Monday, 24 October 2011

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)


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