Difference between revisions of "2021 AIME I Problems/Problem 4"
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==Solution 1== | ==Solution 1== | ||
− | Suppose we have <math>1</math> coin in the first pile. Then <math>(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)</math> all work for a total of <math>31</math> piles. Suppose we have <math>2</math> coins in the first pile, then <math>(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)</math> all work, for a total of <math>29</math>. Continuing this pattern until <math>21</math> coins in the first pile, we have the sum < | + | Suppose we have <math>1</math> coin in the first pile. Then <math>(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)</math> all work for a total of <math>31</math> piles. Suppose we have <math>2</math> coins in the first pile, then <math>(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)</math> all work, for a total of <math>29</math>. Continuing this pattern until <math>21</math> coins in the first pile, we have the sum |
+ | <cmath>\begin{align*} | ||
+ | 31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\ | ||
+ | &= 176+155 \\ | ||
+ | &= \boxed{331}. | ||
+ | \end{align*}</cmath> | ||
− | <math> | + | ==Solution 2== |
+ | |||
+ | We make an equation: <math>a+b+c=66,</math> where <math>a<b<c.</math> We don't have a clear solution, so we'll try complementary counting. First, let's find where <math>a\geq b\geq c.</math> By stars and bars, we have <math>\dbinom{65}{2}=2080</math> to assign positive integer solutions to <math>a + b + c = 66.</math> Now we need to subtract off the cases where it doesn't satisfy the condition. | ||
+ | |||
+ | We start with <math>a = b.</math> We can write that as <math>2b + c = 66.</math> We can find there are 32 integer solutions to this equation. There are <math>32</math> solutions for <math>b=c</math> and <math>a = c</math> by symmetry. We also need to add back <math>2</math> because we subtracted <math>(a,b,c)=(22,22,22)</math> <math>3</math> times. | ||
+ | |||
+ | We then have to divide by <math>6</math> because there are <math>3!=6</math> ways to order <math>a, b,</math> and <math>c.</math> Therefore, we have <math>\dfrac{\dbinom{65}{2}-96+2}{6} = \dfrac{1986}{6} = \boxed{331}.</math> | ||
+ | |||
+ | ~Arcticturn | ||
==Solution 3== | ==Solution 3== | ||
Let the piles have <math>a, b</math> and <math>c</math> coins, with <math>0 < a < b < c</math>. Then, let <math>b = a + k_1</math>, and <math>c = b + k_2</math>, such that each <math>k_i \geq 1</math>. The sum is then <math>a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66</math>. This is simply the number of positive solutions to the equation <math>3x+2y+z = 66</math>. Now, we take cases on <math>a</math>. | Let the piles have <math>a, b</math> and <math>c</math> coins, with <math>0 < a < b < c</math>. Then, let <math>b = a + k_1</math>, and <math>c = b + k_2</math>, such that each <math>k_i \geq 1</math>. The sum is then <math>a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66</math>. This is simply the number of positive solutions to the equation <math>3x+2y+z = 66</math>. Now, we take cases on <math>a</math>. | ||
− | If <math>a = 1</math>, then <math>2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31</math>. Each value of <math>k_1</math> corresponds to a unique value of <math>k_2</math>, so there are <math>31</math> solutions in this case. Similarly, if <math>a = 2</math>, then <math>2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29</math>, for a total of <math>29</math> solutions in this case. If <math>a = 3</math>, then <math>2k_1 + | + | If <math>a = 1</math>, then <math>2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31</math>. Each value of <math>k_1</math> corresponds to a unique value of <math>k_2</math>, so there are <math>31</math> solutions in this case. Similarly, if <math>a = 2</math>, then <math>2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29</math>, for a total of <math>29</math> solutions in this case. If <math>a = 3</math>, then <math>2k_1 + k_2 = 57 \implies 1 \leq k_1 \leq 28</math>, for a total of <math>28</math> solutions. In general, the number of solutions is just all the numbers that aren't a multiple of <math>3</math>, that are less than or equal to <math>31</math>. |
+ | |||
+ | We then add our cases to get | ||
+ | <cmath>\begin{align*} | ||
+ | 1 + 2 + 4 + 5 + \cdots + 31 &= 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) \\ | ||
+ | &= \frac{31(32)}{2} - 3(55) \\ | ||
