If x is a positive integer, is the GCD of x + 3 and x + 5 more than 1?

Solution:

We need to find:
If the GCD of \(x+3\) and \(x+5\) is more than \(1\) or not.
Statement 1
“\(3x\) is the common factor of \(12\) and \(6\)”.
Factors of \(12= 1, 2, 3, 4, 6,12\)
Factors of \(6= 1,2,3,6\)
Common factor of \(12\) and \(6\) which are in the form \(3x\) are, \(3\) and \(6\).
When,

\(3x=3\)

\(x=1\)
\(3x=6\)

\(x=2\)

Thus, we do not have a single value of x.
Therefore, Statement 1 alone is NOT sufficient to answer the question.

Statement 2
“\(2x^n\) has \(1\) prime factor “
We know, \(x^n\) has the same number of prime factors as \(x\) has. Therefore,
\(2x\) also has \(1\) factor.
\(2x= 2*x\)
For \(2x\) to have only \(1\) prime factor, the value of \(x\) can be \(1\) or \(2\).
Thus, we do not have a single value of x.
Therefore, Statement 2 alone is NOT sufficient to answer the question.
We are getting the same value of \(x\) from both the statements. Thus, both statements combined will not give the answer.
Therefore, statement 1 and 2 TOGETHER are not sufficient.

Answer: Option E

Hello all,

I think that the answer should be B. It is mentioned in statement (2) that 2x^n has 1 prime factor. 2 is already there, so x has to be 1 for statement (2) to hold true. How can x be 2 ? If x = 2, then 2x^n has 2 prime factors(2,2) for n=1 ; 3 prime factors (2,2,2) for n=2 ; and so on....

Please correct me if I am wrong.

Hi Samudra1993

You must also consider the possibility that n=0 in which case, x can take any value since any number raised to the power 0 equals 1

So x=1 and x=2 can both satisfy the two statements if we consider n=0

If x=1, then GCD of x+3 and x+5 is not 1

If x=2, then GCD of x+3 and x+5 is 1

Hope this helps!

Posted from my mobile device

Yes. Thank you ! Rookie mistake on my part. Should consider all possibilities.

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I think there is an issue with the first statement. It should be 3x is a common factor of 12 and 6 instead of 'the' common factor.