How many pairs of positive integers $(n, m)$ are there such that $2n+3m=2015$? I know that $m$ must be odd and $m\le671$. Also, $n\le1006$. I can't go any further, any help?
 A: Since $2$ and $3$ are coprime, every integer solution $(a,b)$ of $2a+3b=2015$ is of the form $a=1+3k,b=671-2k$ for some integer $k$. In order that both $a$ and $b$ are positive, $k$ must lie between $0$ and $335$, hence there are $\color{red}{336}$ positive integer solutions.
A: Hint: Any odd $m$ in the range will do it: express $n$ from the equation.
A: $$\begin{array} {|cc|} \hline
 1  &  2014  \\
{\color{red}{ 2 }} & {\color{blue}{ 2013 }} \\
 3  &  2012  \\
{\color{red}{ 4 }} &  2011  \\
 5  & {\color{blue}{ 2010 }} \\
{\color{red}{ 6 }} &  2009  \\
 7  &  2008  \\
{\color{red}{ 8 }} & {\color{blue}{ 2007 }} \\
 9  &  2006  \\
{\color{red}{ 10 }} &  2005  \\
 11  & {\color{blue}{ 2004 }} \\
{\color{red}{ 12 }} &  2003  \\
 13  &  2002  \\
{\color{red}{ 14 }} & {\color{blue}{ 2001 }} \\
 15  &  2000  \\
\vdots & \vdots
\end{array}$$
The ones where both red and blue are highlighted, $2$, $8$, $14$,$\dots$, which are $2$ more than a multiple of $6$.  So it's $\left \lfloor \frac {2015 - 2}{6} \right \rfloor + 1 = 336$.
