AMC10 2025 B

AMC10 2025 B · Q18

AMC10 2025 B · Q18. It mainly tests Inequalities with floors/ceilings (basic), Algebra misc.

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

下列和的个位数是多少 \[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \dots + \lfloor \sqrt{2025} \rfloor?\] （回忆$\lfloor x \rfloor$表示小于或等于$x$的最大整数。）

(A) 1 1

(B) 2 2

(D) 5 5

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

All terms from the 1st to the 3rd will equal 1, because the value inside the floor function will be greater than or equal to $\sqrt{1} = 1$ but less than $\sqrt{4} = 2$. Following this logic, all terms from the 4th to the 8th will equal 2, all terms from the 9th to the 15th will equal 3, and all terms from number $n^2$ to $(n + 1)^2 - 1$ will equal $\sqrt{n}$. Thus, there will be 3 terms equal to 1, 5 terms equal to 2, and so on. Writing out a few terms of the total sum, we have: \[1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots + 44 \cdot 89\] The expression above can be written in summation format as: \[\sum_{n=1}^{44} n(2n + 1).\] We can expand this to the following: \[\sum_{n=1}^{44} 2n^2 + \sum_{n=1}^{44} n\] Using the sum of integers formulas, this becomes \[\frac{2 \cdot 44 \cdot 45 \cdot 89}{6} + \frac{44 \cdot 45}{2}\] The units digit of that expression evaluates to 0. However, we excluded $\lfloor \sqrt{2025} \rfloor = 45$ from our sum because it appears only once in the sequence, so we have to add it back into our expression. Our final answer is $\boxed{\textbf{(D)}\hspace{3pt}5}$.

从第1到第3项的所有项都等于1，因为取整函数内的值大于等于$\sqrt{1} = 1$但小于$\sqrt{4} = 2$。依此类推，从第4到第8项都等于2，从第9到第15项都等于3，从$n^2$到$(n + 1)^2 - 1$的所有项都等于$n$。因此，有3项等于1，5项等于2，依此类推。写出总和的前几项，我们有： \[1 \cdot 3 + 2 \cdot 5 + 3 \cdot 7 + \dots + 44 \cdot 89\] 上面的表达式可以写成求和形式： \[\sum_{n=1}^{44} n(2n + 1).\] 我们可以展开为： \[\sum_{n=1}^{44} 2n^2 + \sum_{n=1}^{44} n\] 使用整数求和公式，这变为 \[\frac{2 \cdot 44 \cdot 45 \cdot 89}{6} + \frac{44 \cdot 45}{2}\] 该表达式的个位数为0。但是，我们从和中排除了$\lfloor \sqrt{2025} \rfloor = 45$，因为它在序列中只出现一次，所以必须加回它。最终答案为$\boxed{\textbf{(D)}\hspace{3pt}5}$。

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