2016 AMC 10A Problems/Problem 14

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

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 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 $(2016-1) C (2-1)$ "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, $2015/6$ $-->$ 335 $-->$ 335+2 = $\boxed{\textbf{(C)}337}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search