ptm30 wrote:

What is the remainder if 7^10 is divided by 100?

A] 1

B] 43

C] 19

D] 70

E] 49

I need a help from those who generally employ binomial thm way to attack these kind of problems, because I feel this is the shortest.

I generally try to bring numerator in a form that denm can divide it using binomial theorem i.e.

Num: (7^2)^5 =49^5 = (50-1)^5

Had the denm been 50, I would have easily got the remainder as ((-1)^5)/50 which is -1 or 50 - 1= 49 (Remainder)

But here, since the denm is 2*50, how should I employ this method?

You have done 99% of the work to get the answer, here is the final bit :

You have established that 7^10 = 50k + 49

Bow there is two possibilities, either k is even or k is odd

If k is even, 7^10 = 100m + 49 (where k=2m) ...Hence remainder is 49

If k is odd, 7^10 = 100m +50 +49 (where k=2m+1) .. Hence remainder is 99

So remainder can only be 49 or 99, but since 99 isnt a choice, by POE, the answer must be 49

Answer : e Can you please tell me where I can find the explanation for this "binomial approach"? Thanks a lot