## anonymous one year ago How do I check 12321 ≡ 111 (mod 3) is true without applying the definition?

1. anonymous

12321 = 111^2 btw

2. mathmath333

as far as i see both are divisible $$3$$

3. anonymous

yeah, but that's more like by inspection. I need to use properties of congruences

4. ganeshie8

12321 and 111 are divisible by 3 is sufficient

5. ganeshie8

$$12321\equiv 0 \pmod {3}$$ and $$111\equiv 0\pmod{3}$$ $$\implies 12321 \equiv 111\pmod{3}$$

6. anonymous

by transitive property?

7. ganeshie8

Yep $$a\equiv c\pmod{n}$$ and $$c\equiv b\pmod{n}$$ $$\implies a\equiv b\pmod{n}$$

8. anonymous

ah yes. I guess this was too easy. Let me pick a different problem. 12345678987654321 ≡ 0 (mod 12345678) ^^

9. anonymous

It's according to wolfram alpha, But 12345678987654321 ≡ 0 (mod 12345679) is true though

10. anonymous

it's *false* according...

11. anonymous

12345678987654321 / 12345679 = 999,999,999

12. anonymous

This problem seems challenging but it's in the section where it covers algebraic properties of congruences (addition,subtract and such...). I'm just assuming the exercise is meant for us to apply those properties but it doesn't seem like it.

13. anonymous

Maybe the exercises are meant to check the readers' understanding of the definition. Who knows.