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

Revision as of 00:01, 13 March 2021

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 $31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+$

$(29+26+23+\ldots+2)=176+155=\boxed{331}$ .

Solution 2

Let the three piles have $a, b, c$ coins respectively. If we disregard order, then we just need to divide by $3! = 6$ at the end.

We know $a + b + c = 66$ . Since $a, b, c$ are positive integers, there are $\binom{65}{2}$ ways from Stars and Bars.

However, we must discard the cases where $a = b$ or $a = c$ or $b = c$ . The three cases are symmetric, so we just take the first case and multiply by 3. We have $2a + c = 66 \implies a = 1, 2, \dots 32$ for 32 solutions. Multiplying by 3, we will subtract 96 from our total.

But we undercounted where $a = b = c = 22$ . 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 $\frac{\binom{65}{2} - 96 + 2}{6} = \boxed{331}.$

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 $1 + 2 + 4 + 5 + \cdots + 31 = 1 + 2 + 3 + \cdots + 31 - 3(1 + 2 + 3 + \cdots + 10) = \frac{31(32)}{2} - 3(55) = 31 \cdot 16 - 165 = 496 - 165 = \boxed{331}$ as our answer.

Video Solution #1

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

@@ Line 3: / Line 3: @@
 ==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)+
+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>
-(29+26+23+\ldots+2)=176+155=\boxed{331}</math>.
+<math>(29+26+23+\ldots+2)=176+155=\boxed{331}</math>.
 ==Solution 2==

2021 AIME I (Problems • Answer Key • Resources)
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