Question

What is the ones digit of the sum $\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor$ $+ \cdots + \lfloor\sqrt{2024}\rfloor$ $+ \lfloor\sqrt{2025}\rfloor?$ (Recall that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x.$ )

$1$

$2$

$3$

$5$

$8$

Answer: D

Concepts:floor and ceiling functions sum of first n squares

Difficulty rating: 1730

Solution:

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.

2025 AMC 10B Problem 18

Answer: D

Difficulty rating: 1730

Solution:

Problem 18 in Other Years