Redundant primes

[lkcr]

J has 2 factors, 12 and 10

12 (3,2,2)

10 (5,2)

Is 24 a factor of J?

Answer: "J is divisible by 12 and 10, its prime factors include 2,2,3,5 but only TWO 2's because the 2 in the prime factorization of 10 may be REDUNDANT i.e. same 2 as on of the 2's in the prime factorization of 12" and thus because 24 = 2x2x2x3, not enough 2's to make 24 hence cannot be decided if 24 is a factor.

Can someone please explain the whole "redundant" prime factor?

Thanks alot!

[Mike@Magoosh]

Hi, there. I'm happy to help with this.

J has 2 factors, 12 and 10

So, J could be 12*10 = 120. In that case

12*10 = (2*2*3)*(2*5) = 2*2*2*3*5 = 120

In that case, the 12 brought its two factors of 2, and separately, the 10 brought its single factor of 2, resulting in the 120 having three factors of 2. No factors of two had to be shared. Notice that 24 goes evenly into 120. (24*5 = 120)

By contrast, J could be the LCM of 12 and 10, which is 60.

60 = 2*2*3*5

Here's where 10 appears as a factor of 60 ---- 2*2*3*5

Here's where 12 appears as a factor of 60 ----- 2*2*3*5

Notice that, unlike in 120, the factors comprising the 10 and the factors comprising the 12 overlap. In particular, one of those factors of 2 is used twice, once to build the 10, and once to build the 12. This is what the source you have is calling a "redundant" factor, because it is used in both the factors of 10 and in the factors of 12. Notice that 24 does not go evenly into 60, because 60 has only two factors of 2, and 24 needs three factors of 2. 60 only has two factors of 2 because the 10 and the 12 are "sharing" a factor of 2, or as your source would say, because one factor of 2 is "redundant."

Does this make sense?

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Please let me know if you have any more questions about anything I have said.

Mike

[ronnie1985]

24 = 2^3*3
12 = 2^2*3
10 = 2*5
24 can be a factor if and only if the number is 2^3*3*5 or 2^3*3.2^3 has to be there. Hence it is not sufficient to say that 24 is a factor of the no.