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

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==Problem==
 
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 1==
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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.
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== Solution 2 (Quadratic Formula)==
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Let <math>a</math> be Alex's peanuts and <math>k</math> the common ratio. Then we have <math>a(k^2+k+1)=444</math>. Adding <math>k</math> to both sides and factoring,
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<cmath>\frac{444}{a}+k=(k+1)^2</cmath>
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For the common difference, <math>ak=5-(a-5)=ak^2-25-(ak-9)</math>. Simplifying, <math>k^2-2k+1=\frac{12}{a}</math>. Factoring,
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<cmath>(k-1)^2=\frac{12}{a}</cmath>
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<cmath>(k+1)^2-(k-1)^2=4k \implies 4k=\frac{444-12}{a}+k \implies k=\frac{144}{a}</cmath>
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Substitute <math>k</math> in the second equation to get <math>(\frac{144-a}{a})^2=\frac{12}{a}</math>. Expanding and applying the quadratic formula,
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<cmath>a=150\pm\frac{\sqrt{300^2-4(144^2)}}{2}</cmath>
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Taking out <math>4^2\cdot3^2</math> from under the radical leaves
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<cmath>a=150\pm6\sqrt{625-576}=108, 192</cmath>
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Since Alex's peanut number was the lowest of the trio, and <math>3*192>444</math>, Alex initially had <math>\boxed{108}</math> peanuts.
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== Solution 3 ==
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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>.
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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>.
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== Solution 4==
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Bashing is not difficult. All we have to consider is the first equation. We can write it as <math>x*(1+r+r^2) = 444</math>. The variable <math>x</math> must be an integer, and after trying all the factors of <math>444</math>, it's clear that <math>r</math> is a fraction smaller than <math>10</math>. When calculating the coefficient of <math>x</math>, we must consider that the fraction produced will very likely have a numerator that divides <math>444</math>. Trying a couple will make it easier to find the fraction, and soon you will find that <math>\frac{4}{3}</math> gives a numerator of <math>37</math>, a rather specific factor of <math>444</math>. Solving for the rest will give you an integer value of <math>\boxed{108}</math>. This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals.
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== See also ==
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{{AIME box|year=2016|n=II|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 19:00, 18 October 2020

Problem

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 2 (Quadratic Formula)

Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$. Adding $k$ to both sides and factoring,

\[\frac{444}{a}+k=(k+1)^2\]

For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$. Simplifying, $k^2-2k+1=\frac{12}{a}$. Factoring, \[(k-1)^2=\frac{12}{a}\]

\[(k+1)^2-(k-1)^2=4k \implies 4k=\frac{444-12}{a}+k \implies k=\frac{144}{a}\]


Substitute $k$ in the second equation to get $(\frac{144-a}{a})^2=\frac{12}{a}$. Expanding and applying the quadratic formula, \[a=150\pm\frac{\sqrt{300^2-4(144^2)}}{2}\] Taking out $4^2\cdot3^2$ from under the radical leaves \[a=150\pm6\sqrt{625-576}=108, 192\] Since Alex's peanut number was the lowest of the trio, and $3*192>444$, Alex initially had $\boxed{108}$ peanuts.

Solution 3

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 $a = 150 - \lambda = \boxed{108}$.

Solution 4

Bashing is not difficult. All we have to consider is the first equation. We can write it as $x*(1+r+r^2) = 444$. The variable $x$ must be an integer, and after trying all the factors of $444$, it's clear that $r$ is a fraction smaller than $10$. When calculating the coefficient of $x$, we must consider that the fraction produced will very likely have a numerator that divides $444$. Trying a couple will make it easier to find the fraction, and soon you will find that $\frac{4}{3}$ gives a numerator of $37$, a rather specific factor of $444$. Solving for the rest will give you an integer value of $\boxed{108}$. This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals.

See also

2016 AIME II (ProblemsAnswer KeyResources)
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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