2025 AMC 10B Problem 18

Attempt Problem 18 of the 2025 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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18.

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

Small hint:

\lfloor\sqrt{n}\rfloor = m

for the

2m + 1

values

m^2 \le n \le (m + 1)^2 - 1

Big hint:

Since

\sqrt{2025} = 45,

sum

m(2m + 1)

for

m = 1

to

44

and add the final term

45

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.