2025 AMC 10B 第 18 题

$\lfloor\sqrt{n}\rfloor = m$ for the $2m + 1$ values $m^2 \le n \le (m + 1)^2 - 1$

Since $\sqrt{2025} = 45,$ sum $m(2m + 1)$ for $m = 1$ to $44$ and add the final term $45$

For each $m,$ $\lfloor\sqrt{n}\rfloor = m$ on the $2m + 1$ integers $m^2 \le n \le (m + 1)^2 - 1.$ Since $\sqrt{2025} = 45,$ the terms with $1 \le m \le 44$ contribute $\sum_{m=1}^{44} m(2m + 1),$ and $n = 2025$ tacks on $45.$ That sum is $\sum_{m=1}^{44}(2m^2 + m)$ $= 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6}$ $+ \tfrac{44 \cdot 45}{2} = 58740$ $+ 990 = 59730,$ so the total is $59775.$ Its ones digit is $5.$ Therefore, the answer is D.