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. | ||
+ | |||
+ | 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 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 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>. | ||
+ | |||
+ | 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 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 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> | ||
+ | |||
+ | ~Technodoggo |
Revision as of 14:21, 15 November 2023
Solution 1
We let the number of -,
-, and
bills be
and
respectively.
We are given that Dividing both sides by
, we see that
We divide both sides of this equation by :
Since
and
are integers,
must also be an integer, so
must be divisible by
. Let
where
is some positive integer.
We can then write Dividinb both sides by
, we have
We divide by
here to get
and
are both integers, so
is also an integer.
must be divisible by
, so we let
.
We now have . 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
and
such that they add to
.
We still have another constraint left, that each of and
must be at least
. For
, let
We are now looking for how many ways we can have
We use a classic technique for solving these sorts of problems: stars and bars. We have things and
groups, which implies
dividers. Thus, the total number of ways is
~Technodoggo