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

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==Solution 1==
 
==Solution 1==
  
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We let the number of <math>\$20</math>-, <math>\$50</math>-, and <math>\$100</math> bills be <math>a,b,</math> and <math>c,</math> respectively.  
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We let the number of <math>\$20</math>, <math>\$50</math>, and <math>\$100</math> bills be <math>a,b,</math> and <math>c,</math> respectively.  
  
 
We are given that <math>20a+50b+100c=800.</math> Dividing both sides by <math>10</math>, we see that <math>2a+5b+10c=80.</math>  
 
We are given that <math>20a+50b+100c=800.</math> Dividing both sides by <math>10</math>, we see that <math>2a+5b+10c=80.</math>  
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We divide both sides of this equation by <math>5</math>: <math>\dfrac25a+b+2c=16.</math> Since <math>b+2c</math> and <math>16</math> are integers, <math>\dfrac25a</math> must also be an integer, so <math>a</math> must be divisible by <math>5</math>. Let <math>a=5d,</math> where <math>d</math> is some positive integer.  
 
We divide both sides of this equation by <math>5</math>: <math>\dfrac25a+b+2c=16.</math> Since <math>b+2c</math> and <math>16</math> are integers, <math>\dfrac25a</math> must also be an integer, so <math>a</math> must be divisible by <math>5</math>. Let <math>a=5d,</math> where <math>d</math> is some positive integer.  
  
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We can then write <math>2\cdot5d+5b+10c=80.</math> Dividinb both sides by <math>5</math>, we have <math>2d+b+2c=16.</math> We divide by <math>2</math> here to get <math>d+\dfrac b2+c=8.</math> <math>d+c</math> and <math>8</math> are both integers, so <math>\dfrac b2</math> is also an integer. <math>b</math> must be divisible by <math>2</math>, so we let <math>b=2e</math>.  
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We can then write <math>2\cdot5d+5b+10c=80.</math> Dividing both sides by <math>5</math>, we have <math>2d+b+2c=16.</math> We divide by <math>2</math> here to get <math>d+\dfrac b2+c=8.</math> <math>d+c</math> and <math>8</math> are both integers, so <math>\dfrac b2</math> is also an integer. <math>b</math> must be divisible by <math>2</math>, so we let <math>b=2e</math>.  
  
 
We now have <math>2d+2e+2c=16\implies d+e+c=8</math>. 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 <math>d,e,</math> and <math>c</math> such that they add to <math>8</math>.  
 
We now have <math>2d+2e+2c=16\implies d+e+c=8</math>. 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 <math>d,e,</math> and <math>c</math> such that they add to <math>8</math>.  
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We still have another constraint left, that each of <math>d,e,</math> and <math>c</math> must be at least <math>1</math>. For <math>n\in\{d,e,c\}</math>, let <math>n'=n-1.</math> We are now looking for how many ways we can have <math>d'+e'+c'=8-1-1-1=5.</math>  
 
We still have another constraint left, that each of <math>d,e,</math> and <math>c</math> must be at least <math>1</math>. For <math>n\in\{d,e,c\}</math>, let <math>n'=n-1.</math> We are now looking for how many ways we can have <math>d'+e'+c'=8-1-1-1=5.</math>  
  
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We use a classic technique for solving these sorts of problems: stars and bars. We have <math>5</math> things and <math>3</math> groups, which implies <math>2</math> dividers. Thus, the total number of ways is <math>\dbinom{5+2}2=\dbinom72=21.</math>  
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We use a classic technique for solving these sorts of problems: stars and bars. We have <math>5</math> stars and <math>3</math> groups, which implies <math>2</math> bars. Thus, the total number of ways is <math>\dbinom{5+2}2=\dbinom72=21.</math>  
  
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~<math>\textbf{Techno}\textcolor{red}{doggo}</math>
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~<math>\textbf{Techno}\textcolor{red}{doggo}</math> ~minoe edits by lucaswujc
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 18:10, 15 November 2023

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 $\dbinom{5+2}2=\dbinom72=21.$

~$\textbf{Techno}\textcolor{red}{doggo}$ ~minoe 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)

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