bettatantalo wrote:

If \(x=p*n^k+p\) where n and k are positive integers, is x divisible by 2?

1) n+kn=915

2)p^35+35^p is even

Source: Exercise given by gmat tutors

OA: A\(x=p*n^k+p=p*(n^k+1)\)

Given \(n\) and \(k\) are positive integers.

Term \(p*(n^k+1)\) will be divisible by \(2\) if either \(p\) or \((n^k+1)\) or both are divisible by \(2\).

if \(n\) is odd and \(k\) can be any positive integer,\((n^k+1)\) will be of form

ODD+ODD= EVEN i.e Term \(p*(n^k+1)\) will be even.

if \(n\) is even and \(k\) can be any positive integer,\((n^k+1)\) will be of form

EVEN+ODD= ODD i.e Term \(p*(n^k+1)\) will be odd.

Statement (1) : \(n+kn=915\)

\(n(1+k)=3*5*61\)

\(n\) can be \(3,5,61,15,183,305\)

this implies that \(n\) will be odd, leading to the term \(p*(n^k+1)\) being even.

So \(x\) will be divisible by \(2\)

Statement \(1\) alone is sufficient.

Statement (2) : \(p^{35}+35^p\) is even

\(p^{35}+35^p\) will be even if \(p^{35}\) is odd as \(35^p\) is odd.

So \(p\) is odd, but term \((n^k+1)\) can be even or odd.

\(x\) can be odd or even, depending upon value of \((n^k+1)\).

Statement \(2\) alone is not sufficient.