The answer is $2$. First of all, redefine the fractions as $\frac{d}{6^{100}}$ where $d$ runs over all divisors of $6^{100}$. So we're looking for the smallest possible number of distinct denominators. Necessity. If it's $1$ and $x_{1}, \dots, x_{c}$ are the divisors written in a certain order then $$\frac{6^{100}}{x_1}=\frac{6^{100}}{(6^{100}, x_{1}+x_{2})}=\cdots$$So $$x_{1}\mid x_{2}\leadsto x_{1}\mid x_{3}\leadsto\cdots$$which is false unless $x_{1}=1$. In that case by assumption $6^{100}$ is the only denominator. But clearly we'll have $x_{1}+\cdots+x_{t}$ even for some $t$, contradiction. $\square$ Sufficiency. Add $1$ first and all the divisors of the form $2^{i}3^{j}$ for $i, j\ge 1$. Then add $2^2, 2^1, 2^4, 2^3, \dots, 2^{100}, 2^{99}$ in this order. After that add $3^2, 3^1, \dots, 3^{100}, 3^{99}$ if the numerator is $1\pmod{4}$ and $3^1, 3^2, \dots, 3^{99}, 3^{100}$ if the numerator is $3\pmod{4}$ (well it's probably easy to find the numerator modulo $4$, it's just ... boring so I put it in a more funny way as a simple algorithm). After all it's easy to see the numerator is always non-zero modulo $3$ and modulo $4$ and the only denominators we have are $6^{100}, 6^{100}/2$. $\blacksquare$