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What is the coefficient of $x^{103}$ in the expansion of $$(1+x+x^2+x^3+x^4)^{199}(x-1)^{201}$$ ?. The answer is an integer between $0-9$. So I wrote the given expression as $(x^5-1)^{199}(x-1)^{2}$. But now I am struggling to do it only 2nd bracket will give $x^{103}$ when multiplied by last term ie $1$ of first bracket. Where is my mistake?

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  • $\begingroup$ $(1+x+x^2+x^3+x^4)^{199}(x-1)^{201}=(x^5-1)^{199}(x-1)^2$, not $200$. $\endgroup$ – Kenny Lau Jun 27 '16 at 12:54
  • $\begingroup$ Ya thanks but how to do it then? $\endgroup$ – Archis Welankar Jun 27 '16 at 12:56
  • $\begingroup$ The terms in the first bracket would have the form of $\displaystyle\binom{199}r(-1)^{199-r}x^{5r}$ $\endgroup$ – Kenny Lau Jun 27 '16 at 12:58
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    $\begingroup$ The indices must be divisible by $5$. However, there's a second bracket $(x-1)^2$, which can promote the indices by $0$ or $1$ or $2$. $\endgroup$ – Kenny Lau Jun 27 '16 at 12:59
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    $\begingroup$ However, $103$ leaves a remainder of $3$, meaning that its coefficient is $0$. $\endgroup$ – Kenny Lau Jun 27 '16 at 12:59
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Community wiki answer so the question can be marked as answered:

As noted in the comments, the second factor contributes either $0$, $1$ or $2$ powers of $x$ and the first factor contributes multiples of $5$; thus the only possible remainders modulo $5$ in the exponents are $0$, $1$ and $2$. Since $103$ has remainder $3$, the coefficient is $0$.

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Note that $1+x+x^2+x^3+x^4={1-x^5\over 1-x}$. It follows that your expression $\Phi(x)$ is equal to $$\Phi(x)=(-1)^{201}(1-x^5)^{199}(1-2x+x^2)\ .$$ Expanding this one obtains only terms of degrees $0$, $1$, $2$ modulo $5$, hence no terms of degree $103$.

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