AMC12 2019 B
AMC12 2019 B · Q4
AMC12 2019 B · Q4. It mainly tests Linear equations, Digit properties (sum of digits, divisibility tests).
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
(C)
10
10
(D)
12
12
(E)
15
15
Answer
Correct choice: (C)
正确答案：（C）
Solution
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 $1+9 = 10$.
将给定方程除以 $n!$，化简并配方，得
$(n+1) + (n+2)(n+1) = 440$，
$n^2 + 4n + 3 = 440$，
$n^2 + 4n + 4 = 441$，
$(n+2)^2 = 21^2$。
因此 $n+2 = 21$，$n = 19$。$n$ 的各位数字之和为 $1+9 = 10$。
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