2016年AMC10竞赛A第20题
若某个特殊的正整数$$N$$使得$$(a+b+c+d+1)^{}N$$ 的展开式中，合并同类项后发现同时含有 $$a$$, $$b$$, $$c$$, and $$d$$四个变量正整数乘幂的项恰好有$$1001$$项，求$$N$$的值？  For some particular value of $$N$$, when $$(a+b+c+d+1)^{}N$$ is expanded and like terms are combined, the resulting expression contains exactly $$1001$$ terms that include all four variables $$a$$, $$b$$, $$c$$, and $$d$$, each to some positive power. What is $$N$$?
All the desired terms are in the form $$a^{}xb^{}yc^{}zd^{}w1^{}t$$, 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 to see that there are $$\left(\frac {(N-4)+4}4\right)$$ or $$\left(\frac N4\right)$$ 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 $$\left(\frac {14}4\right)=1001$$, giving us our answer of $$N=14$$.  An alternative is to instead make the transformation $$6^{}\prime =t+1$$,  so $$x+y+z+w+t^{}\prime =N+1$$, and all variables are positive integers. The solution to this, by Stars and Bars is $$\left(\frac {(N+1)-1}4\right)=\left(\frac N4\right)$$ and we can proceed as above.  <p>By Hockey Stick Identity, the number of terms that have all $$a$$, $$b$$, $$c$$, $$d$$ raised to a positive power is $$\left(\frac {N-1}3\right)+\left(\frac {N-2}3\right)+\cdots +\left(\frac 43\right)+\left(\frac 33\right)=\left(\frac N4\right)$$. We now want to find some $$N$$ such that $$\left(\frac N4\right)=1001$$. As mentioned above, after noticing that $$1001=7\cdot 11\cdot 13$$, and some trial and error, we find that $$\left(\frac {14}4\right)=1001$$, giving us our answer of $$N=14$$.</p>