What are the last two digits of 7^2008?

One can obtain the last two digits of any number by dividing the number by 100 and looking at the remainder. The remainder when divided by 100 gives the last two digits. A very nice way to look at remainders is what is called Modular arithmetic. We will use modular arithmetic and its properties to solve this problem. read more

There is a pattern in the last digits. Since 7^2008 is a ridiculously huge number, we don't bother with computing it. So 7^1=7 7^2=49 7^3=343 7^4=2401 7^5=16807 <--- note how the last digits are repeating 7^6=117649 So the question is how many cycles and at what point in the cycle is 7^2008. 2008 mod 4 = 0, so it would be choice C, 01. read more