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If $p$, $q$, and $r$ are three different odd prime numbers what are the $\operatorname{lcm} (2pq, 2pr, 2qr)$ and $\operatorname{gcf} (2pq, 2pr, 2qr)$. Anyone have any suggestions on how to even start this problem? I am just back in college and really struggling with this stuff. Thank you.

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Hint: For the gcf, is it clear to you that $2$ divides all of them, but nothing bigger than $2$ does? For the lcm, we need a $2$, and also a $p$, and a $q$, and an $r$. – André Nicolas Feb 27 '12 at 23:59

Any multiple of $2pq$, $2pr$, and $2qr$ must contain factors of $2, p, q, r$. Thus, any multiple is at least $2pqr$. What can you say from there?

As far as the greatest common factor of the three, start by finding one common factor of all three parts. Are there any others?

Example (since the letters are confusing you): Let's let $p = 3$, $q = 5$, and $r = 7$? That way there won't be any letters. It's just find the lcm and gcf of 30, 42, and 70.

The easiest way to do this example is list all factors of each number:

The factors of 30 are 2, 3, 5, 6, 10, 15, 30 The factors of 42 are 2, 3, 7, 6, 14, 21, 42 The factors of 70 are 2, 5, 7, 10, 14, 35, 70

What is the greatest common factor? That means, what is the biggest number that appears in each list? The answer is 2.

Now, for the least common multiple. We list the multiples of each number as big lists. Then, we look for the smallest number that appears in each list.

Multiples of 30 are 30, 60, 90, 120, 150, 210, 240, ... Multiples of 42 are 42, 84, 126, 168, 210, 252, ... Multiples of 70 are 70, 140, 210, 280, 350, ...

What's the smallest number that appears in all lists? It's 210. That's the least common multiple.

That just gave us the answer for this one specific case of the problem. If you do the problem as stated, you'll find the answer that works for any 3 odd primes.

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I am so frustrated, I just don't understand how you can find multiples and factors of letters. – SNS Feb 28 '12 at 0:21
We're not working with letters. Letters are just ways to help us represent a number. I know that $p$ represents some odd prime. It may be 5, or 13, or 67, or some 1 million digit prime number. But, if we did this problem with $p = 5$ and $q = 13$ and $r = 67$, that would do the problem for those specific numbers, but as soon as another problem like this came along, we'd have to do it again. If we did it with letters representing the numbers, we can prove it's true for all numbers with those specified properties. – Graphth Feb 28 '12 at 2:34
So for the GCF of 2pr, 2qr, and 2pq is 2? For the LCM of 2pr, 2qr, and 2pq is 2pqr? – SNS Feb 28 '12 at 3:44
@SNS: Since $p$, $q$, and $r$ are distinct primes, and all are different from $2$, the only number that can divide all of $2pr$, $2qr$, and $2pq$ is $2$: the only divisors of $2pr$ are $1$, $2$, $p$, $r$, $2p$, $2r$, $pr$, and $2pr$. But $p$ does not divide $2qr$; and $r$ does not divide $2pq$; however, $2$ does divide all of them, so that's the GCF. Any common multiple of $2pr$ and $2qr$ must be divisible by $2$, $p$, $q$, and $r$, and since they are all distinct primes, therefore by $2pqr$; and now you can check that $2pqr$ is also a multiple of $2pq$; so it works, and it's the smallest. – Arturo Magidin Feb 28 '12 at 5:17
  • $\mathrm GCF(2pr,2qr,2pq)$

    The divisors of "$2pr$" are: $1$, $2$, $p$, $r$, $2p$, $2r$, $pr$ and $2pr$ (while $p$, $q$ and $r$ are odd primes)

    The divisors of "$2pq$" are: $1$, $2$, $p$, $q$, $2p$, $2q$, $pq$ and $2pq$ (while $p$, $q$ and $r$ are odd primes)

    The divisors of "$2qr$" are: $1$, $2$, $r$, $q$, $2r$, $2q$, $qr$ and $2qr$ (while $p$, $q$ and $r$ are odd primes)

    Hence, what's the greatest common factor of the three?

  • $\mathrm LCM(2pq,2pr,2qr)$

    Multiples of "$2pr$" are: $2pr$, $\dots$, $2pqr$, $\dots$

    Multiples of "$2pq$" are: $2pq$, $\dots$, $2pqr$, $\dots$

    Multiples of "$2qr$" are: $2qr$, $\dots$, $2pqr$, $\dots$

What's the least common multiple of the three?

See also: Greatest Common Divisor and Least Common Multiple

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