2018 AMC 12A Problems/Problem 19

Revision as of 00:49, 25 January 2020 by Mprincess0229 (talk | contribs) (Solution 2)

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 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 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 $\frac{15}{4}$, but we are asked to add the numerator and denominator, which sums up to $19$, which is the answer.

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 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

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