Tricky subsets question

Create a generating function for each of the conditions:

If $Z(x)$ = no. integers, then $Z(x)=1+x^5+x^{10}+... = \frac {1}{1-x^5}$

If $Q(x)$ =no. irrationals, then $Q(x)=1+x^6+x^{12}+... =\frac {1}{1-x^6}$

If $C(x)$ =no. complex numbers, then $C(x)=1+x+x^2+x^3+x^4+x^5 =\frac {1-x^6}{1-x}$ using the geometric sum formula.

The generating function $T(x)$ describing the total number of elements is just the product of these functions, which is $(\frac {1}{1-x^5}) \cdot (\frac {1}{1-x^6}) \cdot (\frac {1-x^6}{1-x}) = \frac {1}{(1-x^5)(1-x)}.$

Expanding $T(x)$ gives $T(x)=(1+x^5+x^{10}+x^{15}+...)(1+x+x^2+x^3+x^4+...)$ The number of subsets is simply the coefficient of $x^{100}$ in the expansion of $T(x).$ Therefore, the coefficient of $x^{100}$ is:

$\displaystyle \sum_{i+j=100} (a_i b_j),$ where $a_i$ is the coefficient of $x^i$ of $1+x^5+x^{10}+...$ and $b_j$ is the coefficient of $x^j$ of $1+x+x^2+x^3+...$ , which $=1$ for all $j$ .

For $1+x^5+x^{10}+...$ , the coefficient of $x^i=1$ where $i=5k$ for some integer $k$ , and therefore there are 21 terms with coefficient 1. The other 80 terms all have a coefficient of 0.

Therefore, the coefficient of $x^{100}=21(1 \cdot 1) + 80(0 \cdot 1) = \boxed {21}.$