+ | &= 31 \cdot 16 - 165 \\ | ||
+ | &= 496 - 165 \\ | ||
+ | &= \boxed{331} | ||
+ | \end{align*}</cmath> | ||
+ | as our answer. | ||
+ | |||
+ | ==Solution 4== | ||
+ | Let the first pile have <math>a</math> coins, the second <math>a+b</math> coins, and the third <math>a+b+c</math> coins, where <math>a</math>, <math>b</math>, and <math>c</math> are strictly positive integers. Thus the total number of coins is <math>3a+2b+c=66</math>. Perform the substitution <math>x=a-1</math>, <math>y=b-1</math>, and <math>z=c-1</math> to yield the equation <math>3x+2y+z=60</math>, where <math>x</math>, <math>y</math>, and <math>z</math> are instead nonnegative integers. | ||
− | + | From here we can set up the [[generating function]] <math>(x^0+x^3+\cdots+x^{60})(x^0+x^2+\cdots+x^{60})(x^0+x^1+\cdots+x^{60})</math>. We need to find the coefficient of <math>x^{60}</math>. Multiplying the second and third polynomials with clever reasoning returns | |
− | = | + | <math>(x^0+x^3+\cdots+x^{60})(x^0+x^1+2x^2+2x^3+\cdots+31x^{60}+\cdots+2x^{117}+2x^{118}+x^{119}+x^{120})</math> |
+ | |||
+ | where in the second polynomial, for every two terms, the coefficient increases or decreases by one (depending on which side of the polynomial the term resides). | ||
+ | |||
+ | One can notice here that for every term in the first polynomial there exists one and only one term in the second polynomial that, when multiplied, yield <math>x^{60}</math>. Furthermore, we need only consider the <u>coefficients</u> of the second polynomial. | ||
+ | |||
+ | The corresponding coefficient for <math>x^{60}</math> is <math>1</math>, for <math>x^{57}</math> is <math>2</math>, and for <math>x^{54}</math> is <math>4</math>. We notice the pattern: increase by one, increase by two, and so on. When does this pattern stop? For <math>x^0</math>, the corresponding coefficient is <math>31</math>, and we notice that <math>31\equiv1\mod3</math>. As a result, we know that the pattern has <math>21</math> terms, and we can take advantage of the first <math>20</math> by symmetry. | ||
+ | |||
+ | The answer is simply <math>10(1+29)+31=\boxed{331}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
+ | |||
+ | Solution 4 | ||
− | |||
− | + | ==Video Solution == | |
− | + | https://youtu.be/f7KtovZ4GAk | |
− | |||
− | ~ | + | ~MathProblemSolvingSkills.com |
+ | ** | ||
==Video Solution == | ==Video Solution == |
Latest revision as of 20:09, 7 August 2024
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 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
Solution 2
We make an equation: where
We don't have a clear solution, so we'll try complementary counting. First, let's find where
By stars and bars, we have
to assign positive integer solutions to
Now we need to subtract off the cases where it doesn't satisfy the condition.
We start with We can write that as
We can find there are 32 integer solutions to this equation. There are
solutions for
and
by symmetry. We also need to add back
because we subtracted
times.
We then have to divide by because there are
ways to order
and
Therefore, we have
~Arcticturn
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.
Solution 4
Let the first pile have coins, the second
coins, and the third
coins, where
,
, and
are strictly positive integers. Thus the total number of coins is
. Perform the substitution
,
, and
to yield the equation
, where
,
, and
are instead nonnegative integers.
From here we can set up the generating function . We need to find the coefficient of
. Multiplying the second and third polynomials with clever reasoning returns
where in the second polynomial, for every two terms, the coefficient increases or decreases by one (depending on which side of the polynomial the term resides).
One can notice here that for every term in the first polynomial there exists one and only one term in the second polynomial that, when multiplied, yield . Furthermore, we need only consider the coefficients of the second polynomial.
The corresponding coefficient for is
, for
is
, and for
is
. We notice the pattern: increase by one, increase by two, and so on. When does this pattern stop? For
, the corresponding coefficient is
, and we notice that
. As a result, we know that the pattern has
terms, and we can take advantage of the first
by symmetry.
The answer is simply .
~eevee9406
Solution 4
Video Solution
~MathProblemSolvingSkills.com
Video Solution
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 |
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