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$\mathrm{GCF}(a,b)=4$ and $\mathrm{LCM}(a,b)=96$. Find all pairs of whole numbers $a$ and $b$ for which both statements are true.

I have no clue where to even start with this problem. Thank you for any help!

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No, I don't take orders from you. But I do suggest you (1) rephrase your command into a question, and (2) tell us what progress you have made on it, what you have tried, what you don't understand, where you get stuck, etc. – Gerry Myerson Feb 27 '12 at 23:32
I am not sure I get what I did wrong...I was asking for help. I have no clue where to start on this question. – SNS Feb 27 '12 at 23:45
@SNS: Rephrase your question and ask for help. No one would help you here asking a question like that. For more information, have a look at this meta post: How to ask a homework question. – Gigili Feb 27 '12 at 23:47
Ok, I did. Sorry I did not know I offended anyone. Sorry again. – SNS Feb 27 '12 at 23:49
@SNS There are some readers (not I) who take offense when questions are posed imperatively. To remedy that you can instead write "how can I find...". More importantly, the more context that you supply, the more likely that you'll receive helpful answers. – Bill Dubuque Feb 27 '12 at 23:50

Both $a$ and $b$ must contain 4 as a factor, since 4 is the greatest common factor. Thus, $a = 4i$ for some $i$ and $b = 4j$ for some $j$. And, since it's the greatest common factor, $i$ and $j$ contain no common factors.

Now, the least common multiple of $a$ and $b$ is 96, which means 96 contains all the factors of $a$ and all the factors of $b$. So, what does this tell us about $i$ and $j$? Can you take it from there?

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I don't understand where i and j are coming from. – SNS Feb 28 '12 at 0:17
Here's the deal with $i$ and $j$. We know that 4 is a factor of $a$. That means $a$ is 4 times "something". We don't know what that something is, so we just call it $i$. Putting that all together, $a = 4i$. With the exact same thinking, $b = 4j$. – Graphth Feb 28 '12 at 2:31
Ok, so am I trying to figure out what i and j are? – SNS Feb 28 '12 at 3:33
Exactly. And, you can use the lcm being 96 to figure those out. – Graphth Feb 28 '12 at 13:39

I use $\gcd(a,b)$ for what you call "greatest common factor" (in my experience, it is more generally called the "greatest common divisor"; go figure).

For any positive integers $a$ and $b$, $\gcd(a,b)\mathrm{lcm}(a,b)=ab$. This is not hard to prove, and you may already know it.

If you do already know it, then you have that $ab = 4\times 96 = 384 = 2^7\times 3$. Now, you want to find values of $a$ and $b$ whose product is $2^7\times 3$, but where the largest common factor they have is exactly $2^2=4$. So $a$ will account for $2^2$; $b$ will account for $2^2$; that leaves you with $2^3\times 3$ still to distribute. Now, the $3$ can go into either $a$ or $b$ and that will not affect the gcd; but you have to be careful with the three remaining factors of $2$: they must all go into $a$ or into $b$ (can you see why?). That limits the possibilities rather strongly.

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