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

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==Solution 2==
 
==Solution 2==
  
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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 3 subtracted from 16 will be an even number. We can see if <math>y</math> = 0, <math>x</math> = 1008. All the way to <math>x</math> = 0, and <math>y</math> = <math>672</math>.Therefore, between <math>0</math> and <math>672</math>, the number of multiples of 2 is \makebox[2 in]{(C) 337}
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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 3 subtracted from 16 will be an even number. We can see if <math>y</math> = 0, <math>x</math> = 1008. All the way to <math>x</math> = 0, and <math>y</math> = <math>672</math>.Therefore, between <math>0</math> and <math>672</math>, the number of multiples of 2 is (C) 337.

Revision as of 18:43, 3 February 2016

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 $2 \cdot 3 = lcm(2, 3)$.

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 a equation $2x$ + $3y$ = $2016$. Therefore the question is just how many multiples of 3 subtracted from 16 will be an even number. We can see if $y$ = 0, $x$ = 1008. All the way to $x$ = 0, and $y$ = $672$.Therefore, between $0$ and $672$, the number of multiples of 2 is (C) 337.

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