AMC10 2023 A

AMC10 2023 A · Q21

AMC10 2023 A · Q21. It mainly tests Quadratic equations, Functions basics.

Let $P(x)$ be the unique polynomial of minimal degree with the following properties: The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime integers. What is $m+n$?

设 $P(x)$ 是具有以下性质的最低次数唯一多项式： $P(x)$ 的根都是整数，除了一个例外。那个非整数根可以写成 $\frac{m}{n}$，其中 $m$ 和 $n$ 是互质的整数。求 $m+n$？

(A) 41 41

(B) 43 43

(D) 47 47

(E) 49 49

Answer

Correct choice: (D)

正确答案：（D）

Solution

From the problem statement, we find $P(2-2)=0$, $P(9)=0$ and $4P(4)=0$. Therefore, we know that $0$, $9$, and $4$ are roots. So, we can factor $P(x)$ as $x(x - 9)(x - 4)(x - a)$, where $a$ is the unknown root. Since $P(x) - 1 = 0$, we plug in $x = 1$ which gives $1(-8)(-3)(1 - a) = 1$, so $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$. Therefore, our answer is $23 + 24 =\boxed{\textbf{(D) }47}$

从题目描述中，我们得知 $P(2-2)=0$，$P(9)=0$ 和 $4P(4)=0$。因此，我们知道 $0$、$9$ 和 $4$ 是根。所以，可以将 $P(x)$ 分解为 $x(x - 9)(x - 4)(x - a)$，其中 $a$ 是未知根。由于 $P(x) - 1 = 0$，代入 $x = 1$ 得到 $1(-8)(-3)(1 - a) = 1$，所以 $24(1 - a) = 1 \implies 1 - a = 1/24 \implies a = 23/24$。因此，答案是 $23 + 24 =\boxed{\textbf{(D) }47}$

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