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

(Solution 2)
(Solution 2)
Line 16: Line 16:
 
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.  
 
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.  
  
The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</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</math> and <math>b</math> are greater than or equal to 1.  
  
The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1.  
+
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.  
  
The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1.  
+
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.  
  
The seventh infinite series is all real numbers in the form <math> 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 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.  
  
 
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>.  
Line 28: Line 28:
 
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{1}{30}(a_{8})</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{1}{30}(a_{8})=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+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, which when we solve for, we get <math>29/8=29/30(a_{8})</math>, <math>1/8=1/30(a_{8})</math>, <math>30/8=(a_{8})</math>, <math>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.
+
Finally, for the seventh sequence, we see <math>a_{7}=\frac{1}{30}(a_{8})</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{1}{30}(a_{8})=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{1}{30}(a_{8})=a_{8}</math>, which when we solve for, we get <math>\frac{29}{8}=\frac{29(a_{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.
  
 
~~Edited by mprincess0229~~
 
~~Edited by mprincess0229~~

Revision as of 01:00, 25 January 2020

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 $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{30}(a_{8})=a_{8}$, but when we separated the sequence into its parts, we ignored the $1/1$, so adding in the $1$, we get $1+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{30}(a_{8})=a_{8}$, which when we solve for, we get $\frac{29}{8}=\frac{29(a_{8})}{30}$, $\frac{1}{8}=\frac{(a_{8})}{30}$, $\frac{30}{8}=(a_{8})$, $\frac{15}{4}=(a_{8})$. 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~~

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

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

Invalid username
Login to AoPS