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 3
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 4
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 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
~not_slay
Solution 6 (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
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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