Minimum number of cuts required to cut a cylindrical cake into $16$ equal parts? This question was asked in a competitive exam.
If we have a cylindrical cake and we want cut it into sixteen equal parts what is the minimum number of cuts required to do so?
Surely we can place $3$ equidistant horizontal cuts parallel to the circular base of cylinder plus two vertical perpendicular diametric cuts  to get $16$ equal parts. So the number of cuts is less than or equal to $6$. But $6$ was not an option instead one of the options was $5$. So can we have less than $6$ cuts??
 A: Think of it this way:
A cake with no slices has $1$ piece. When you draw the first line cut, you cut it into $2$ pieces. After that, in order to maximize the number of slices, each line cut that you make should intersect each other line cut previously made. If you calculate the number of pieces made this way, you end up with
$$p(n)=\frac{n(n+1)}{2}+1$$
Where $p$ is the number of pieces and $n$ is the number of cuts made. As you can see, the first $p(n)$ that exceeds or equals $16$ is $p(5)=16$. So the answer is $5$.
If you would like a thorough explanation of the derivation of this formula, see this page.
As for the equality of the parts, I am assuming that there is some way in which you can adjust these lines in order to make $16$ equal pieces. But we need not prove it, since you cannot use any fewer than $5$ cuts to make $16$ pieces.
A: You can cut the cake into eight equal slices using four vertical cuts (each cut after the first creates two further slices) and then make one horizontal cut.
You don't say whether you are allowed to rearrange pieces between cuts - if so you can double the number of pieces with each cut, and sixteen pieces take just four cuts.
A: This should work (left, my intended solution; right, the one by Mark Bennet):


(the red cuts are parallel to the base of the cylinder; the green cuts are perpendicular to it)
