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Difference between revisions of "2005 AIME II Problems/Problem 3"

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==Solution 2==
 
==Solution 2==
 
We can write the sum of the original series as <math>a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005</math>, where the common ratio is equal to <math>\dfrac{m}{n}</math>. We can also write the sum of the second series as <math>a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050</math>. Using the formula for the sum of an infinite geometric series <math>S=\dfrac{a}{1-r}</math>, where <math>S</math> is the sum of the sequence, <math>a</math> is the first term of the sequence, and <math>r</math> is the ratio of the sequence, the sum of the original series can be written as <math>\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}</math>, and the second sequence can be written as <math>\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}</math>. Dividing <math>\text{(2)}</math> by <math>\text{(1)}</math>, we obtain <math>\dfrac{a\cdot n}{m+n}=10</math>, which can also be written as <math>a\cdot n=10(m+n)</math>. Substitute this value for <math>a\cdot n</math> back into <math>\text{(1)}</math>, we obtain <math>10\cdot \dfrac{n+m}{n-m}=2005</math>. Dividing both sides by 10 yields <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> we can now write a system of equations with <math>n+m=401</math> and <math>n-m=2</math>, but this does not output integer solutions. However, we can also write <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> as <math>\dfrac{n+m}{n-m}=\dfrac{802}{4}</math>. This gives the system of equations <math>m+n=802</math> and <math>n-m=4</math>, which does have integer solutions. Our answer is therefore <math>m+n=\boxed{802}</math> (Solving for <math>m</math> and <math>n</math> gives us <math>399</math> and <math>403</math>, respectively, which are co-prime).
 
We can write the sum of the original series as <math>a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005</math>, where the common ratio is equal to <math>\dfrac{m}{n}</math>. We can also write the sum of the second series as <math>a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050</math>. Using the formula for the sum of an infinite geometric series <math>S=\dfrac{a}{1-r}</math>, where <math>S</math> is the sum of the sequence, <math>a</math> is the first term of the sequence, and <math>r</math> is the ratio of the sequence, the sum of the original series can be written as <math>\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}</math>, and the second sequence can be written as <math>\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}</math>. Dividing <math>\text{(2)}</math> by <math>\text{(1)}</math>, we obtain <math>\dfrac{a\cdot n}{m+n}=10</math>, which can also be written as <math>a\cdot n=10(m+n)</math>. Substitute this value for <math>a\cdot n</math> back into <math>\text{(1)}</math>, we obtain <math>10\cdot \dfrac{n+m}{n-m}=2005</math>. Dividing both sides by 10 yields <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> we can now write a system of equations with <math>n+m=401</math> and <math>n-m=2</math>, but this does not output integer solutions. However, we can also write <math>\dfrac{n+m}{n-m}=\dfrac{401}{2}</math> as <math>\dfrac{n+m}{n-m}=\dfrac{802}{4}</math>. This gives the system of equations <math>m+n=802</math> and <math>n-m=4</math>, which does have integer solutions. Our answer is therefore <math>m+n=\boxed{802}</math> (Solving for <math>m</math> and <math>n</math> gives us <math>399</math> and <math>403</math>, respectively, which are co-prime).
 +
 +
==Video Solution==
 +
https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=4500
 +
- AMBRIGGS
  
 
== See also ==
 
== See also ==

Revision as of 14:47, 30 July 2022

Problem

An infinite geometric series has sum 2005. A new series, obtained by squaring each term of the original series, has 10 times the sum of the original series. The common ratio of the original series is $\frac mn$ where $m$ and $n$ are relatively prime integers. Find $m+n.$

Solution 1

Let's call the first term of the original geometric series $a$ and the common ratio $r$, so $2005 = a + ar + ar^2 + \ldots$. Using the sum formula for infinite geometric series, we have $\;\;\frac a{1 -r} = 2005$. Then we form a new series, $a^2 + a^2 r^2 + a^2 r^4 + \ldots$. We know this series has sum $20050 = \frac{a^2}{1 - r^2}$. Dividing this equation by $\frac{a}{1-r}$, we get $10 = \frac a{1 + r}$. Then $a = 2005 - 2005r$ and $a = 10 + 10r$ so $2005 - 2005r = 10 + 10r$, $1995 = 2015r$ and finally $r = \frac{1995}{2015} = \frac{399}{403}$, so the answer is $399 + 403 = \boxed{802}$.

(We know this last fraction is fully reduced by the Euclidean algorithm -- because $4 = 403 - 399$, $\gcd(403, 399) | 4$. But 403 is odd, so $\gcd(403, 399) = 1$.)

Solution 2

We can write the sum of the original series as $a + a\left(\dfrac{m}{n}\right) + a\left(\dfrac{m}{n}\right)^2 + \ldots = 2005$, where the common ratio is equal to $\dfrac{m}{n}$. We can also write the sum of the second series as $a^2 + a^2\left(\dfrac{m}{n}\right)^2 + a^2\left(\left(\dfrac{m}{n}\right)^2\right)^2 + \ldots = 20050$. Using the formula for the sum of an infinite geometric series $S=\dfrac{a}{1-r}$, where $S$ is the sum of the sequence, $a$ is the first term of the sequence, and $r$ is the ratio of the sequence, the sum of the original series can be written as $\dfrac{a}{1-\frac{m}{n}}=\dfrac{a}{\frac{n-m}{n}}=\dfrac{a \cdot n}{n-m}=2005\;\text{(1)}$, and the second sequence can be written as $\dfrac{a^2}{1-\frac{m^2}{n^2}}=\dfrac{a^2}{\frac{n^2-m^2}{n^2}}=\dfrac{a^2\cdot n^2}{(n+m)(n-m)}=20050\;\text{(2)}$. Dividing $\text{(2)}$ by $\text{(1)}$, we obtain $\dfrac{a\cdot n}{m+n}=10$, which can also be written as $a\cdot n=10(m+n)$. Substitute this value for $a\cdot n$ back into $\text{(1)}$, we obtain $10\cdot \dfrac{n+m}{n-m}=2005$. Dividing both sides by 10 yields $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ we can now write a system of equations with $n+m=401$ and $n-m=2$, but this does not output integer solutions. However, we can also write $\dfrac{n+m}{n-m}=\dfrac{401}{2}$ as $\dfrac{n+m}{n-m}=\dfrac{802}{4}$. This gives the system of equations $m+n=802$ and $n-m=4$, which does have integer solutions. Our answer is therefore $m+n=\boxed{802}$ (Solving for $m$ and $n$ gives us $399$ and $403$, respectively, which are co-prime).

Video Solution

https://youtu.be/z4-bFo2D3TU?list=PLZ6lgLajy7SZ4MsF6ytXTrVOheuGNnsqn&t=4500 - AMBRIGGS

See also

2005 AIME II (Problems • Answer Key • Resources)
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Problem 4
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All AIME Problems and Solutions

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