2025 AMC 10B Problem 18

1. Look at the first few terms.

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.

1. 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 $\boxed{\textbf{(D)}\ 5}$.

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.

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 $0 + 0 + 5 \to 5$.

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 $\boxed{\textbf{(D)}\ 5}$.