Question

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

• A. 21
• B. 61
• C. 1
• D. 41
• E. 81

Answer 1

The correct option is C 1

Ans: (c) 01

The last two digits of a number is nothing but the remainder obtained when the number is divided by 100.This number leaves a remainder 1 when divided by 4 as well as 25. Hence the remainder obtained when this number is divided by 100 is also 1. Hence the last two digits of this number are 01.

Alternatively

Using Chinese remainder theorem: as we have to divide by 100 and find the reminder by 4 and then by 25 ,

i.e., 4A+1=25B+1, finding integer solutions we see that A=25 and B=4 will hence give a remainder of 01.

Alternatively, :

Use the last 2 digit rule for 7. Even here u will get 01 as the last 2 digits.