Primes & Divisibility

[silpau]

In Number Properties strategy guide, chapter 1 Q5
If j is divisible by 12 and 10 is j divisible by 24

I have not understood the explanation given

My approach:

Prime Factors of 12: 2 * 2 * 3

Prime Factors of 10: 2*5

Prime factors of j would be : 2 * 2 * 3 & 2 * 5

But the book gives it as 2 * 2 * 3 * 5

Didnot understand why one of the 2 is considered as a duplicate

Is there any generic rule when one of the prime factors need to be considered as a duplicate

[2amitprakash]

The rule can be explained more subtly but I have this understanding. It is not saying that "j" is divisible by product of 10 and 12. "j" is divisible by 10 and also divisible by 12. Hence you have to take the maximum of each unique prime factor of both numbers.
2: 2
3: 1
5: 1
So j must be divisible by 2*2*3*5. For example, j = 60 and it is divisible by 10 and 12, but not by 24.

[Ben Ku]

When we have j divisible by 12 and 10, since 2 is a prime factor of both 12 and 10, we only need to count it once (not separately).

We can see this with an example. If the prime factors of j are 2*2*3*2*5, that means the smallest number divisible j could be is 120. However, we see that the number 60 (2*2*3*5) is also divisible by 12 and 10. This must mean that we've double counted the "2" from both 12 and 10.

Let me know if you have further questions. Thanks.

[rockrock]

So does it hold true what the previous post stated, that if the question asked whether or not "j" is divisible by product of 10 and 12...then we would double count the factors, because we are simply asking if "j" is divisible by 120. In other words factor foundation rule that if j is divisble by 12, and divisible by 10. then j MUST be divisible by 12*10.

[tim]

Your last sentence did not make sense and was incorrect, but as to the rest of your post, yes. If we know that j is divisible by the product of 12 and 10, then we take the prime factors of 12 and 10 and set them side by side - including all three 2's - because that's what you will need in order to build a 120..

[plg_cp]

When we have j divisible by 12 and 10, since 2 is a prime factor of both 12 and 10, we only need to count it once (not separately).

Let's pretend the question was instead "If j is divisible by 24 and 20..."

How would we set up the prime factors of j?

We have 24 (2, 2, 2, 3) and 20 (2, 2, 5).

j's prime factors would be (3, 5, [how many 2's?])? Would we put two or three 2's and why?

Does a Venn diagram come into play in understanding the number of 2's to include?

[tim]

a Venn diagram would be useful for this one. you could also set up a prime box for J, and start dumping in primes. your goal is to create the SMALLEST set of prime factors that will allow you to build both 24 and 20. start with the 3 and 5 as you suggest. then we know we need to add 2s, so add them one at a time until you can build a 24 and a 20. once you have three 2s, your prime box contains 2 2 2 3 5; from this you can build a 24 and you can also build a 20, so that's all you need..