# 2023 AMC 10B Problems/Problem 11

## Contents

## Problem

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

## Solution 1

We let the number of , , and bills be and respectively.

We are given that Dividing both sides by , we see that

We divide both sides of this equation by : Since and are integers, must also be an integer, so must be divisible by . Let where is some positive integer.

We can then write Dividing both sides by , we have We divide by here to get and are both integers, so is also an integer. must be divisible by , so we let .

We now have . 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 and such that they add to .

We still have another constraint left, that each of and must be at least . For , let We are now looking for how many ways we can have

We use a classic technique for solving these sorts of problems: stars and bars. We have stars and groups, which implies bars. Thus, the total number of ways is

~Technodoggo ~minor edits by lucaswujc

## Solution 2

Denote by , , the amount of $20 bills, $50 bills and $100 bills, respectively. Thus, we need to find the number of tuples with that satisfy

First, this equation can be simplified as

Second, we must have . Denote . The above equation can be converted to

Third, we must have . Denote . The above equation can be converted to

Denote , and . Thus, the above equation can be written as

Therefore, the number of non-negative integer solutions is .

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

## Solution 3

To start, we simplify things by dividing everything by , the resulting equation is , and since the problem states that we have at least one of each, we simplify this to . Note that since the total is odd, we need an odd number of dollar bills. We proceed using casework.

Case 1: One dollar bill

, we see that can be or . Ways

Case 2: Three dollar bills

, like before we see that can be , so way.

Now we should start to see a pattern emerges, each case there is less way to sum to , so the answer is just , or

~andyluo

## Solution 4

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

~not_slay

## Solution 5 (generating functions)

The problem is equivalent to the number of ways to make from bills, bills, and bills. We can use generating functions to find the coefficient of :

The bills provide

The bills provide

The bills provide

Multiplying, we get

## Video Solution 1 by OmegaLearn

## Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=sfZRRsTimmE

## Video Solution 3 by paixiao

https://youtu.be/EvA2Nlb7gi4?si=fVLG8gMTIC5XkEwP&t=89s

## Video Solution 4

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