AMC10 2019 B

AMC10 2019 B · Q6

AMC10 2019 B · Q6. It mainly tests Quadratic equations, Manipulating equations.

A positive integer $n$ satisfies the equation $(n+1)!+(n+2)!=440 \cdot n!$. What is the sum of the digits of $n$?

一个正整数 $n$ 满足方程 $(n+1)!+(n+2)!=440 \cdot n!$. $n$ 的各位数字之和是多少？

(A) 2 2

(B) 5 5

(D) 12 12

(E) 15 15

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Dividing the given equation by $n!$, simplifying, and completing the square yields \[ (n+1) + (n+2)(n+1) = 440 \] \[ n^2 + 4n + 3 = 440 \] \[ n^2 + 4n + 4 = 441 \] \[ (n+2)^2 = 21^2. \] Thus $n+2=21$ and $n=19$. The requested sum of the digits of $n$ is $1+9=10$.

答案（C）：将所给等式除以 $n!$，化简并配方得到 \[ (n+1) + (n+2)(n+1) = 440 \] \[ n^2 + 4n + 3 = 440 \] \[ n^2 + 4n + 4 = 441 \] \[ (n+2)^2 = 21^2. \] 因此 $n+2=21$，$n=19$。所求 $n$ 的各位数字之和为 $1+9=10$。

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