Question

Sean $A,$ $M,$ y $C$ dígitos tales que $\begin{aligned} &(100A + 10M + C) \\ &\quad {}\cdot (A + M + C) = 2005. \end{aligned}$ ¿Cuánto vale $A$ ?

Let $A,$ $M,$ and $C$ be digits with $\begin{aligned} &(100A + 10M + C) \\ &\quad {}\cdot (A + M + C) = 2005. \end{aligned}$ What is $A?$

$1$

$2$

$3$

$4$

$5$

Solución:

Como $A + M + C \le 9 + 9 + 9 = 27,$ y $2005 = 5 \cdot 401,$ la suma de dígitos debe ser el factor menor: $\begin{aligned} &100A + 10M + C = 401, \\ &\quad A + M + C = 5. \end{aligned}$

Leyendo los dígitos, $A = 4,$ $M = 0,$ y $C = 1.$

Así, la respuesta correcta es D.

Since $A + M + C \le 9 + 9 + 9 = 27,$ and $2005 = 5 \cdot 401,$ the digit sum must be the smaller factor: $\begin{aligned} &100A + 10M + C = 401, \\ &\quad A + M + C = 5. \end{aligned}$

Reading off the digits, $A = 4,$ $M = 0,$ and $C = 1.$

Thus, the correct answer is D.

2005 AMC 12A Problema 8

Solución:

El Problema 8 en otros años