How do I check 12321 ≡ 111 (mod 3) is true without applying the definition?

Question

anonymous · Answer

12321 = 111^2 btw

mathmath333 · Answer

as far as i see both are divisible $3$

anonymous · Answer

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

ganeshie8 · Answer

12321 and 111 are divisible by 3 is sufficient

ganeshie8 · Answer

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

anonymous · Answer

by transitive property?

ganeshie8 · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

it's *false* according...

anonymous · Answer

12345678987654321 / 12345679 = 999,999,999

anonymous · Answer

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.

anonymous · Answer

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