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 B) 
.
~yourmomisalosinggame (a.k.a. Aaron)
