2023 AMC 10B Problems/Problem 11: Difference between revisions

Latest revision as of 10:47, 29 October 2025

Problem

Suzanne went to the bank and withdrew $\$800$ . The teller gave her this amount 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

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)

Solution 3

To start, we simplify things by dividing everything by $10$ , the resulting equation is $2x+5y+10z=80$ , and since the problem states that we have at least one of each, we simplify this to $2x+5y+10z=63$ . Note that since the total is odd, we need an odd number of $5$ dollar bills. We proceed using casework.

Case 1: One $5$ dollar bill

$2x+10z=58$ , we see that $10z$ can be $0,10,20,30,40,50$ or $6$ ways

Case 2: Three $5$ dollar bills

$2x+10z=48$ , like before we see that $10z$ can be $0,10,20,30,40$ , so $5$ ways

Now we should start to see a pattern emerge, each case there is $1$ less way to sum to $80$ , so the answer is just $\frac{6(6+1)}{2}=\boxed{\textbf{(B)}~21}$ .

~andyluo

Solution 4

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 $\binom{5+2}{2}=\boxed{\textbf{(B) 21}}$

~not_slay

Solution 5 (generating functions)

The problem is equivalent to the number of ways to make $\$80$ from $\$2$ bills, $\$5$ bills, and $\$10$ bills. We can use generating functions to find the coefficient of $x^{80}$ :

The $\$2$ bills provide $1+x^2+x^4+x^6 \cdots = \frac{1}{1-x^2},$

The $\$5$ bills provide $1+x^5+x^{10}+x^{15} \cdots = \frac{1}{1-x^5},$

The $\$10$ bills provide $1+x^{10}+x^{20}+x^{30} \cdots = \frac{1}{1-x^{10}}.$

Multiplying, we get $(1-x^{2})^{-1}(1-x^{5})^{-1}(1-x^{10})^{-1}.$

Solution 6 (easy logic)

There aren't dollar signs because the $\text{[math]}$ thinks they're latex symbols. If you find how to override this error, please edit this. There's no $\text{[math]}$ here but feel free to add some! ~SwordAxe

We can see in the problem that the teller gave her at least one of 20, 50, and 100. Therefore, she has 800 - 20 - 50 - 100 = 630 "left over".

Since all bills and 630 are multiples of 10, we can divide by ten. ==> Question becomes: How many different collections of 2, 5, and 10 could she get if her total was 63?

We notice that because 63 is odd, we need an odd amount of 5 bills. (2 and 10 are both even, and 63 is not a multiple of 5, so we need 2 and/or 10 bills. PM SwordAxe if you don't get this.)

We can do casework.

1: She gets one 5 (50) dollar bill. She has 58 (580) left.

 1) She is given only 2 dollar (20) bills => ONE COLLECTION (all 20 bills with one 50)
 2) She is given one 10 dollar (100) bill
      1. The rest of the money is given in 2 dollar(20) dollar bills. => ONE COLLECTION (one 100 and rest 20 with one 50)
      2. She is given another 10 dollar (100) bill
            I) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (two 100, one 50, and rest 20)
            II) She is given another 10 dollar (100) bill
                   a) The rest of the money is given in 2 dollar (20) dollar bills. => ONE COLLECTION (three 100, one 50, and rest 20)
                   b) She is given another 10 dollar (100) bill.
                         1) The rest of the money is given in 2 dollar (20) dollar bills  => ONE COLLECTION (following same pattern)
                         2)  AND SO ON...

This looks very tedious, but draw a simple tree diagram, and you'll see that its very easy!

If she gets one 50, there are 6 ways If she gets three 50, there are 5 ways ... If she gets nine 50, there are 2 ways If she gets eleven 50, there is one way

We can add them all up, with a grand sum of 6+5+4+3+2+1 = 21.

Therefore, we get the answer (B) 21.

~SwordAxe (PM me if you have any questions! :)) EDIT: use /$ for the dollar signs

Video Solution 1 by OmegaLearn

https://youtu.be/UeX3eEwRS9I

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=sfZRRsTimmE

Video Solution 3 by paixiao

https://youtu.be/EvA2Nlb7gi4?si=fVLG8gMTIC5XkEwP&t=89s This video link is invalid now.

Video Solution 4

https://youtu.be/D-ZvFBiZsaY