Prove $n \in Z$ if $3|n$ and $8|n$ then $24|n$

Question

I want to prove this by saying if 3 divides n and divides then then $8\times 3$ must divide $n$ which is 24, but that doesn't seem like a proof.
Or maybe a proof by contradiction that if 3 divides n and 8 divides n then 24 does not divide $n$, but that creates a contradiction because 3 and 8 divides $n$ so $3\times8$ must.

I'm hoping there's a theorem that says if $x$ and $y$ divide $n$ then $xy|n$

this is true: $x|n$ and $y|n$, then $\lcm(x,y)|n$ – MafPrivate Oct 05 '19 at 03:59 — MafPrivate, Oct 05 '19 at 03:59
It should be the special thing of common multiples – MafPrivate Oct 05 '19 at 04:16 — MafPrivate, Oct 05 '19 at 04:16

score 2 · Accepted Answer · answered Oct 05 '19 at 04:05

2

$3\mid n\implies n=3k$ for some integer $k$.

$8\mid n\implies n=8l$ for some integer $l$.

Now note that $(8,3)=1$. This implies $\exists x,y \in \mathbb{Z}$ such that $$8x+3y=1$$.

Multiplying $n$ to both sides of the above equation we get, $$8nx+3ny=n$$ This implies $$8(3k)x+3(8l)y=n$$

I.e $24(kx+ly)=n$. As $k,x,l$ and $y$ are all integers, $24\mid n$.

answered Oct 05 '19 at 04:05

very nice I like it. Using the theorem that the GCD can be represented as a linear combination of the two numbers. – Gooby Oct 05 '19 at 04:12
Regarding your last sentence...If $x\mid n$ and $y\mid n$ then $lcm(x,y)\mid n$. – Oct 05 '19 at 04:15
The last statement is true only when $(x,y)=1$. – Oct 05 '19 at 04:16
See @Martund 's answer below . – Oct 05 '19 at 04:18

Martund · Answer 2 · 2019-10-05T05:46:18.353

Now, regarding your case, if $ x$ and $y$ divide $n$, then $n$ is a common multiple of both of them, but there must be a least element among the set of positive common multiples of $x$ and $y$. Let that least element be $k$. Then $n=kq+r$ for some $q$ and $r$ by division algorithm. Now, $x|k$ and $x|n\Rightarrow x|(n-kq)\Rightarrow x|r$. Similarly $y|r$. Bu then $r$ is a common multiple of $x$ and $y$, which is less than $k$. Hence, $r=0$ and $k|n$.

Alternatively, if you know group theory, define LCM of $x$ and $y$ to be the generator of intersection of groups generated by $x$ and $y$, i.e. if $x\mathbb Z\cap y\mathbb Z=k\mathbb Z$, define $lcm(x,y)=k$. Then all properties follow.

score 0 · Answer 3 · answered Oct 05 '19 at 05:48

0

$8|n$ then $n=8k$;

$3|n$ then $3|8k$;

Euclid's lemma: $3|k$ or $3|8$;

Since $3 \not | 8$, it follows that $3|k$.

$k=3l$;

$n=8k=8(3l)=24l$.

Recall Euclid's lemma:

If a prime $p$ divides $ab$ then $p|a$ or $p|b$.

answered Oct 05 '19 at 05:48

Peter Szilas

21,123

Prove $n \in Z$ if $3|n$ and $8|n$ then $24|n$

3 Answers3