🔥 New! 2025 AMC 10 B

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Problem 18 Easy

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.)

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    5

  • E.

    8

Answer:D
Solution

\sum_{n=1}^{2025} \lfloor \sqrt{n} \rfloor = \sum_{\substack{1 \leq n \leq 2025 \\ k^2 \leq n < (k+1)^2}} k

For 1 \leq k \leq 44, there are (k+1)^2 - k^2 = 2k + 1 terms of value k.

For k = 45, there is exactly one term.

Thus:

\sum_{k=1}^{44} k \cdot (2k+1) + 45 = 2 \sum_{k=1}^{44} k^2 + \sum_{k=1}^{44} k + 45

= 2 \times \frac{1}{6} \times 44 \times 45 \times 89 + \frac{1}{2} \times 44 \times 45 + 45

Both of the first 2 terms have 0 as ones digit, so the answer is 5.

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