Difference between revisions of "2016 AIME II Problems/Problem 1"
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Revision as of 22:32, 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 |