Difference between revisions of "2021 AIME I Problems/Problem 4"
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We first by starting out with a = b. We can write that as 2b + c = 66. We can find there are 32 integer solutions to this equation. There are 32 solutions for b=c, and a = c by symmetry. We also need to subtract 2 from the 96, because we counted 22, 22, 22 3 times. | We first by starting out with a = b. We can write that as 2b + c = 66. We can find there are 32 integer solutions to this equation. There are 32 solutions for b=c, and a = c by symmetry. We also need to subtract 2 from the 96, because we counted 22, 22, 22 3 times. | ||
− | We then have to divide by 6 because there are 3! ways to order the a, b, and c. Therefore, we have <math>\dfrac{\dbinom{65}{2}-94}{6}</math> = <math>\dfrac{1986}{6} = </math> \boxed{331} | + | We then have to divide by 6 because there are 3! ways to order the a, b, and c. Therefore, we have <math>\dfrac{\dbinom{65}{2}-94}{6}</math> = <math>\dfrac{1986}{6} = </math><math>\boxed{331}</math> |
==Solution 1== | ==Solution 1== |
Revision as of 19:42, 24 March 2021
Contents
Problem
Find the number of ways identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.
Solution 4
We make an equation for how this is true. a+b+c = 66, where a is less than b, is less than c. We don't have a clear solution, so we'll try complimentary counting. First, let's find where a doesn't have to be less than b, which doesn't have to be less than c. We have =2080 by stars and bars. Now we need to subtract off the cases where it doesn't satisfy the condition.
We first by starting out with a = b. We can write that as 2b + c = 66. We can find there are 32 integer solutions to this equation. There are 32 solutions for b=c, and a = c by symmetry. We also need to subtract 2 from the 96, because we counted 22, 22, 22 3 times.
We then have to divide by 6 because there are 3! ways to order the a, b, and c. Therefore, we have =
Solution 1
Suppose we have coin in the first pile. Then
all work for a total of
piles. Suppose we have
coins in the first pile, then
all work, for a total of
. Continuing this pattern until
coins in the first pile, we have the sum
.
(Minor edit to make everything fit in the page made by KingRavi)
Solution 2
Let the three piles have coins respectively. If we disregard order, then we just need to divide by
at the end.
We know . Since
are positive integers, there are
ways from Stars and Bars.
However, we must discard the cases where or
or
. The three cases are symmetric, so we just take the first case and multiply by 3. We have
for 32 solutions. Multiplying by 3, we will subtract 96 from our total.
But we undercounted where . This is first counted 1 time, then we subtract it 3 times, so we add it back twice. There is clearly only 1 way, for a total of 2.
Hence, the answer is
Solution 3
Let the piles have and
coins, with
. Then, let
, and
, such that each
. The sum is then
. This is simply the number of positive solutions to the equation
. Now, we take cases on
.
If , then
. Each value of
corresponds to a unique value of
, so there are
solutions in this case. Similarly, if
, then
, for a total of
solutions in this case. If
, then
, for a total of
solutions. In general, the number of solutions is just all the numbers that aren't a multiple of
, that are less than or equal to
.
We then add our cases to get as our answer.
Video Solution #1
https://youtu.be/M3DsERqhiDk?t=1073
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.