Prime Confusion

[Bueno]

Question: If J is divisible by 12 and 10, is J divisible by 24?

Ok so i understand that this question is testing the rules for prime factors so i build my little box and get:

Primes of 12: 2,2,3

Primes of 10: 2,5

I thought that any combination of the product of these primes would also yield factor of J.
Thus 2x2x2x3 = 24

So why is the number properties prep guide (Question 5/In Action/Divisibility & Primes) say that it CANNOT BE DETERMINED.

[mdhamil]

The problem is, you don't have 3 twos as prime factors, you only have two.

Primes of 12: 2,2,3

Primes of 10: 2,5

You don't know for sure if the "2" in the primes of 10 is also one of the "2"s in the primes of 12.

So all you can say for sure is the primes are 2,2,3,5.

Mike

[rfernandez]

I find using the phrase "at least" helps for these problems.

If we are told that j is divisible by 12, then that means that j has at least two 2s and one 3 in its prime box. There could be other prime factors in j, including some more 2s and 3s, but we're not sure about them.

Similarly, j is divisible by 10 tells us that there are at least one 2 and one 5 in j's prime box.

If you combine the evidence, then what's the most that we can say about the number of 2s that must be in j's prime box? Well, what can you accurately conclude if you combine "j has at least two 2s in its prime box" with "j has at least one 2 in its prime box?" The most you could say with certainty is that j has at least two 2s in its prime box.

Rey