Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AMC 12A Problems/Problem 19"

m (→‎Solution 3)
m (→‎Solution 3)
Line 43: Line 43:
 
Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>\frac{1}{2}</math>. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>\frac{1}{4}</math> and <math>\frac{1}{8}</math>.  
 
Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>\frac{1}{2}</math>. If we do similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>\frac{1}{4}</math> and <math>\frac{1}{8}</math>.  
  
−
Finally, for the seventh sequence, we see <math>a_{7}=\frac{a_{8}}{30}</math>, where <math>a_{8}</math>  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 <math>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>\frac11</math>, so adding in the <math>1</math>, we get <math>1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}</math>, which when we solve for, we get <math>\frac{29}{8}=\frac{29a_{8}}{30}</math>, <math>\frac{1}{8}=\frac{a_{8}}{30}</math>, <math>\frac{30}{8}=(a_{8})</math>, <math>\frac{15}{4}=(a_{8})</math>. So our answer is <math>\frac{15}{4}</math>, but we are asked to add the numerator and denominator, which sums up to <math>\boxed{\textbf{(C) } 19}</math>.
+
Finally, for the seventh sequence, we see <math>a_{7}=\frac{a_{8}}{30}</math>, where <math>a_{8}</math>  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 <math>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>\frac11</math>, so adding in the <math>1</math>, we get <math>1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}</math>, which when we solve for, we get <math>\frac{29}{8}=\frac{29a_{8}}{30}</math>, or <math>\frac{15}{4}=a_{8}</math>. So our answer is <math>\frac{15}{4}</math>, but we are asked to add the numerator and denominator, which sums up to <math>\boxed{\textbf{(C) } 19}</math>.
  
 
~~Edited by mprincess0229~~
 
~~Edited by mprincess0229~~

Revision as of 02:06, 15 December 2022

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 $m+n$?

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

Solution 1

Note that the fractions of the form $\frac{1}{2^a3^b5^c},$ where $a,b,$ and $c$ are nonnegative integers, span all terms of the infinite sum.

Therefore, the infinite sum becomes \begin{align*} \sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty}\frac{1}{2^a3^b5^c} &= \left(\sum_{a=0}^{\infty}\frac{1}{2^a}\right)\cdot\left(\sum_{b=0}^{\infty}\frac{1}{3^b}\right)\cdot\left(\sum_{c=0}^{\infty}\frac{1}{5^c}\right) \\ &= \frac{1}{1-\frac12}\cdot\frac{1}{1-\frac13}\cdot\frac{1}{1-\frac15} \\ &= 2\cdot\frac32\cdot\frac54 \\ &= \frac{15}{4} \end{align*} by a product of geometric series, from which the answer is $15+4=\boxed{\textbf{(C) } 19}.$

~athens2016 ~MRENTHUSIASM

Solution 2

This solution is the same as Solution 1 but potentially clearer.

Clearly this is just summing over the reciprocals of the numbers of the form $2^i3^j5^k$, where $i,j,k\in [0,\infty)$. SO our desired sum is $\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}$. By the infinite geometric series formula, $\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}$ is just $\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}$. Applying the infinite geometric series formula again gives that $\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}$. Applying the infinite geometric series formula again yields $\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}$. Hence our final answer is $15+4=\boxed{\textbf{(C) } 19}$.

~vsamc

Solution 3

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 $a,b\geq1$.

The fifth is all real numbers in the form $\frac{1}{2^a5^b}$, where $a,b\geq1$.

The sixth is all real numbers in the form $\frac{1}{3^a5^b}$, where $a,b\geq1$.

The seventh infinite series is all real numbers in the form $\frac{1}{2^a3^b5^c}$, where $a,b,c\geq1$.

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 $a_{7}=\frac{a_{8}}{30}$, 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 $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{a_{8}}{30}=a_{8}$, but when we separated the sequence into its parts, we ignored the $\frac11$, 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 $\frac{29}{8}=\frac{29a_{8}}{30}$, or $\frac{15}{4}=a_{8}$. So our answer is $\frac{15}{4}$, but we are asked to add the numerator and denominator, which sums up to $\boxed{\textbf{(C) } 19}$.

~~Edited by mprincess0229~~

Video Solution by LetsSolveMathProblems

https://www.youtube.com/watch?v=woXlEargLpI&ab_channel=LetsSolveMathProblems

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&oldid=184032"