4 what is the answer and how ?

We have:

#(("",2,x,""),("","",y,""),("",4,z,""),(5,6,"",12))#

with each row being equal and the numbers 1-16 increasing from left to right and no number repeated.

We can place 1 to the left of 2. We can place 3 to the left of 4.

#((1,2,x,""),("","",y,""),(3,4,z,""),(5,6,"",12))#

In the bottom row there is a blank between 6 and 12. Let's drop a value in there for now - say 10 - which gives a rough estimate to the sum of the rows to be 33.

#((1,2,x,""),("","",y,""),(3,4,z,""),(5,6,10,12))#

The top row is currently summing to 3 and so we need 30, which is 14 and 16:

#((1,2,14,16),("","",y,""),(3,4,z,""),(5,6,10,12))#

The third row sums to 7, meaning we need 26, which is 15 and 11:

#((1,2,14,16),("","",y,""),(3,4,11,15),(5,6,10,12))#

We're left with #7, 8, 9, 13# and they sum to 37. We know we're not right but let's drop in the values as place holders:

#((1,2,14,16),(7,8,9,13),(3,4,11,15),(5,6,10,12))#

We need to increase the first three rows we did by 1 each and decrease that last row by 3, so we'll have sums of 34.

~~~~~

As an extra exercise, let's get the box filled out. In the bottom row, trade the 10 out for the 11.

#((1,2,14,16),(7,8,9,13),(3,4,10,15),(5,6,11,12))#

Then in the top row trade out the 14 for the 15:

#((1,2,15,16),(7,8,9,13),(3,4,10,14),(5,6,11,12))#

Lastly swap the 10 and 13:

#((1,2,15,16),(7,8,9,10),(3,4,13,14),(5,6,11,12))#