For positive integers k and m, [fraction]k/3[/fraction] and [fraction]

\( \frac{k}{3}\) and \( \frac{k}{2}\) are both integers, hence k is a multiple of \(6\)

Also \(\frac{k}{3}\) and \( \frac{k}{2}\) should be greater than \(4\) and \(3\) respectively.

For remainder to be \(4\), divisor (m+1 ) must be more than \(4\). Hence \( m+1 \geq 5\)­

So \(k\) is a multiple of \(6\) and \(\frac{k}{3} > 4 \) hence \(k >12 \)

Let's test a couple of values:

\(k=18\) then \(\frac{k}{3} = 6\) , then \(m+1 \) needs to be \(2\) for remainder \(4\), but we know \( m+1 \geq 5\)­ : Reject

\(k=24\), then \(\frac{k}{3} = 8\), then \(m+1\) needs to be \(4\) for remainder \(4\), but we know \( m+1 \geq 5\)­ : Reject

\(k=30\) then \(\frac{k}{3} = 10\) , then \(m+1\) needs to be \(6\) for remainder \(4\) this is possible

Let's see if it satisfies other criteria :\( \frac{k}{2} =15\), \(\frac{15}{6} =\) remainder \(3 \)

Thus \(k = 30 \) and \(m+1= 6 \) and \(\frac{30}{6} =\) remainder \(0\)

Ans A

Hope it helped.­