2023 AMC 10B Problems/Problem 11

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 Dividinb 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 things and groups, which implies dividers. Thus, the total number of ways is

~Technodoggo

Solution 2

First, we note that there can only be an even number of dollar bills.

Next, since there is at least one of each bill, we find that the amount of dollar bills is between and . Doing some casework, we find that the amount of dollar bills forms an arithmetic sequence: + + + + + .

Adding these up, we get .

~yourmomisalosinggame (a.k.a. Aaron)