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

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Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\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</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
 
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\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</cmath>of the reciprocals of the elements of <math>A</math> can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?
  
<math>\textbf{(A)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math>
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<math>\textbf{(A) } 16 \qquad \textbf{(B) } 17 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 36</math>
  
== Solution ==
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== Solution 1 ==
It's just <cmath>
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Note that the fractions of the form <math>\frac{1}{2^a3^b5^c},</math> where <math>a,b,</math> and <math>c</math> are nonnegative integers, span all terms of the infinite sum.
\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)}.
 
</cmath> since this represents all the numbers in the denominator.
 
(athens2016)
 
  
== Solution 2==
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Therefore, the infinite sum becomes
Separate into 7 separate infinite series's so we can calculate each and find the original sum:
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<cmath>\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} \\
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&= 2\cdot\frac32\cdot\frac54 \\
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&= \frac{15}{4}
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\end{align*}</cmath>
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by a product of geometric series, from which the answer is <math>15+4=\boxed{\textbf{(C) } 19}.</math>
  
The first infinite sequence shall be all the reciprocals of the powers of <math>2</math>, the second shall be reciprocals of the powers of <math>3</math>, and the third will consist of reciprocals of the powers of 5. We can easily calculate these to be <math>1, \frac{1}{2}, \frac{1}{4}</math> respectively.
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~athens2016 (Solution)
  
The fourth infinite series shall be all real numbers in the form <math> \frac{1}{2^a3^b}</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1.
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~MRENTHUSIASM (Revision)
  
The fifth is all real numbers in the form <math> \frac{1}{2^a5^b}</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1.
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== Solution 2 ==
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Separate into <math>7</math> separate infinite series's so we can calculate each and find the original sum:
  
The sixth is all real numbers in the form <math> \frac{1}{3^a5^b}</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1.  
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The first infinite sequence shall be all the reciprocals of the powers of <math>2</math>, the second shall be reciprocals of the powers of <math>3</math>, and the third will consist of reciprocals of the powers of <math>5</math>. We can easily calculate these to be <math>1, \frac{1}{2}, \frac{1}{4}</math> respectively.  
  
The seventh infinite series is all real numbers in the form <math> \frac{1}{2^a3^b5^c}</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1.  
+
The fourth infinite series shall be all real numbers in the form <math> \frac{1}{2^a3^b}</math>, where <math>a,b\geq1</math>.
 +
 
 +
The fifth is all real numbers in the form <math> \frac{1}{2^a5^b}</math>, where <math>a,b\geq1</math>.
 +
 
 +
The sixth is all real numbers in the form <math> \frac{1}{3^a5^b}</math>, where <math>a,b\geq1</math>.
 +
 
 +
The seventh infinite series is all real numbers in the form <math> \frac{1}{2^a3^b5^c}</math>, where <math>a,b,c\geq1</math>.  
  
 
Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=\frac{1}{2}</math>, <math>a_{3}=\frac{1}{4}</math>, let us find <math>a_{4}</math>. factoring out <math>\frac{1}{6}</math> from the terms in this subsequence, we would get <math>a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})</math>.  
 
Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=\frac{1}{2}</math>, <math>a_{3}=\frac{1}{4}</math>, let us find <math>a_{4}</math>. factoring out <math>\frac{1}{6}</math> from the terms in this subsequence, we would get <math>a_{4}=\frac{1}{6}(1+a_{1}+a_{2}+a_{4})</math>.  
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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>1/1</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>19</math>, which is the answer.
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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>1/1</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>.
  
 
~~Edited by mprincess0229~~
 
~~Edited by mprincess0229~~
==Solution 3==
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Clearly this is just summing over the reciprocals of the numbers of the form <math>2^i3^j5^k</math>, where <math>i,j,k\in [0,\infty)</math>. SO our desired sum is <math>\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}</math>. By the infinite geometric series formula, <math>\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}</math> is just <math>\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}</math>. Applying the infinite geometric series formula again gives that <math>\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}</math>. Applying the infinite geometric series formula again yields <math>\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}</math>. Hence our final answer is <math>15+4=\boxed{19}</math>.
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== Solution 3 ==
 +
Clearly this is just summing over the reciprocals of the numbers of the form <math>2^i3^j5^k</math>, where <math>i,j,k\in [0,\infty)</math>. SO our desired sum is <math>\sum_{k=0}^{\infty}\sum_{j=0}^{\infty}\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}</math>. By the infinite geometric series formula, <math>\sum_{i=0}^{\infty}\frac{1}{2^i3^j5^k}</math> is just <math>\frac{\frac{1}{3^j5^k}}{1-\frac{1}{2}}=\frac{2}{3^j5^k}</math>. Applying the infinite geometric series formula again gives that <math>\sum_{j=0}^{\infty}\frac{2}{3^j5^k}=\frac{\frac{2}{5^k}}{1-\frac{1}{3}}=\frac{3}{5^k}</math>. Applying the infinite geometric series formula again yields <math>\sum_{k=0}^{\infty}\frac{3}{5^k}=\frac{3}{1-\frac{1}{5}}=\frac{15}{4}</math>. Hence our final answer is <math>15+4=\boxed{\textbf{(C) } 19}</math>.
  
 
-vsamc
 
-vsamc
 +
 +
== Video Solution by LetsSolveMathProblems ==
 +
 +
https://www.youtube.com/watch?v=woXlEargLpI&ab_channel=LetsSolveMathProblems
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}
 
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:19, 21 August 2021

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 (Solution)

~MRENTHUSIASM (Revision)

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 $a,b\geq1$.

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 $1/1$, 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 $\boxed{\textbf{(C) } 19}$.

~~Edited by mprincess0229~~

Solution 3

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

Video Solution by LetsSolveMathProblems

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

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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