+26367 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
