2023 AMC 10B Problems/Problem 11
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[hide]Problem
Suzanne went to the bank and withdrew . The teller gave her this ammount 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
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)
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 .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution 1 by OmegaLearn
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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