# 2023 AMC 10B Problems/Problem 11

## Problem

Suzanne went to the bank and withdrew $800$. The teller gave her this amount using $20$ bills, $50$ bills, and $100$ bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?

$\textbf{(A) } 45 \qquad \textbf{(B) } 21 \qquad \text{(C) } 36 \qquad \text{(D) } 28 \qquad \text{(E) } 32$

## 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 $2\cdot5d+5b+10c=80.$ Dividing 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 $5$ stars and $3$ groups, which implies $2$ bars. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$

~Technodoggo ~minor edits by lucaswujc

Denote by $x$, $y$, $z$ the amount of $20 bills,$50 bills and $100 bills, respectively. Thus, we need to find the number of tuples $\left( x , y, z \right)$ with $x, y, z \in \Bbb N$ that satisfy $$20 x + 50 y + 100 z = 800.$$ First, this equation can be simplified as $$2 x + 5 y + 10 z = 80.$$ Second, we must have $5 |x$. Denote $x = 5 x'$. The above equation can be converted to $$2 x' + y + 2 z = 16 .$$ Third, we must have $2 | y$. Denote $y = 2 y'$. The above equation can be converted to $$x' + y' + z = 8 .$$ Denote $x'' = x' - 1$, $y'' = y' - 1$ and $z'' = z - 1$. Thus, the above equation can be written as $$x'' + y'' + z'' = 5 .$$ Therefore, the number of non-negative integer solutions $\left( x'', y'', z'' \right)$ is $\binom{5 + 3 - 1}{3 - 1} = \boxed{\textbf{(B) 21}}$. ~stephen chen (Professor Chen Education Palace, www.professorchenedu.com) ## Solution 4 To start, we simplify things by dividing everything by $10$, the resulting equation is $2x+5y+10z=80$, and since the problem states that we have at least one of each, we simplify this to $2x+5y+10z=63$. Note that since the total is odd, we need an odd number of $5$ dollar bills. We proceed using casework. Case 1: One $5$ dollar bill $2x+10z=58$, we see that $10z$ can be $10,20,30,40,50$ or $0$. $6$ Ways Case 2: Three $5$ dollar bills $2x+10z=48$, like before we see that $10z$ can be $0,10,20,30,40$, so $5$ way. Now we should start to see a pattern emerges, each case there is $1$ less way to sum to $80$, so the answer is just $\frac{6(6+1)}{2}$, $21$ or $(B)$ ~andyluo ## Solution 5 We notice that each$100 can be split 3 ways: 5 $20 dollar bills, 2$50 dollar bills, or 1 $100 dollar bill. There are 8 of these$100 chunks in total--take away 3 as each split must be used at least once.

Now there are five left--so we use stars and bars.

5 chunks, 3 categories or 2 bars. This gives us $\binom{5+2}{2}=\boxed{\textbf{(B) 21}}$

~not_slay

## Solution 6 (generating functions)

The problem is equivalent to the number of ways to make $80$ from $2$ bills, $5$ bills, and $10$ bills. We can use generating functions to find the coefficient of $x^{80}$:

The $2$ bills provide $1+x^2+x^4+x^6...+x^{78}+x^{80} = \frac{1-x^{82}}{1-x^2},$

The $5$ bills provide $1+x^5+x^{10}+x^{15}...+x^{75}+x^{80} = \frac{1-x^{85}}{1-x^5},$

The $10$ bills provide $1+x^{10}+x^{20}+x^{30}...+x^{70}+x^{80} = \frac{1-x^{90}}{1-x^{10}}.$

Multiplying, we get $(x^{82}-1)(x^{85}-1)(x^{90}-1)(x^2-1)^{-1}(x^5-1)^{-1}(x^{10}-1)^{-1}.$

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)