Guessing Won't Work-2

We will be using a simple yet useful identity throughout the problem which is

$a^{2}$ - $b^{2}$ = (a+b)(a-b)

$2^{48}$ - 1

= ( $2^{24}$ - 1)( $2^{48}$ +1)

=( $2^{12}$ - 1)( $2^{12}$ +1)( $2^{12}$ - 1)( $2^{12}$ +1)

=( $2^{6}$ - 1)( $2^{6}$ +1)( $2^{6}$ - 1)( $2^{6}$ +1)( $2^{6}$ - 1)( $2^{6}$ +1)( $2^{6}$ - 1)( $2^{6}$ +1)

= d

Therefore, the factors between 5 and 10 are 7 and 9.

⇒ m =16

$(99999)^{99984}$ $\equiv ?$ (mod 17)

99999 $\equiv 5$ (mod 17)

but according to fermat's little remainder theorem

$a^{p-1}$ $\equiv 1$ (mod p) where p is prime and a is any positive integer...

$(99999)^{16}$ $\equiv 1$ (mod 17)

$((99999)^{16})^{6249}$ $\equiv 1$ (mod 17)

⇒ n =1

Therefore m+n = 16+1

= $\huge{17}$