15) What is the coefficient of $x^{199}$ in $\left(x^{2}+x+1\right)^{100}$?
(A) 100
(B) 298
(C) 4950
(D) 5050
(E) $99^{2}$
15. The answer is $(\mathbf{A})$.

Let's write:

$$
\left(x^{2}+x+1\right)^{100}=\underbrace{\left(x^{2}+x+1\right) \cdot\left(x^{2}+x+1\right) \cdots\left(x^{2}+x+1\right)}_{100 \text { times }} .
$$

The result of this product is a sum of monomials, each of which is the product of one of the three monomials from the first parenthesis, one of the three monomials from the second, one of the three from the third, and so on until one of the three monomials from the last parenthesis. For example, if we choose the monomial $x^{2}$ from each parenthesis, we get the term $x^{200}$. If instead we choose the monomial $x$ from the first parenthesis and $x^{2}$ from all the others, we get $x^{199}$. This happens every time we choose the monomial 1 from one of the 100 parentheses and $x^{2}$ from all the others; since we have 100 different ways to do this, in the end, the term $x^{199}$ appears in the sum at least 100 times.

Finally, observe that there are no other choices that lead to terms $x^{199}$; indeed, if we do not choose the monomial $x^{2}$ from at least two parentheses, no matter how we choose the monomials from all the others, the product will have a degree less than or equal to 198. Therefore, the coefficient of $x^{199}$ is 100. [Problem proposed by F. Poloni]