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 "2023 AMC 10B Problems/Problem 11"

(→‎Solution 2)
(→‎Solution 2)
Line 19: Line 19:
 
== Solution 2 ==
 
== Solution 2 ==
  
−
First, we note that there can only be an even number of <math>50</math> dollar bills. Next, since there is at least one of each bill, we find that the amount of <math>50</math> dollar bills is between <math>2</math> and <math>14</math>. Doing some casework, we find that the amount of <math>100</math> dollar bills forms an arithmetic sequence: <math>6</math>+<math>5</math>+<math>4</math>+<math>3</math>+<math>2</math>+<math>1</math>. Adding these up, we get B) <math>21</math>.
+
First, we note that there can only be an even number of <math>50</math> dollar bills. Next, since there is at least one of each bill, we find that the amount of <math>50</math> dollar bills is between <math>2</math> and <math>14</math>. Doing some casework, we find that the amount of <math>100</math> dollar bills forms an arithmetic sequence: <math>6</math>+<math>5</math>+<math>4</math>+<math>3</math>+<math>2</math>+<math>1</math>. Adding these up, we get <math>21</math>.
  
 
~yourmomisalosinggame (a.k.a. Aaron)
 
~yourmomisalosinggame (a.k.a. Aaron)

Revision as of 16:18, 15 November 2023

Solution 1

We let the number of $$20$-, $$50$-, and $$100$ bills be $a,b,$ and $c,$ respectively.

We are given that $20a+50b+100c=800.$ Dividing both sides by $10$, we see that $2a+5b+10c=80.$

We divide both sides of this equation by $5$: $\dfrac25a+b+2c=16.$ Since $b+2c$ and $16$ are integers, $\dfrac25a$ must also be an integer, so $a$ must be divisible by $5$. Let $a=5d,$ where $d$ is some positive integer.

We can then write $2\cdot5d+5b+10c=80.$ Dividinb both sides by $5$, we have $2d+b+2c=16.$ We divide by $2$ here to get $d+\dfrac b2+c=8.$ $d+c$ and $8$ are both integers, so $\dfrac b2$ is also an integer. $b$ must be divisible by $2$, so we let $b=2e$.

We now have $2d+2e+2c=16\implies d+e+c=8$. Every substitution we made is part of a bijection (i.e. our choices were one-to-one); thus, the problem is now reduced to counting how many ways we can have $d,e,$ and $c$ such that they add to $8$.

We still have another constraint left, that each of $d,e,$ and $c$ must be at least $1$. For $n\in\{d,e,c\}$, let $n'=n-1.$ We are now looking for how many ways we can have $d'+e'+c'=8-1-1-1=5.$

We use a classic technique for solving these sorts of problems: stars and bars. We have $5$ things and $3$ groups, which implies $2$ dividers. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$

~Technodoggo

Solution 2

First, we note that there can only be an even number of $50$ dollar bills. Next, since there is at least one of each bill, we find that the amount of $50$ dollar bills is between $2$ and $14$. Doing some casework, we find that the amount of $100$ dollar bills forms an arithmetic sequence: $6$+$5$+$4$+$3$+$2$+$1$. Adding these up, we get $21$.

~yourmomisalosinggame (a.k.a. Aaron)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_11&oldid=203230"