Question

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

$2$

$5$

$10$

$12$

$15$

从左边提出因式 $(n+1)!$

Factor $(n+1)!$ out of the left side

除以 $n!$ 后得到 $(n+1)(n+3)=440$

Dividing by $n!$ gives $(n+1)(n+3)=440$

分解左边： $\begin{gathered} (n+1)!+(n+2)! \\ =(n+1)!\,[1+(n+2)] \\ =(n+1)!\,(n+3) \end{gathered}\text{。}$

两边同除以 $n!$ 并使用 $(n+1)!=(n+1)\,n!$ 得到 $(n+1)(n+3)=440\text{。}$

所以 $n^2+4n-437=0$ ，可分解为 $(n-19)(n+23)=0$ ，得 $n=19$ 。它的数字和为 $1+9=10$ 。

所以正确答案是 C。

Factor the left side: $\begin{gathered} (n+1)!+(n+2)! \\ =(n+1)!\,[1+(n+2)] \\ =(n+1)!\,(n+3). \end{gathered}$

Dividing both sides by $n!$ and using $(n+1)!=(n+1)\,n!$ gives $(n+1)(n+3)=440.$

So $n^2+4n-437=0,$ which factors as $(n-19)(n+23)=0,$ giving $n=19.$ Its digit sum is $1+9=10.$

Thus, C is the correct answer.