Question

What are the last two digits of 7^2008?

Answer 1

Let us find the value for the power of 7. $7^1= 07\ 7^2= 49\  7^3= 343\ 7^4= 2401\ 7^5= 16807\ 7^6= 117649\ 7^7= 823543\ 7^8= 5764801$

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$7^{12}$=13841287201  

From the above values, we know that for every 4th power of 7, the value ends with 01.  

So, for 7 to the power of an exponent which is divisible by 4, the last two digits are 01.

2008 is also divisible by 4 and hence $7^{2008}$ will also end with 01.