Coefficients Party, Part 5

Any number divisible by $12$ is divisible by $2^2\times3.$ There are three places where this isn't achieved by direct application of the powers of the coefficients in the expression: the $x^{2014}$ term, the $y^{2014}$ term, and the $xy^{2013}$ term. You might think the answer is $12,$ but remember that because of the Binomial Theorem, the coefficient of the $xy^{2013}$ term is $\binom{2014}{2013}\times2\times3^{2013}=4028\times3^{2013}.$ This is divisible by $12,$ so the only two terms that aren't divisible by $12$ are the $x^{2014}$ term and the $y^{2014}$ term. Subtracting this from the $2015$ terms in the expression, there are $\boxed{2013}$ terms with coefficients divisible by $12.$

no of term will be n+1 ..... and first and last term is not divided by 12 so remaining term are 2013.

its so simple its made easy