Question

Un entero positivo $n$ satisface la ecuación $(n+1)!+(n+2)!=440\cdot n!.$ ¿Cuánto vale la suma de los dígitos de $n$ ?

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

$2$

$5$

$10$

$12$

$15$

Factoriza el lado izquierdo: $\begin{gathered} (n+1)!+(n+2)! \\ =(n+1)!\,[1+(n+2)] \\ =(n+1)!\,(n+3). \end{gathered}$

Dividir ambos lados entre $n!$ y usar $(n+1)!=(n+1)\,n!$ da $(n+1)(n+3)=440.$

Así $n^2+4n-437=0,$ que se factoriza como $(n-19)(n+23)=0,$ dando $n=19.$ Su suma de dígitos es $1+9=10.$

Por lo tanto, C es la respuesta correcta.

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.