$21x \equiv 49 \pmod{10}$?
21 % 10 = 1
and
49 % 10 = 9
So we can rewrite this as
$1x \equiv 9 \pmod{10}$
But which theorem shows that we can substitute the LHS & RHS with the remainder?
Asked
Active
Viewed 165 times
0
-
That's the definition of congruence. You can substitute an element by any element with the same remainder. – Anurag A Aug 26 '20 at 08:22
-
@AnuragA - where do I find this definition? – user93353 Aug 26 '20 at 08:32
-
You used the Congruence Product Rule (see the dupe), and the the fact that congruence is an equivalence relation. – Bill Dubuque Aug 26 '20 at 15:56
1 Answers
2
If $a \equiv b \pmod{n}$, then $ax \equiv bx \pmod{n}$
Since $21 \equiv 1 \pmod{10}$, we have $21x \equiv x \pmod{10}$.
Also, note that it is an equivalence relation, we have symmetric and transitivity.
Since $21x \equiv x \pmod{10}$ and $21x \equiv 49 \pmod{10}$,
We have $$ x \equiv 49\pmod{10}$$
Again, since $x \equiv 49 \pmod{10}$ and $49 \equiv 9 \pmod{10}$, we have $$x \equiv 9 \pmod{10}.$$
Siong Thye Goh
- 149,520
- 20
- 88
- 149