2016 AIME II Problems/Problem 1

Revision as of 14:57, 23 March 2016 by Mathymath (talk | contribs) (Solution)

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

Let $r$ be the common ratio, where $r>1$. We than 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

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