Difference between revisions of "2016 AMC 10A Problems/Problem 14"

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<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math>
 
<math>\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672</math>
  
== Solution ==
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==Solution 1==
  
===Solution 1 ===
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The amount of twos in our sum ranges from <math>0</math> to <math>1008</math>, with differences of <math>3</math> because <math>2 \cdot 3 = \operatorname{lcm}(2, 3)</math>.
  
The amount of twos in our sum ranges from <math>0</math> to <math>1008</math>, with differences of <math>3</math> because <math>2 \cdot 3 = lcm(2, 3)</math>.
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The possible amount of twos is <math>\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{\textbf{(C)} 337}</math>.
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==Solution 2==
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You can also see that you can rewrite the word problem into an equation <math>2x</math> + <math>3y</math> = <math>2016</math>. Therefore the question is just how many multiples of <math>3</math> subtracted from 2016 will be an even number. We can see that <math>x = 1008</math>, <math>y = 0</math> all the way to <math>x = 0</math>, and <math>y = 672</math> works, with <math>y</math> being incremented by <math>2</math>'s.Therefore, between <math>0</math> and <math>672</math>, the number of multiples of <math>2</math> is <math>\boxed{\textbf{(C)}337}</math>.
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 +
== Solution 3==
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We can utilize the stars-and-bars distribution technique to solve this problem.
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We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have <math>\binom{2016-1}{2-1}</math> "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in <math>2015/6</math>. We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0.
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So, <math>2015/6\implies335\implies335+2=\boxed{\textbf{(C)}337}</math>.
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 +
==Video Solution==
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https://youtu.be/dHY8gjoYFXU?t=1058
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 +
~IceMatrix
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 +
https://youtu.be/99wDp0JcSUE
  
The possible amount of twos is <math>\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{\textbf{(C)} 337}</math>.
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~savannahsolver
  
===Solution 2===
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https://youtu.be/ZhAZ1oPe5Ds?t=2959
  
You can also see that you can rewrite the word problem into a equation <math>2x</math> + <math>3y</math> = <math>2016</math>. Therefore the question is just how many multiples of <math>3</math> subtracted from 2016 will be an even number. We can see that <math>x = 1008</math>, <math>y = 0</math> all the way to <math>x = 0</math>, and <math>y = 672</math> works, with <math>y</math> being incremented by <math>2</math>'s.Therefore, between <math>0</math> and <math>672</math>, the number of multiples of <math>2</math> is <math>\boxed{\textbf{(C)}337}</math>.
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~ pi_is_3.14
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=A|num-b=13|num-a=15}}
 
{{AMC10 box|year=2016|ab=A|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:24, 17 January 2021

Problem

How many ways are there to write $2016$ as the sum of twos and threes, ignoring order? (For example, $1008\cdot 2 + 0\cdot 3$ and $402\cdot 2 + 404\cdot 3$ are two such ways.)

$\textbf{(A)}\ 236\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 337\qquad\textbf{(D)}\ 403\qquad\textbf{(E)}\ 672$

Solution 1

The amount of twos in our sum ranges from $0$ to $1008$, with differences of $3$ because $2 \cdot 3 = \operatorname{lcm}(2, 3)$.

The possible amount of twos is $\frac{1008 - 0}{3} + 1 \Rightarrow \boxed{\textbf{(C)} 337}$.

Solution 2

You can also see that you can rewrite the word problem into an equation $2x$ + $3y$ = $2016$. Therefore the question is just how many multiples of $3$ subtracted from 2016 will be an even number. We can see that $x = 1008$, $y = 0$ all the way to $x = 0$, and $y = 672$ works, with $y$ being incremented by $2$'s.Therefore, between $0$ and $672$, the number of multiples of $2$ is $\boxed{\textbf{(C)}337}$.

Solution 3

We can utilize the stars-and-bars distribution technique to solve this problem. We have 2 "buckets" in which we will distribute parts of our total sum, 2016. By doing this, we know we will have $\binom{2016-1}{2-1}$ "total" answers. We want every third x and second y, so we divide our previous total by 6, which will result in $2015/6$. We have to round down to the nearest integer, and we have to add 2 because we did not consider the 2 solutions involving x or y being 0. So, $2015/6\implies335\implies335+2=\boxed{\textbf{(C)}337}$.

Video Solution

https://youtu.be/dHY8gjoYFXU?t=1058

~IceMatrix

https://youtu.be/99wDp0JcSUE

~savannahsolver

https://youtu.be/ZhAZ1oPe5Ds?t=2959

~ pi_is_3.14

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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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