Question

Sea $A$ el conjunto de enteros positivos que no tienen factores primos distintos de $2,$ $3,$ o $5.$ La suma infinita $\begin{aligned} &\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ &\quad {}+ \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} \\ &\quad {}+ \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} \\ &\quad {}+ \frac{1}{18} + \frac{1}{20} + \cdots \end{aligned}$ de los recíprocos de todos los elementos de $A$ se puede expresar como $\tfrac{m}{n},$ donde $m$ y $n$ son enteros positivos primos entre sí. ¿Cuánto vale $m + n$ ?

Let $A$ be the set of positive integers that have no prime factors other than $2,$ $3,$ or $5.$ The infinite sum $\begin{aligned} &\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \\ &\quad {}+ \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} \\ &\quad {}+ \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} \\ &\quad {}+ \frac{1}{18} + \frac{1}{20} + \cdots \end{aligned}$ of the reciprocals of all the elements of $A$ can be expressed as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m + n?$

$16$

$17$

$19$

$23$

$36$

Solución:

Cada elemento de $A$ es de forma única $2^i 3^j 5^k$ con $i, j, k \ge 0,$ así que sumar todos los recíprocos se factoriza como $\begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}. \end{aligned}$ Esto es igual a $2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.$ Con $\gcd(15, 4) = 1,$ $m + n = 15 + 4 = 19.$

Por lo tanto, la respuesta correcta es C.

Each element of $A$ is uniquely $2^i 3^j 5^k$ with $i, j, k \ge 0,$ so summing all reciprocals factors as $\begin{aligned} &\left(\sum_{i \ge 0} \tfrac{1}{2^i}\right) \\ &\quad {}\cdot \left(\sum_{j \ge 0} \tfrac{1}{3^j}\right) \\ &\quad {}\cdot \left(\sum_{k \ge 0} \tfrac{1}{5^k}\right) \\ &= \frac{1}{1 - \frac12} \cdot \frac{1}{1 - \frac13} \\ &\quad {}\cdot \frac{1}{1 - \frac15}. \end{aligned}$ This equals $2 \cdot \tfrac32 \cdot \tfrac54 = \tfrac{15}{4}.$ With $\gcd(15, 4) = 1,$ $m + n = 15 + 4 = 19.$

Thus, the correct answer is C.

2018 AMC 12A Problema 19

Solución:

El Problema 19 en otros años