vaibhav101 wrote:Is the integer n odd?
1) n is divisible by 3.
2) 2n is divisible by twice as many integers as n.
All nonzero integers are divisors of zero, therefore statement (2) would not have sense.
(Cardinality - in particular dealing with "infinity sizes" - is out-of-GMAT´s scope.)
We will assume that n is positive without loss of generality.
(Reason: m is a factor of n if, and only if, m is a factor of the opposite of n.)
(1) Insufficient:
> If n = 3 , we have YES
> If n = 6 , we have NO
(2) Sufficient: this statement is equivalent to saying that n is odd, as we shall see below:
> If n = 1 , n has 2 factors (1, -1) and 2n=2 has 4 factors (1, -1, 2, -2) , hence n=1 satisfies (2) and is odd.
> If n is odd and greater than 1, n may be written as a product of odd primes, and we have:
\[n \geqslant 3\,\,\,{\text{odd}}\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
n = {p_1}^{{k_{\,1}}} \cdot {p_2}^{{k_{\,2}}} \cdot \ldots \cdot {p_m}^{{k_{\,m}}}\,\,\,\,,\,\,\,3 \leqslant {p_1} < {p_2} < \ldots < {p_m}\,\,\, \Rightarrow \,\,\,2\left[ {\left( {{k_1} + 1} \right)\left( {{k_2} + 1} \right) \ldots \left( {{k_m} + 1} \right)} \right]\,\,{\text{factors}} \hfill \\
2n = 2 \cdot {p_1}^{{k_{\,1}}} \cdot {p_2}^{{k_{\,2}}} \cdot \ldots \cdot {p_m}^{{k_{\,m}}}\,\,\,\, \Rightarrow \,\,\,\,\,2\left[ {\left( {1 + 1} \right)\left[ {\left( {{k_1} + 1} \right)\left( {{k_2} + 1} \right) \ldots \left( {{k_m} + 1} \right)} \right]} \right] = 4\left[ {\left( {{k_1} + 1} \right)\left( {{k_2} + 1} \right) \ldots \left( {{k_m} + 1} \right)} \right]\,\,{\text{factors}} \hfill \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\left( 2 \right)\,\,\,{\text{satisfied}}\]
> If n=2 , n has 4 factors (1,-1,2,-2) but 2n=4 has "only" 6 factors (1, -1, 2, -2, 4, -4) , hence n=2 does not satisfy (2).
> If n is even and greater than 2, we may write: n = (2^k)*M, where M is odd (we have put all 2´s together), hence:
\[n = {2^k} \cdot M\,,\,\,M\,\,{\text{odd}}\,\,\, \Rightarrow \,\,\,n\,\,{\text{has}}\,\,\,2\left[ {\left( {k + 1} \right) \cdot \left( {\# \,\,{\text{positive}}\,\,{\text{factors}}\,\,{\text{of}}\,\,M} \right)} \right]\,\,\,{\text{factors}}\]
\[2n = {2^{k + 1}} \cdot M\,,\,\,M\,\,{\text{odd}}\,\,\, \Rightarrow \,\,\,n\,\,{\text{has}}\,\,\,2\left[ {\left( {k + 1 + 1} \right) \cdot \left( {\# \,\,{\text{positive}}\,\,{\text{factors}}\,\,{\text{of}}\,\,M} \right)} \right]\,\,\,{\text{factors}}\]
Note that
\[2\left[ {\left( {k + 1 + 1} \right)} \right] \ne 2\left( {k + 1} \right)\]
guarantees that no positive even integer satisfies statement (2)... we are done.
Regards,
fskilnik.
