2018 AMC 12A Problems/Problem 19

Problem

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

$\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}$

Solution

It's just $\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.$ since this represents all the numbers in the denominator. (ayushk)

Solution 2

Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be $1, 1/2, 1/4$ respectively. The fourth infinite series shall be all real numbers in the form $1/(2^a3^b)$

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

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2018 AMC 12A Problems/Problem 19

Contents

Problem

Solution

Solution 2

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