What is the minimum amount of 1bit Full Adders required to implement the equation 4X + 3Y + 13?
The problem is in \$z_4\$. You want to add four bits, which could make 100, so you'd have to go to \$z_2\$ with the carry.

There is a normal approach, where you see 13 as a variable like x and y, so basically this is able to solve equations \$4x+3y+a\$ with \$a = (a_1,a_2,0,a_4)\$. In this approach, you can easily calculate the number of full adders needed by working from the right to the left:


At \$z_6\$ you add two bits, so you require 1 adder.
At \$z_5\$ you add two bits and the carry \$c_6\$ from \$z_6\$, so you require 1 adder.
At \$z_4\$ you add four bits and \$c_5\$, so you require 2 adders.
At \$z_3\$ you add three bits and two carries, so you require 2 adders.
At \$z_2\$ you add one bit and two carries, so you require 1 adder.
At \$z_1\$ you'd only 'add' the carry, which makes 0 adders.


This requires 7 adders.



However, because 13 isn't really a variable but a constant, you might be able to do something smart. In that case, you can work from the normal setup for \$4x+3y\$:


\$z_6 = y_3\$ (1 bit, 1 adder, 2 bits open)
\$z_5 = c_6 + y_3 + y_2\$ (3 bits, 1 adder, 0 bits open)
\$z_4 = c_5 + y_2 + y_1 + x_3\$ (4  bits, 2 adders, 2 bits open)
\$z_3 = c_{4_1} + c_{4_2} + y_1 + x_2\$ (4 bits, 2 adders, 2 bits open)
\$z_2 = c_{3_1} + c_{3_2} + x_1\$ (3 bits, 1 adder, 0 bits open)
\$z_1 = c_2\$ (1 bit, 0 adders, 0 bits open)


However, this basic setup for \$4x+3y\$ already requires 7 adders. Since \$z_4\$ uses two adders, you will have two carries that you have to add to \$z_3\$, which makes that you have there two adders as well - and thus two carries.

It might be possible to do something smart to lose one adder, but it's not going to work with 5 adders, and here is why: 



\$z_1\$ is just an overflow and requires no adders. \$z_2\$ through \$z_6\$ though, there are adders needed. This makes 5 adders. Now, you need at least two adders at \$z_4\$, since you are adding \$x_3\$, \$y_2\$, \$y_1\$ and the carry from \$z_5\$, \$c_5\$. So at \$z_4\$ you have to add four bits, which makes that you need two adders instead of one, making the total at least 6.

Please note that you need at least 6 for the simplified equation, \$4x+3y\$, so it's not relevant whether you can do something smart with the 13 or not - this is the basic setup and you will need at least 6 adders for it.