Difference between revisions of "2016 AIME II Problems/Problem 1"
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Let <math>k > 1</math> be the common ratio. Then <cmath>\frac{144}{k} + 144 + k \cdot 144 = 444 \implies \frac{144}{k} + k \cdot 144 = 300\implies \frac{12}{k} + k \cdot 12 = 25\implies k = \frac{4}{3} \implies \frac {144\cdot 3}{4} = \boxed {108}.</cmath> | Let <math>k > 1</math> be the common ratio. Then <cmath>\frac{144}{k} + 144 + k \cdot 144 = 444 \implies \frac{144}{k} + k \cdot 144 = 300\implies \frac{12}{k} + k \cdot 12 = 25\implies k = \frac{4}{3} \implies \frac {144\cdot 3}{4} = \boxed {108}.</cmath> | ||
− | ''' | + | '''vladimir.shelomovskii@gmail.com, vvsss''' |
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2016|n=II|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:46, 31 August 2022
Contents
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 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,
Substitute in the second equation to get
. Expanding and applying the quadratic formula,
Taking out
from under the radical leaves
Since Alex's peanut number was the lowest of the trio, and
, Alex initially had
peanuts.
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
.
Solution 4
Bashing is not difficult. All we have to consider is the first equation. We can write it as . The variable
must be an integer, and after trying all the factors of
, it's clear that
is a fraction smaller than
. When calculating the coefficient of
, we must consider that the fraction produced will very likely have a numerator that divides
. Trying a couple will make it easier to find the fraction, and soon you will find that
gives a numerator of
, a rather specific factor of
. Solving for the rest will give you an integer value of
. This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals.
Solution 5
Let be the finish number of Betty's peanuts. Then
Let be the common ratio. Then
vladimir.shelomovskii@gmail.com, vvsss
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.