2016 AIME II Problems/Problem 1
Problem
Initially Alex, Betty, and Charlie had a total of 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 of his peanuts, Betty eats of her peanuts, and Charlie eats 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 be the common ratio, where . We then have . We now have, letting, subtracting the 2 equations, , so we have or , which is how much Betty had. Now we have , or , or , which solving for gives , since , so Alex had peanuts.
Solution by Shaddoll
Solution 2 (Quadratic Formula)
Let be Alex's peanuts and the common ratio. Then we have . Adding to both sides and factoring,
For the common difference, . Simplifying, . Factoring,
, so . Then, . Substitute in to get . Simplifying, expanding, and applying the quadratic formula,
\[a=150\pm\frac{\sqrt{300^2-4(144^2)}{2}\] (Error compiling LaTeX. )
Taking out from under the radical leaves Since Alex's peanut number was the lowest of the trio, and , Alex initially had peanutes.
(Solution by BJHHar)
Solution 3
Let the initial numbers of peanuts Alex, Betty and Charlie had be , , and respectively. Let the final numbers of peanuts, after eating, be , , and .
We are given that . Since a total of peanuts are eaten, we must have . Since , , and form an arithmetic progression, we have that and for some integer . Substituting yields and so . Since Betty ate peanuts, it follows that .
Since , , and form a geometric progression, we have that and . Multiplying yields . Since , it follows that and for some integer . Substituting yields , which expands and rearranges to . Since , we must have , and so .
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 |