CAT 2008 — QA Question 11
Last 2 digitsEasy
What are the last two digits of 72008?
Answer & solution
Correct answer: 01
- A
21
- B
61
01
- D
41
- E
81
Solution
Last 2 digits of
71 = 07
72 = last 2 digits of (07 × 7) = 49
73 = last 2 digits of (49 × 7) = 43
74 = last 2 digits of (43 × 7) = 01
75 = last 2 digits of (01 × 7) = 07
76 = last 2 digits of (07 × 7) = 49
77 = last 2 digits of (49 × 7) = 43
78 = last 2 digits of (43 × 7) = 01
As we can see, for every 4th power of 7, the last two digits are 01.
Since 2008 is divisible by 4, we can conclude that last two digits of 72008 are 01.
Hence, option (c).