Greatest Prime Number

Using the fact that $(2k)^{20n}\equiv 76\pmod{100}$ where $n \in N , k \in N ,5\nmid k$ .

$2^{74207281}\equiv2^{20\times3710364 + 1}\equiv2^{20\times3710364}\times2\equiv76^{3710364}\times2\pmod{100}$

Now , since $76^m \equiv76\pmod{100}$ where $m \in N$ ,

$\Rightarrow 76^{3710364}\times2\equiv76\times2\equiv152\equiv52\pmod{100}$

Therefore $2^{74207281} - 1 \equiv52 - 1\equiv51\pmod{100}$

Hence our answer is $\boxed{51}$ .

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$\underline{NOTE :}$

By Euler's Theorem , $n^{\phi(25)}\equiv1\pmod{25}$ , where $gcd(n , 25) = 1$ .

Substituting $\phi(25) = 20$ , $(2k)^{20}\equiv1\pmod{25}$ , where $k \in N$ Hence possible last two digits are $\boxed{01 , 26 , 51 , 76}$ .

But $01 ; 51$ aren't possible since even number raised to the power anything is even.

$26$ is also not possible because even number raised to the power $20$ is obviously divisible by $4$ .

Hence possible last two digits is $\boxed{76}$ only.

We start off by noticing that ${ 2 }^{ 8 }\equiv 56\quad (mod\quad 100)$ .

By modular exponentiation, we will realize that $({ { 2 }^{ 8 }) }^{ 16^{n} }\equiv 56\quad (mod\quad 100)$ .

Then the problem can be solved.