Question

What are the last two digits of $7^{2008}$?

Answer 1

Find out the last two digits of the given number.

The given number is $7^{2008}$.

Now analyze the pattern of the last two digits of exponent powers of $7$.

$\begin{array}{rcl}{7}^1& =& 7\\ 7^2& =& 49\\ 7^3& =& 343\\ 7^4& =& 2401\\ 7^5& =& 16807\\ 7^6& =& 117649\\ 7^7& =& 823543\\ 7^8& =& 5764801\end{array}$

From the above pattern, It is observed that the last two digits are repeated after every exponent power of $4$.

Thus, for every power that is a multiple of 4, the last two digits will be $01$.

Now, $2008$ is completely divisible by $4$.

Since the exponent power is $2008$, the last two digits of the given number will be $01$.

Hence, the last two digits of $7^{2008}$ will be $01$.