Arithmetic-8 (CAT-2008)

What are the last two digits of 7²⁰⁰⁸
(1) 21      (2) 61          (3) 01           (4) 41             (5) 81
Solution follows here:

Solution:

Let us observe the pattern of last two digits for lower powers of 7:

71 = 07	75 = 16807
72 = 49	76 = 117649
73 = 343	77 = 823543
74 = 2401	78 = 5764801

Observe that it is following a pattern--- 07,49,43,01, this repeats for every 4 power increments.

Given exponent 2008, which is multiple of 4 and hence the last two digits are “01”

Answer(3)