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

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==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 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.
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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 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>
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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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~Technodoggo

Revision as of 14:21, 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 $2\cdot5d+5b+10c=80.$ Dividinb 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 $5$ things and $3$ groups, which implies $2$ dividers. Thus, the total number of ways is $\dbinom{5+2}2=\dbinom72=21.$

~Technodoggo

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