Question

Please answer this alone very fastly

This is from congruence of numbers

Find the last two digits of 3¹⁹⁹⁷

Answer 1

3^1997

3*3^1996

3*(3^4)^499

3*(81)^499

Now we can do 81^499

Here unit place is 1 so 1st place will be 1

After we will multiply 8*9=72

But we can write only unit place because we need only last 2 digits

So the two digit is 21

But here 3 was in front of this factor.

So we have to multiply by 3

So the last two digit will be 3*21=63