# simple problem on divides [closed]

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

• because 24 is the lowest common multiple of 6 and 8. – Tito Eliatron Dec 2 '18 at 17:21
• 24 is the lowest common multiple of $6$ and $8$. – user3482749 Dec 2 '18 at 17:21
• Thanks i figured that out but dont know why it works – hitherematey Dec 2 '18 at 17:21

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 $$m$$.

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 $$x.$$

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 $$24$$