only the ones digit is wanted, so track only ones digits. The ones digit of $n(2n+1)$ depends only on the ones digit of $n$ . For $n = 1$ to $10$ the pieces are $3, 10, 21, 36, 55, 78, 105, 136, 171, 210$ , with ones digits $3, 0, 1, 6, 5, 8, 5, 6, 1, 0$ , adding to $35$ .

$n = 1$ to $40$ : four such blocks, $4 \times 35 = 140$ , ones digit $0$ .
$n = 41$ to $44$ : same ones digits as $1$ to $4$ , which are $3 + 0 + 1 + 6 = 10$ , ones digit $0$ .
The last term: $45$ , ones digit $5$ .

So the ones digit is $0 + 0 + 5 \to 5$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Group the terms by their common value. For a positive integer $n$ ,
$\lfloor \sqrt{k} \rfloor = n \iff n^2 \le k \le (n+1)^2 - 1,$
and that block contains $(n+1)^2 - n^2 = 2n + 1$ integers.

Since $2025 = 45^2$ , the values $k = 1, 2, \dots, 2024$ are exactly the complete blocks for $n = 1, 2, \dots, 44$ (the last one ends at $45^2 - 1 = 2024$ ), and $k = 2025$ is left over on its own, contributing $45$ . Hence
$S = \sum_{n=1}^{44} n(2n+1) + 45 = 2\sum_{n=1}^{44} n^2 + \sum_{n=1}^{44} n + 45 .$

Now $\displaystyle\sum_{n=1}^{44} n^2 = \frac{44 \cdot 45 \cdot 89}{6} = 29370$ and $\displaystyle\sum_{n=1}^{44} n = \frac{44 \cdot 45}{2} = 990$ , so
$S = 58740 + 990 + 45 = 59775 .$

Only the ones digit is wanted, and it can be read off before the additions are finished: $2 \cdot 29370$ ends in $0$ , $990$ ends in $0$ , so the ones digit is the ones digit of $45$ , namely $5$ .

The answer is $\boxed{\textbf{(D)}\ 5}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 1: Integers: The Basics
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)