- #1

tellmesomething

- 263

- 24

- Homework Statement
- What is the remainder when 5^99 is divided by 13?

- Relevant Equations
- !!

The book solution is to first take one 5 out

5(5^98)= 5(25^49)=5(26-1)^49

And then when we expand it using Binomial theorem we get a number which isnt a multiple of 13, we get -5 as the remainder. But since remainders have to be positive we add 13 to it (this i generalised by dividing numbers with divisors which generate a negative remainder and then added the divisor to get the "correct remainder", i dont know exactly why this works)

But can i not write 5^99 as (13-8)^99 and expand it and expect the same answer? After expanding it the only number which doesnt have a multiple of 13 in it is the last term which is = (-8)^99 which is very far from the remainder, why's that?

5(5^98)= 5(25^49)=5(26-1)^49

And then when we expand it using Binomial theorem we get a number which isnt a multiple of 13, we get -5 as the remainder. But since remainders have to be positive we add 13 to it (this i generalised by dividing numbers with divisors which generate a negative remainder and then added the divisor to get the "correct remainder", i dont know exactly why this works)

But can i not write 5^99 as (13-8)^99 and expand it and expect the same answer? After expanding it the only number which doesnt have a multiple of 13 in it is the last term which is = (-8)^99 which is very far from the remainder, why's that?