Question

\[5^m =3q+r \] and \(0\leq r\leq 3\)
determine r
Please, help

Answer 1

Sorry, it is \(0\leq r<3\)

Answer 2

I'm assuming m, q, and r are integers.

If m = 1, then r = 2 (5/3 = 1 remainder 2)

If m = 2, then r = 1 (25/3 = 8 remainder 1)

It turns out that any power of 5 will leave a remainder of 1 or 2, but NOT 0. Why not? Because 5^m doesn't have 3 as a factor, so 5^m is not divisible by 3.

Answer 3

Then?

Answer 4

r = 1 or r = 2, but r = 0 is not possible

Answer 5

it depends on what m is

Answer 6

Oh, so that in this argument, we have to argue if m is even, then r =1, if m is odd, r =2, right?

Answer 7

That sounds about right

Answer 8

One more question: How to argue that 5^m has no factor of 3?
For example : \(4 \cancel{|} 6\) but 4 |6^2 that is 4|36

Answer 9

I mean \(3\cancel{|} 5\) , it doesn't lead to \(3\cancel {|}5^m\)

Answer 10

@jim_thompson5910

Answer 11

4 and 6 share the common factor 2, so that's why 4 | 36 is true

however, 3 and 5 don't share any common factors (other than 1), so 3 | 5 and 3 | 25, etc are false

Answer 12

5^2 = 5*5
5^3 = 5*5*5
5^4 = 5*5*5*5
...
...
etc

there are no factors of 3 anywhere in 5^m (m is an integer, m > 0)

Answer 13

So, for this problem, we just try and error to argue , right?
don't we have to use modulo, right?

Answer 14

There might be another way to prove it (more rigorously), but idk

Answer 15

Since I see it as 5^m divided by 3 and r is remainder .

Answer 16

I need the argument to have the conclusion. Suppose m is even, then
\(5^m= 3q +r\) , how to argue to get r =1?

Answer 17

@ganeshie8 Give me your hand, please. :)

Answer 18

\[5^m\equiv 2^m \equiv (-1)^m \pmod{3}\]

Answer 19

the remainder alternates between +1 and -1

Answer 20

OK, got it. Thanks a lot.:)