Difference between revisions of "2023 AMC 10B Problems/Problem 11"
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+ | ==Solution 6 (kind of bash)== | ||
+ | Casework on if there is one 100. There are 6 ways. Casework on if there are two 100s. There are 5 ways. Notice that this continues all the way until there are 6 100s. Our sum is <math>1+2+3+...+5+6=\frac{6\cdot7}{2}=\frac{42}{2}=21</math>, or <math>\boxed{B}</math>. | ||
+ | ~MC413551 | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 16:30, 16 November 2023
Contents
[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)
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 (kind of bash)
Casework on if there is one 100. There are 6 ways. Casework on if there are two 100s. There are 5 ways. Notice that this continues all the way until there are 6 100s. Our sum is , or . ~MC413551
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.