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Difference between revisions of "2023 AMC 10B Problems/Problem 11"
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Case 2: Three <math>5</math> dollar bills
 
Case 2: Three <math>5</math> dollar bills
  
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<math>2x+10z=48</math>, like before we see that <math>10z</math> can be <math>0,10,20,30,40</math>, so <math>5</math> cases.
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<math>2x+10z=48</math>, like before we see that <math>10z</math> can be <math>0,10,20,30,40</math>, so <math>5</math> way.
  
 
Now we should start to see a pattern emerges, each case there is <math>1</math> less way to sum to <math>80</math>, so the answer is just <math>\frac{6(6+1)}{2}</math>, <math>21</math> or <math>(B)</math>
 
Now we should start to see a pattern emerges, each case there is <math>1</math> less way to sum to <math>80</math>, so the answer is just <math>\frac{6(6+1)}{2}</math>, <math>21</math> or <math>(B)</math>
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~andyluo
  
 
==Video Solution 1 by OmegaLearn==
 
==Video Solution 1 by OmegaLearn==

Revision as of 09:10, 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

Video Solution 1 by OmegaLearn

https://youtu.be/UeX3eEwRS9I

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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