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If 6 divides x and 8 divides x how do you deduce 24 divides x

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  • $\begingroup$ because 24 is the lowest common multiple of 6 and 8. $\endgroup$ – Tito Eliatron Dec 2 '18 at 17:21
  • $\begingroup$ 24 is the lowest common multiple of $6$ and $8$. $\endgroup$ – user3482749 Dec 2 '18 at 17:21
  • $\begingroup$ Thanks i figured that out but dont know why it works $\endgroup$ – hitherematey Dec 2 '18 at 17:21
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In general, this is due to prime factorisation. If $6, 8$ both divide $x$, then $3$ divides $x$ (as $3$ divides $6$) and $2^3=8$ divides $x$. So it must be that $2^3\cdot 3=24$ will divide $x$, as we get $x= 2^3\cdot 3\cdot m$ for some integer $m$.

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Given:

1)$x=6m$, and

2)$x=8n$;

$x=6m=3(2m)$; i .e. $3|x.$

Using 2): $3$ divides $x= 8n.$

Euclid's lemma:

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

$3|8n$ then $3|n$ (why?) , i.e. $n=3k$.

Combining

$x= 8n = 8(3k)=(24)k$,

$24$ divides $x.$

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what we know is:

$\frac{x}{6}$ is an integer so : $\frac{x}{2}$ and $\frac{x}{3}$ are also integers. Also:

$\frac x8$ is an integer so :$\frac x4$ and $\frac x2$ are also integers.

But, for it to be divisible by $8$ and $6$ that means it must be divisible by $1,2,3,4,6,8$. The smallest a number can be whilst divisible by $8,6$ is the LCM, 24. so we know that $x$ must be a multiple of $24$ and is therefore divisible by $24$

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