Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page with "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...")
 
Line 1: Line 1:
 
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.
 +
 +
==Solution==
 +
Let <math>r</math> be the common ratio, where <math>r>1</math>. We than 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, or 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}{16}</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

Revision as of 18:23, 17 March 2016

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 $r+\dfrac{1}{r}=\dfrac{25}{16}$, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_1&oldid=77652"