2016 AIME II Problems/Problem 1: Difference between revisions
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== See also == | == See also == | ||
{{AIME box|year=2016|n=II| | {{AIME box|year=2016|n=II|before=First Problem|num-a=2}} | ||
Revision as of 21:32, 16 May 2016
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
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
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 | ||
