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

Revision as of 17:30, 16 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)

Solution 4

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 $10,20,30,40,50$ or $0$ . $6$ Ways

Case 2: Three $5$ dollar bills

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

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

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

~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 $1+2+3+...+5+6=\frac{6\cdot7}{2}=\frac{42}{2}=21$ , or $\boxed{B}$ . ~MC413551

Video Solution 1 by OmegaLearn

https://youtu.be/UeX3eEwRS9I

@@ Line 104: / Line 104: @@
 ~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 <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==

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

Page

Toolbox

Search