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 $\sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} =\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty\frac1{2^a3^b5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.$ since this represents all the numbers in the denominator. (athens2016)

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 will consist of reciprocals of the powers of 5. We can easily calculate these to be $1, \frac{1}{2}, \frac{1}{4}$ respectively.

The fourth infinite series shall be all real numbers in the form $\frac{1}{2^a3^b}$ , where $a$ and $b$ are greater than or equal to 1.

The fifth is all real numbers in the form $\frac{1}{2^a5^b}$ , where $a$ and $b$ are greater than or equal to 1.

The sixth is all real numbers in the form $\frac{1}{3^a5^b}$ , where $a$ and $b$ are greater than or equal to 1.

The seventh infinite series is all real numbers in the form $\frac{1}{2^a3^b5^c}$ , 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 $a_{1}=1$ , $a_{2}=\frac{1}{2}$ , $a_{3}=\frac{1}{4}$ , let us find $a_{4}$ . factoring out $\frac{1}{6}$ from the terms in this subsequence, we would get $a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})$ .

Knowing $a_{1}$ and $a_{2}$ , we can substitute and solve for $a_{4}$ , and we get $\frac{1}{2}$ . If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them $\frac{1}{4}$ and $\frac{1}{8}$ .

Finally, for the seventh sequence, we see $a_{7}=\frac{1}{30}(a_{8})$ , 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 $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}(a_{8})=a_{8}$ , but when we separated the sequence into its parts, we ignored the $1/1$ , so adding in the $1$ , we get $1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}$ , which when we solve for, we get $\frac{29}{8}=\frac{29a_{8}}{30}$ , $\frac{1}{8}=\frac{a_{8}}{30}$ , $\frac{30}{8}=(a_{8})$ , $\frac{15}{4}=(a_{8})$ . So our answer is $\frac{15}{4}$ , but we are asked to add the numerator and denominator, which sums up to $19$ , which is the answer.

~~Edited by mprincess0229~~

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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2018 AMC 12A Problems/Problem 19

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Problem

Solution

Solution 2

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