Observe that $29,31,37,41,43,47$ are primes that are over $49/2$ which means they all should be in one row($k=4$). See that the last number on the $4$th row should be $13$ because numbers that $13$ divides in the set of number from $1$ to $49$ are $13,2*13,3*13$, we pick the number $13$ for the last spot because the product of all rows except $4$th should be a quadratic number. So the sum of the $4$th row is $241$ and $241$ is a prime number and we are done. Example of such board is pretty simple to construct,so I will leave it to the reader.