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If A is a factor of BC , and GCD(A,B)=1 , then A is a factor

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If A is a factor of BC , and GCD(A,B)=1 , then A is a factor [#permalink]

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New post 28 Jan 2014, 16:59
If A is a factor of BC , and GCD(A,B)=1 , then A is a factor of C.

If A is a factor of B and B is a factor of A, A = B then or A=-B.

These statements are found on GMATclub Math workbook. Could someone please explain the statements as I have trouble understanding them, could someone also please give examples to both statements. Thanks in advance.
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Re: If A is a factor of BC , and GCD(A,B)=1 , then A is a factor [#permalink]

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New post 29 Jan 2014, 03:36
The first statement means: if A is a factor of BC and A and B have no common factors than A divide C.
For example A=2 B=3 and C=4. A is a factor (=divide) BC which is 12. But A and B have no common factor. Therefore A divide C (2 divides 4).

The second statements is more straightforward. It just means that if a number divides another one for example 2 divides 4, then the only way it can be divided by this number as well is that it is the same number or the opposite.

Hope this clarify things
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Re: If A is a factor of BC , and GCD(A,B)=1 , then A is a factor [#permalink]

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smartyman wrote:
If A is a factor of BC , and GCD(A,B)=1 , then A is a factor of C.

If A is a factor of B and B is a factor of A, A = B then or A=-B.

These statements are found on GMATclub Math workbook. Could someone please explain the statements as I have trouble understanding them, could someone also please give examples to both statements. Thanks in advance.


If a is a factor of bc, and gcd(a,b)=1, then a is a factor of c.

Say \(a=2\), \(b=3\) (\(gcd(a,b)=gcd(2,3)=1\)), and \(c=4\).

\(a=2\) IS a factor of \(bc=12\), and \(a=2\) IS a factor of \(c\).

OR: if \(a\) is a factor of \(bc\) and NOT a factor of \(b\), then it must be a factor of \(c\).

Hope it's clear.
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Re: If A is a factor of BC , and GCD(A,B)=1 , then A is a factor [#permalink]

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New post 11 Jun 2015, 04:56
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Re: If A is a factor of BC , and GCD(A,B)=1 , then A is a factor   [#permalink] 11 Jun 2015, 04:56
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