AMC10 2024 B

AMC10 2024 B · Q24

AMC10 2024 B · Q24. It mainly tests Fractions, Remainders & modular arithmetic.

Let \[P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}\] How many of the values $P(2022)$, $P(2023)$, $P(2024)$, and $P(2025)$ are integers?

设 \[P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}\] 其中$P(2022)$、$P(2023)$、$P(2024)$、$P(2025)$中有多少个是整数？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (E)

正确答案：（E）

Solution

First, we know that $P(2022)$ and $P(2024)$ must be integers since they are both divisible by $2$. Then let’s consider the remaining two numbers. Since they are not divisible by $2$, the result of the first term must be a certain number $+\frac{1}{2}$. The divisibility rule for $4$ states that the last two digits of the number must be divisible by $4$, and we know that the last two digits of ${2023}^2$ are the last two digits of ${23}^2$, which is $29$. And $29$ is $1$ greater than a multiple of $4$. Therefore, the result of the second term must be $\frac{1}{4}$. Similarly, the remaining two terms must each be $\frac{1}{8}$. Their sum is $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1$, so $P(2023)$ and $P(2025)$ are also integers. Therefore, the answer is $\boxed{\textbf{(E) }4}$.

首先，$P(2022)$和$P(2024)$必为整数，因为它们都被$2$整除。再考虑另外两个数。由于它们不被$2$整除，第一项结果一定是某个数$+\frac{1}{2}$。 $4$的整除规则是数的最后两位数字必须被$4$整除，$2023^2$的最后两位数字是$23^2$的最后两位，即$29$。$29$比$4$的倍数大$1$。因此，第二项结果为$\frac{1}{4}$。类似地，其余两项各为$\frac{1}{8}$。它们的和为$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{8} = 1$，所以$P(2023)$和$P(2025)$也是整数。因此，答案为$\boxed{\textbf{(E) }4}$。

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