Question

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

• A. $01$
• B. $21$
• C. $61$
• D. $71$

Answer 1

The correct option is A $01$

$7^1=7$

$7^2=49$

$7^3=343$

$7^4=2401$

$7^5=16807$

$7^6=117649$

Observe the above pattern,

The last two digits are repeating.

For every power of the form $4n,$ the last two digits are $01.$

$\therefore$ The last two digits of $7^{2008}$ is $01.$