2025 AMC 10B Problem 18

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

A floor function is constant on long stretches, so the right move is always to ask how long each stretch is rather than to evaluate terms one by one. Here the stretch lengths $2n+1$ are the odd numbers, which is why the sum reduces to a sum of squares. The same grouping handles $\lfloor \sqrt[3]{k} \rfloor$, $\lfloor k/m \rfloor$ and similar sums; the only care needed is at the top end, where the last block is usually incomplete.

Count each term by how many squares it dominates: $\lfloor \sqrt{k} \rfloor = \#\{n \ge 1 : n^2 \le k\}$. Swapping the order of summation,
$$S = \sum_{n=1}^{45} \#\{k \le 2025 : k \ge n^2\} = \sum_{n=1}^{45} \left(2026 - n^2\right) = 45 \cdot 2026 - \frac{45 \cdot 46 \cdot 91}{6} = 91170 - 31395 = 59775 .$$
This version avoids having to notice that the last block is a single term.

• Using $2n$ terms per block instead of $2n+1$, which loses the perfect squares themselves and changes the ones digit.
• Running the blocks up to $n = 45$, which would cover $k$ up to $45^2 + 90 = 2115$ instead of stopping at $2025$.
• Computing the whole sum with the digit-by-digit arithmetic done carelessly; tracking only the last digit of each of the three pieces is far safer.