preform using reduction first

Question

preform usi0ng reduction first

1.$(273+147) \pmod{10}$

2.$(148+14432)\pmod{12}$

I don't know what they mean by reduction? I just perform the operation and i got zero

Answer 1

They mean reduction modulo the modulus. For the first one, if you add and then reduce, you get$$ 273+147\equiv420\pmod{10}\equiv 0\pmod{10}$$ but if you reduce first you get $$273+147\equiv3+7\pmod{10}\equiv0\pmod{10}$$

Of course, you get the same answer either way, but the second way you work with smaller numbers.

Answer 2

Maybe you still don't know the meaning of reducing using the "mod", so here what it means, when you're working with (mod $a$) for example you can replace a number $x$ by a number $y$ if they both have the same remainder when divided by $a$; (we write $x\equiv y$ (mod $a$))

In your example in part $2$, you can replace $148$ by $4$ because $148=12×12+4$ (so the remainder of the division of $148$ by $12$ is $4$) and ofcourse the remainder of the division of $4$ by $12$ is also $4$ ($4=0×12+4$)

Now you what can you replace the number $14432$ by?