Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 2: Deleted repetitive solutions. Will make some adjustments.)
m (→‎Solution 4)
Line 14: Line 14:
 
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.
 
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 4==
+
==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> We have <math>\dbinom{65}{2}=2080</math> by stars and bars for <math>a>b>c.</math> Now we need to subtract off the cases where it doesn't satisfy the condition.
 
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> We have <math>\dbinom{65}{2}=2080</math> by stars and bars for <math>a>b>c.</math> Now we need to subtract off the cases where it doesn't satisfy the condition.
  
−
We first by starting out 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 subtract <math>2</math> from the <math>96,</math> because we counted <math>(a,b,c)=(22,22,22)</math> for <math>3</math> times.
+
We first by starting out 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> for <math>3</math> times.
 
   
 
   
−
We then have to divide by <math>6</math> because there are <math>3!=6</math> ways to order the <math>a, b,</math> and <math>c.</math> Therefore, we have <math>\dfrac{\dbinom{65}{2}-94}{6} = \dfrac{1986}{6} = \boxed{331}</math>
+
We then have to divide by <math>6</math> because there are <math>3!=6</math> ways to order the <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
 
~Arcticturn

Revision as of 13:19, 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 $31+29+28+26+25+\ldots+4+2+1=(31+28+25+22+\ldots+1)+$

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

Solution 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.

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_I_Problems/Problem_4&oldid=170330"