Number Theory

What is the product of the two smallest prime factors of $2^{100} - 1$?

First prime factor 2 ?
$2^{100} - 1$ is an odd number so 2 is no prime factor in $2^{100} - 1$

next prime factor 3 ?
If 3 is a prime factor in $2^{100} - 1$, then $2^{100} - 1 \equiv 0 \pmod{3}$
$\begin{array}{|rcll|} \hline 2^{100} - 1 &\equiv& 0 \pmod{3} \quad ? \quad | \quad 2 \equiv -1 \pmod{3} \\ (-1)^{100} - 1 &\equiv& 0 \pmod{3} \quad ? \\ 1-1 &\equiv& 0 \pmod{3} \quad ? \\ 0&\equiv& 0 \pmod{3}~ \checkmark \quad | \quad {\color{red}3} ~\text{is a prime factor in}~ 2^{100} - 1 \\ \hline \end{array}$

next prime factor 5 ?
If 5 is a prime factor oin $2^{100} - 1$, then $2^{100} - 1 \equiv 0 \pmod{5}$
$\begin{array}{|rcll|} \hline 2^{100} - 1 &\equiv& 0 \pmod{5} \quad ? \\ 2^{2*50} - 1 &\equiv& 0 \pmod{5} \quad ? \\ \left(2^2\right)^{50} - 1 &\equiv& 0 \pmod{5} \quad ? \quad | \quad 2^2=4 \equiv -1 \pmod{5} \\ (-1)^{50} - 1 &\equiv& 0 \pmod{5} \quad ? \\ 1-1 &\equiv& 0 \pmod{5} \quad ? \\ 0&\equiv& 0 \pmod{5} ~ \checkmark \quad | \quad {\color{red}5} ~\text{is a prime factor in}~ 2^{100} - 1 \\ \hline \end{array}$

$3*5 = 15$

The product of the two smallest prime factors of $2^{100} - 1$ is 15