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# Difference between revisions of "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 $1/(2^a3^b)$, where $a$ and $b$ are greater than or equal to 1. The fifth is all real numbers in the form $1/(2^a5^b)$, where $a$ and $b$ are greater than or equal to 1. The sixth is all real numbers in the form $1/(3^a5^b)$, where $a$ and $b$ are greater than or equal to 1. The seventh infinite series is all real numbers in the form $1/(2^a3^b5^c)$, where $a$ and $b$ and $c$ are greater than or equal to 1. Let us denote the first sequence as $a_{1}$, the second as $a_{2}$, etc. We know $a_{1}=1$, $a_{2}=1/2$, $a_{3}=1/4$, let us find $a_{4}$. factoring out $1/6$ from the terms in this subsequence, we would get $a_{4}=1/6(1+a_{1}+a_{2}+a_{4})$. Knowing $a_{1}$ and $a_{2}$, we can substitute and solve for $a_{4}$, and we get $1/2$. If we do the similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them $1/4$ and $1/8$. Finally, for the seventh sequence, we see $a_{7}=1/30(a_{8})$, where $a_{8}$ is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get $1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}$, but when we separated the sequence into its parts, we ignored the $1/1$, so adding in the $1$, we get $1+1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}$, which when we solve for, we get $29/8=29/30(a_{8})$, $1/8=1/30(a_{8})$, $30/8=(a_{8})$, $15/4=(a_{8})$. So our answer is 15/4, but we are asked to add the numerator and denominator, which sums up to 19.

## See Also

 2018 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 18 Followed byProblem 20 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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