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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