**Quote:**

How do you jump from: *m* + 2*n* = (*m* + *n*) + *n* = 3*k* + *n* to saying that there is a remainder of 2?

If we consider divisibility of

*x* by

*t*, then adding or subtracting a multiple of

*t* does not affect the remainder.

What we have:

(

*m* + 2

*n*) = 3

*k* +

*n*So when we add 3

*k* to

*n*, the remainder when dividing by 3 stays the same. We know that the remainder for (

*m* + 2

*n*) is 2, so it must be 2 for

*n* as well.

**Here is another way of explanation**:

"(1) When (

*m* + 2

*n*) is divided by 3 the remainder is 2" results in

(

*m* + 2

*n*) = 3

*b* + 2, where

*b* is an integer.

Combine it with the result we have,

*m* + 2

*n* = 3

*k* +

*n*, to get

3

*b* + 2 = 3

*k* +

*n* *|rearrange**n* = 3(

*b* –

*k*) + 2

(

*b* –

*k*) is an integer, so

*n* has 2 as a remainder, when divided by 3.