$m$ is some natural number. Why is $15m=0\pmod {18}$ equivalent to $3m=0 \pmod {18}$?
This is my question basically. I don't understand why that's true.
Asked
Active
Viewed 167 times
1
-
$18m\equiv 0\pmod{18}$ for sure. Consider the difference between $18m$ and $15m$. – Jack D'Aurizio Jul 25 '16 at 16:25
-
Hint $\ 18\mid 15m\iff 18\mid 18m!-!15m = 3m.\ $ Generally $$\ a\mid bm\iff a\mid am,bm\iff a\mid (am,bm) = (a,b)m\ $$ – Bill Dubuque Jul 25 '16 at 16:40
-
$15\equiv -3 \pmod{18}$. – celtschk Jul 25 '16 at 21:29
3 Answers
5
Because you get one equation by multiplying the other by $5$, and $5$ is invertible modulo $18$ (namely, $5\cdot 11\equiv 1 \pmod{18}$).
So if $3m\equiv 0\pmod{18}$ then also $5\cdot 3m\equiv 5\cdot 0\pmod{18}$ which is the same as $15m\equiv 0\pmod{18}$. And if $15m\equiv 0\pmod{18}$ then also $11\cdot 15m\equiv 11\cdot 0\pmod{18}$ which is the same as $3m\equiv 0\pmod{18}$.
hmakholm left over Monica
- 286,031
2
The statement: $$15m\equiv 0 \bmod 18$$ means that 18 divides $15m$, i.e., there's an integer number $k$ with $$ 18k = 15m. $$ (That's straight from the definition of modular equivalence.)
If you subtract $18m$ from both sides, you get $$ 18(k-m) = -3m $$ and if you then negate, you get $$ 18(m-k) = 3m $$ That shows that $3m$ is also divisible by 18, which is the same as $$ 3m \equiv 0 \bmod 18.$$
When you really understand what I've written, look at @HenningMakholm's answer for possible deeper insight.
John Hughes
- 93,729
1
We have to show that $18|15m\iff 18|3m$
Clearly if $18|3m$ then $18|5(3m)$
The hard part is showing that if $18|15m$ then $18|3m$.
Suppose that $18|15m$, then there exists an integer $k$ with $5\times 3\times m=18\times k$.
This implies that $5$ divides $18\times k$, since $5$ is a prime we conclude $5$ divides $k$ by Euclid's lemma, so $k=5j$.
Finally notice $5\times3\times m=18\times 5\times 5\implies 3m=18j\implies 18|3m$
Asinomás
- 105,651