What are the last two digits of 7^2008
What are the last two digits of 72008?
- 01
- 21
- 41
- 61
Answer
Observe the pattern:
71 = 7
72 = 49
73 = 343
74 = 2401
75 = 16807
76 = 117649
The last two digits are repeating.
For every power of the form 4n, the last two digits are 01.
The correct option is A.