How is (a mod b +a)mod b equal to (2*a)mod b

You can work it out by assuming a = b*k + c. Then perform the given operations.

```
let a mod b =k
(k+a) mod b
k mod b + a mod b
as k is already modded w.r.t b so k mod b=k
so k+k=2k.
now on rhs
2*a=a+a
(a+a)mod b
a mod b + a mod b
k+k=2k.
```