# Pretty Large

Since $x ^ { ab } - 1 = (x^a -1 ) \times \left( x^{ a (b-1) } + x^ { a (b-2) } + \cdots + x^ a + 1 \right)$, we conclude that the number $2^ {105} -1$ has factors of $2^3 - 1 = 7$, $2^5 - 1 = 31$ and $2^{7} - 1 = 127$. Is $\frac{ 2 ^ { 105} - 1 } { \big( 2^3 -1 \big) \times \big( 2^5 -1 \big) \times \big(2 ^ 7 -1 \big) }$ prime?

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