Difference between revisions of "2023 AMC 10B Problems/Problem 11"

Revision as of 20:42, 15 November 2023

Problem

Suzanne went to the bank and withdrew $$800$ . The teller gave her this ammount using $$20$ bills, $$50$ bills, and $$100$ bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?

$\textbf{(A) } 45 \qquad \textbf{(B) } 21 \qquad \text{(C) } 36 \qquad \text{(D) } 28 \qquad \text{(E) } 32$

Solution 1

We let the number of $$20$ , $$50$ , and $$100$ bills be $a,b,$ and $c,$ respectively.

We are given that $20a+50b+100c=800.$ Dividing both sides by $10$ , we see that $2a+5b+10c=80.$

We divide both sides of this equation by $5$ : $\dfrac25a+b+2c=16.$ Since $b+2c$ and $16$ are integers, $\dfrac25a$ must also be an integer, so $a$ must be divisible by $5$ . Let $a=5d,$ where $d$ is some positive integer.

We can then write $2\cdot5d+5b+10c=80.$ Dividing both sides by $5$ , we have $2d+b+2c=16.$ We divide by $2$ here to get $d+\dfrac b2+c=8.$ $d+c$ and $8$ are both integers, so $\dfrac b2$ is also an integer. $b$ must be divisible by $2$ , so we let $b=2e$ .

We now have $2d+2e+2c=16\implies d+e+c=8$ . 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 $d,e,$ and $c$ such that they add to $8$ .

We still have another constraint left, that each of $d,e,$ and $c$ must be at least $1$ . For $n\in\{d,e,c\}$ , let $n'=n-1.$ We are now looking for how many ways we can have $d'+e'+c'=8-1-1-1=5.$

We use a classic technique for solving these sorts of problems: stars and bars. We have $5$ stars and $3$ groups, which implies $2$ bars. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$

~Technodoggo ~minor edits by lucaswujc

Solution 2

First, we note that there can only be an even number of $50$ dollar bills.

Next, since there is at least one of each bill, we find that the amount of $50$ dollar bills is between $2$ and $12$ . Doing some casework, we find that the amount of $100$ dollar bills forms an arithmetic sequence: $6$ + $5$ + $4$ + $3$ + $2$ + $1$ .

Adding these up, we get $21$ .

~yourmomisalosinggame (a.k.a. Aaron)

Solution 3

Denote by $x$ , $y$ , $z$ the amount of $20 bills, $50 bills and $100 bills, respectively. Thus, we need to find the number of tuples $\left( x , y, z \right)$ with $x, y, z \in \Bbb N$ that satisfy $20 x + 50 y + 100 z = 800.$

First, this equation can be simplified as $2 x + 5 y + 10 z = 80.$

Second, we must have $5 |x$ . Denote $x = 5 x'$ . The above equation can be converted to $2 x' + y + 2 z = 16 .$

Third, we must have $2 | y$ . Denote $y = 2 y'$ . The above equation can be converted to $x' + y' + z = 8 .$

Denote $x'' = x' - 1$ , $y'' = y' - 1$ and $z'' = z - 1$ . Thus, the above equation can be written as $x'' + y'' + z'' = 5 .$

Therefore, the number of non-negative integer solutions $\left( x'', y'', z'' \right)$ is $\binom{5 + 3 - 1}{3 - 1} = \boxed{\textbf{(B) 21}}$ .

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

Video Solution 1 by OmegaLearn

https://youtu.be/UeX3eEwRS9I

@@ Line 76: / Line 76: @@
 ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+==Video Solution 1 by OmegaLearn==
+https://youtu.be/UeX3eEwRS9I
 ==See also==
 {{AMC10 box|year=2023|ab=B|num-b=10|num-a=12}}
 {{MAA Notice}}

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