2023 AMC 10B Problems/Problem 11

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 ways

Now we should start to see a pattern emerge, 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

Solution 6 (easy logic)

There aren't dollar signs because the thinks they're latex symbols. If you find how to override this error, please edit this. There's no here but feel free to add some! ~SwordAxe

We can see in the problem that the teller gave her at least one of 20, 50, and 100. Therefore, she has 800 - 20 - 50 - 100 = 630 "left over".

Since all bills and 630 are multiples of 10, we can divide by ten. ==> Question becomes: How many different collections of 2, 5, and 10 could she get if her total was 63?

We notice that because 63 is odd, we need an odd amount of 5 bills. (2 and 10 are both even, and 63 is not a multiple of 5, so we need 2 and/or 10 bills. PM SwordAxe if you don't get this.)

We can do casework.

1: She gets one 5 (50) dollar bill. She has 58 (580) left.

 1) She is given only 2 dollar (20) bills => ONE COLLECTION (all 20 bills with one 50)
 2) She is given one 10 dollar (100) bill
      1. The rest of the money is given in 2 dollar(20) dollar bills. => ONE COLLECTION (one 100 and rest 20 with one 50)
      2. She is given another 10 dollar (100) bill
            I) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (two 100, one 50, and rest 20)
            II) She is given another 10 dollar (100) bill
                   a) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (three 100, one 50, and rest 20)
                   b) She is given another 10 dollar (100) bill.
                         1) The rest of the money is given in 2 dollar (20) dollar bills  => ONE COLLECTION (following same pattern)
                         2)  AND SO ON...

This looks very tedious, but draw a simple tree diagram, and you'll see that its very easy!

If she gets one 50, there are 6 ways If she gets three 50, there are 5 ways ... If she gets nine 50, there are 2 ways If she gets eleven 50, there is one way

We can add them all up, with a grand sum of 6+5+4+3+2+1 = 21.

Therefore, we get the answer (B) 21.

~SwordAxe (PM me if you have any questions! :))

Video Solution 1 by OmegaLearn

https://youtu.be/UeX3eEwRS9I

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

https://youtu.be/D-ZvFBiZsaY

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

Video Solution 5 by Lucas637

https://www.youtube.com/watch?v=kXLHjclTD44&t=27s