Explanation:

\(2\) has a repeating pattern of \(4\) i.e. the unit digit values repeat after every 4 powers
\(2^1 = 2\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 16\)
\(2^5 = 32\)

Now,\( 2^{32}\) will have a unit digit as \(6\)

How - Divide \(32\) by \(4\) and we get a remainder \(0\)

We are looking for a number which will have a unit digit as \(6\) again!

A. \(2^{16}\) has unit digit as \(6\) but the number is less than \(2^{32}\)
B. \(2^{16}\) same reason as above
C. \(2^{96}\) has unit digit as \(6\) and is greater than \(2^{32}\)
D. \(2^{33}\) has unit digit as \(2\)
E. None of the above

Hence, option C

Another Approach:

We can also use the algebraic identity \((a^3 + b^3) = (a + b)(a^2 + b^2 - ab)\)
\(2^{96} + 1 = (2^{32})^3 + 1^3 = (2^{32} + 1)(2^{64} + 1 - 2^{32})\)

Since, \(2^{96} + 1\) is divisible by \(2^{32} + 1\), it will also be divisible by that number