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Difference between revisions of "2016 AIME II Problems/Problem 1"

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Initially Alex, Betty, and Charlie had a total of <math>444</math> peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats <math>5</math> of his peanuts, Betty eats <math>9</math> of her peanuts, and Charlie eats <math>25</math> of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
 
Initially Alex, Betty, and Charlie had a total of <math>444</math> peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats <math>5</math> of his peanuts, Betty eats <math>9</math> of her peanuts, and Charlie eats <math>25</math> of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
  
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==Solution==
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==Solution 1==
 
Let <math>r</math> be the common ratio, where <math>r>1</math>. We then have <math>ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16</math>. We now have, letting, subtracting the 2 equations, <math>ar^{2}+-2ar+a=12</math>, so we have <math>3ar=432,</math> or <math>ar=144</math>, which is how much Betty had. Now we have <math>144+\dfrac{144}{r}+144r=444</math>, or <math>144(r+\dfrac{1}{r})=300</math>, or <math>r+\dfrac{1}{r}=\dfrac{25}{12}</math>, which solving for <math>r</math> gives <math>r=\dfrac{4}{3}</math>, since <math>r>1</math>, so Alex had <math>\dfrac{3}{4} \cdot 144=\boxed{108}</math> peanuts.
 
Let <math>r</math> be the common ratio, where <math>r>1</math>. We then have <math>ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16</math>. We now have, letting, subtracting the 2 equations, <math>ar^{2}+-2ar+a=12</math>, so we have <math>3ar=432,</math> or <math>ar=144</math>, which is how much Betty had. Now we have <math>144+\dfrac{144}{r}+144r=444</math>, or <math>144(r+\dfrac{1}{r})=300</math>, or <math>r+\dfrac{1}{r}=\dfrac{25}{12}</math>, which solving for <math>r</math> gives <math>r=\dfrac{4}{3}</math>, since <math>r>1</math>, so Alex had <math>\dfrac{3}{4} \cdot 144=\boxed{108}</math> peanuts.
  
 
Solution by Shaddoll
 
Solution by Shaddoll
  
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== Solution 2 ==
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Let the initial numbers of peanuts Alex, Betty and Charlie had be <math>a</math>, <math>b</math>, and <math>c</math> respectively.
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Let the final numbers of peanuts, after eating, be <math>a'</math>, <math>b'</math>, and <math>c'</math>.
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We are given that <math>a + b + c = 444</math>. Since a total of <math>5 + 9 + 25 = 39</math> peanuts are eaten, we must have <math>a' + b' + c' = 444 - 39 = 405</math>.
 +
Since <math>a'</math>, <math>b'</math>, and <math>c'</math> form an arithmetic progression, we have that <math>a' = b' - x</math> and <math>c' = b' + x</math> for some integer <math>x</math>.
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Substituting yields <math>3b' = 405</math> and so <math>b' = 135</math>. Since Betty ate <math>9</math> peanuts, it follows that <math>b = b' + 9 = 144</math>.
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Since <math>a</math>, <math>b</math>, and <math>c</math> form a geometric progression, we have that <math>a = \frac{b}{r}</math> and <math>c = br</math>.
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Multiplying yields <math>ac = b^2 = 144^2</math>.
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Since <math>a + c = 444 - b = 300</math>, it follows that <math>a = 150 - \lambda</math> and <math>c = 150 + \lambda</math> for some integer <math>\lambda</math>.
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Substituting yields <math>(150-\lambda)(150+\lambda) = 144^2</math>, which expands and rearranges to <math>\lambda^2 = 150^2-144^2 = 42^2</math>.
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Since <math>\lambda > 0</math>, we must have <math>\lambda = 42</math>, and so <math>a = 150 - \lambda = \boxed{108}</math>.
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2016|n=II|before=First Problem|num-a=2}}

Revision as of 18:23, 26 May 2017

Initially Alex, Betty, and Charlie had a total of $444$ peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.

Solution 1

Let $r$ be the common ratio, where $r>1$. We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$. We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$, so we have $3ar=432,$ or $ar=144$, which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$, or $144(r+\dfrac{1}{r})=300$, or $r+\dfrac{1}{r}=\dfrac{25}{12}$, which solving for $r$ gives $r=\dfrac{4}{3}$, since $r>1$, so Alex had $\dfrac{3}{4} \cdot 144=\boxed{108}$ peanuts.

Solution by Shaddoll

Solution 2

Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$, $b$, and $c$ respectively. Let the final numbers of peanuts, after eating, be $a'$, $b'$, and $c'$.

We are given that $a + b + c = 444$. Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$. Since $a'$, $b'$, and $c'$ form an arithmetic progression, we have that $a' = b' - x$ and $c' = b' + x$ for some integer $x$. Substituting yields $3b' = 405$ and so $b' = 135$. Since Betty ate $9$ peanuts, it follows that $b = b' + 9 = 144$.

Since $a$, $b$, and $c$ form a geometric progression, we have that $a = \frac{b}{r}$ and $c = br$. Multiplying yields $ac = b^2 = 144^2$. Since $a + c = 444 - b = 300$, it follows that $a = 150 - \lambda$ and $c = 150 + \lambda$ for some integer $\lambda$. Substituting yields $(150-\lambda)(150+\lambda) = 144^2$, which expands and rearranges to $\lambda^2 = 150^2-144^2 = 42^2$. Since $\lambda > 0$, we must have $\lambda = 42$, and so $a = 150 - \lambda = \boxed{108}$.

See also

2016 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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