imc

All the desired terms are in the form $a^x{b}^y{c}^z{d}^w{1}^t$, where $x+y+z+w+t=N$ (the $1^t$ part is necessary to make stars and bars work better.) Since $x$, $y$, $z$, and $w$ must be at least $1$ ($t$ can be $0$), let $x^{\prime }=x-1$, $y^{\prime }=y-1$, $z^{\prime }=z-1$, and $w^{\prime }=w-1$, so $x^{\prime }+y^{\prime }+z^{\prime }+w^{\prime }+t=N-4$. Now, we use stars and bars (also known as ball and urn) to see that there are $\left(\genfrac{}{}{0ex}{}{(N-4)+4}{4}\right)$ or $\left(\genfrac{}{}{0ex}{}{N}{4}\right)$ solutions to this equation. We notice that $1001=7\cdot 11\cdot 13$, which leads us to guess that $N$ is around these numbers. This suspicion proves to be correct, as we see that $\left(\genfrac{}{}{0ex}{}{14}{4}\right)=1001$, giving us our answer of $\boxed{14}$ Note: An alternative is instead of making the transformation, we "give" the variables $x,y,z,w$ 1, and then proceed as above.