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Difference between revisions of "2021 AIME I Problems/Problem 4"

m (→‎Solution 4)
m (Fixed LaTeX throughout)
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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 <math>31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+</math>
+
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*}
−
<math>(29+26+23+\ldots+2)=176+155=\boxed{331}</math>.
+
31+29+28+26+25+\cdots+4+2+1 &= (31+28+25+22+\cdots+1)+(29+26+23+\cdots+2) \\
−
 
+
&= 176+155 \\
−
==Solution 3==
+
&= \boxed{331}.
−
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>.
+
\end{align*}</cmath>
−
 
 
−
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_1 = 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>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}</cmath> as our answer.
 
  
 
==Solution 2==
 
==Solution 2==
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~Arcticturn
 
~Arcticturn
 +
 +
==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>.
 +
 +
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_1 = 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.
  
 
==Video Solution ==
 
==Video Solution ==

Revision as of 13:25, 27 January 2022

Problem

Find the number of ways $66$ 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 $1$ coin in the first pile. Then $(1, 2, 63), (1, 3, 62), \ldots, (1, 32, 33)$ all work for a total of $31$ piles. Suppose we have $2$ coins in the first pile, then $(2, 3, 61), (2, 4, 60), \ldots, (2, 31, 33)$ all work, for a total of $29$. Continuing this pattern until $21$ coins in the first pile, we have the sum \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*}

Solution 2

We make an equation: $a+b+c=66,$ where $a<b<c.$ We don't have a clear solution, so we'll try complementary counting. First, let's find where $a\geq b\geq c.$ We have $\dbinom{65}{2}=2080$ by stars and bars for $a>b>c.$ 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 add back $2$ because we subtracted $(a,b,c)=(22,22,22)$ for $3$ times.

We then have to divide by $6$ because there are $3!=6$ ways to order the $a, b,$ and $c.$ Therefore, we have $\dfrac{\dbinom{65}{2}-96+2}{6} = \dfrac{1986}{6} = \boxed{331}.$

~Arcticturn

Solution 3

Let the piles have $a, b$ and $c$ coins, with $0 < a < b < c$. Then, let $b = a + k_1$, and $c = b + k_2$, such that each $k_i \geq 1$. The sum is then $a + a+k_1 + a+k_1+k_2 = 66 \implies 3a+2k_1 + k_2 = 66$. This is simply the number of positive solutions to the equation $3x+2y+z = 66$. Now, we take cases on $a$.

If $a = 1$, then $2k_1 + k_2 = 63 \implies 1 \leq k_1 \leq 31$. Each value of $k_1$ corresponds to a unique value of $k_2$, so there are $31$ solutions in this case. Similarly, if $a = 2$, then $2k_1 + k_2 = 60 \implies 1 \leq k_1 \leq 29$, for a total of $29$ solutions in this case. If $a = 3$, then $2k_1 + k_1 = 57 \implies 1 \leq k_1 \leq 28$, for a total of $28$ solutions. In general, the number of solutions is just all the numbers that aren't a multiple of $3$, that are less than or equal to $31$.

We then add our cases to get \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*} as our answer.

Video Solution

https://youtu.be/M3DsERqhiDk?t=1073

See Also

2021 AIME I (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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