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Factorization

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30 Jan 2013, 07:51
Q 1. If x is divisible by 12 and 10, is x divisible by 24?

Sol: By Factor Foundation Rule:

X = 2, 2, 3, ..... ?
x = 2, 5, ........ ?

and then combining both above, X = 2, 2, 3, 5, ........... ?

If x is divisible by 12 and 10, its prime factors include 2, 2, 3, and 5 as indicated by the prime box above. There are only TWO 2’s that are definitely in the prime factorization of x, because the 2 in prime factorization of 10 may be redundant – that is, it may be the same 2 as one of the 2’s in the prime factorization of 12.

24 = 2 × 2 × 2 × 3. There are only two 2’s in the prime box of x; 24 requires three 2’s . Therefore, 24 is not necessarily a factor of x.

Q 2. If 24 is a factor of h and 28 is a factor of k, must 21 be a factor of hk?

Sol: By Factor Foundation Rule,

h = 2, 2, 2, 3, ..........?
k = 2, 2, 7, ........ ?
and hence hk = 2, 2, 2, 2, 2, 3, 7, ........ ?

By the Factor Foundation Rule, all the factors of h and k must be factors of the product, hk. Therefore the factors of hk include 2, 2, 2, 2, 2, 3, and 7 as shown in the prime box. Both 3 and 7 are in the prime box. Therefore, 21 is a factor of hk.

My question now is “In Ques 1, one 2 is excluded however in Ques 2, all 2 is taken” Why is is so?
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30 Jan 2013, 09:03
ranjeet75 wrote:
Q 1. If x is divisible by 12 and 10, is x divisible by 24?

Sol: By Factor Foundation Rule:

X = 2, 2, 3, ..... ?
x = 2, 5, ........ ?

and then combining both above, X = 2, 2, 3, 5, ........... ?

If x is divisible by 12 and 10, its prime factors include 2, 2, 3, and 5 as indicated by the prime box above. There are only TWO 2’s that are definitely in the prime factorization of x, because the 2 in prime factorization of 10 may be redundant – that is, it may be the same 2 as one of the 2’s in the prime factorization of 12.

24 = 2 × 2 × 2 × 3. There are only two 2’s in the prime box of x; 24 requires three 2’s . Therefore, 24 is not necessarily a factor of x.

Q 2. If 24 is a factor of h and 28 is a factor of k, must 21 be a factor of hk?

Sol: By Factor Foundation Rule,

h = 2, 2, 2, 3, ..........?
k = 2, 2, 7, ........ ?
and hence hk = 2, 2, 2, 2, 2, 3, 7, ........ ?

By the Factor Foundation Rule, all the factors of h and k must be factors of the product, hk. Therefore the factors of hk include 2, 2, 2, 2, 2, 3, and 7 as shown in the prime box. Both 3 and 7 are in the prime box. Therefore, 21 is a factor of hk.

My question now is “In Ques 1, one 2 is excluded however in Ques 2, all 2 is taken” Why is is so?

Case 1 :

a)x is divisible by 12. This means x = 2*2*3*n. Where n can be any whole number. (such as 5)
b)x is divisible by 10. This means x = 2*5*m. Where m can be any whole number. (such as 6 i.e 2*3)

So, what do we now know about x. We know that
1) x has AT LEAST two 2s. (Hence, encompassing our inference from b that x has at least one 2)
2) x has AT LEAST one 3.
3) x has AT LEAST one 5.

Note : We are NOT multiplying x by x.

We are asked whether x is divisible by 24 or in other words, we are asked whether x has AT LEAST three 2s and one 3. Which can now be found.

Case 2 :

a) 24 is a factor of h. This means h = 2*2*2*3*v. Where v can be any whole number.
b) 28 is a factor of k. This means k = 2*2*7*w. Where w can be any whole number.

Now, what we are doing is multiplying h and k. So, hk = 2*2*2*3*v*2*2*7*w.

So, what do we now know about hk. We know that
1) hk has AT LEAST five (not three) 2s.
2) hk has AT LEAST one 3.
3) hk has AT LEAST one 7.

We are asked whether hk is divisible by 21 or in other words, we are asked whether hk has AT LEAST one 3 and one 7. Which can now be found.
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30 Jan 2013, 09:18
Hi Ranjeet,

Good question. In the first question we are using the least amount of prime factors that can represent the two numbers, 12 and 10. If x is divisible by 10 and 12 then x is divisible by the UNIQUE (meaning that any primes in 10 that can be paired off with primes in 12 should only be counted once) prime factors of the two numbers. If you were to add the extra 2 in there you would be assuming that x is divisible by 20 and 12 or by 10 and 24 both of which would produce an extra 2.

In the second question we know that at a minimum hk has 24 and 28 as factors. We cannot assume more than that. Now we have all of the prime factors of 24 and 28 to play with when considering whether 21 is a factor of hk.

So the difference is that in the first question we have a combination of primes that can form a number that is divisible by 12 and 10. Remember: the number does not have to divisible by 10 and 12 at the same time! In the second question we actually have the numbers 24 and 28 (and all the associated primes).

I hope that this makes sense!

HG.
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"It is a curious property of research activity that after the problem has been solved the solution seems obvious. This is true not only for those who have not previously been acquainted with the problem, but also for those who have worked over it for years." -Dr. Edwin Land

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Re: Factorization   [#permalink] 30 Jan 2013, 09:18
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