Making $100$ ordered lists with the numbers $1 - n$ $5$ times such that no two lists are the same at more than one index. What is the minimum $n$ such that it is possible to make at least $100$ ordered lists with the numbers $1-n$ (can repeat numbers) of length $5$ with the property that any two ordered lists are the same at one index?
Ex:
If we considered $n =10$, then it would be impossible to have the lists $\{1, 2, 6, 8, 3 \}$ and $\{1, 8, 6, 6, 4 \}$ because they both share a $1$ at the first index and a $6$ at the third index.
However, we could have the lists $\{1, 2, 3, 4, 5 \}$ and $\{6, 7, 8, 9, 5\}$ because they only share a $5$ at the last index and one common element is allowed.
I honestly have no idea how to begin with this problem. I was thinking of derangements at first but the casework got very complex. Does anyone have any ideas?

The minimum $n$ is either $10$ or $11$, and the answer is unknown by the greater math community.
Obviously, you need $n\ge 10$; if you look at the first two entries of each list, you will have a sequence of $100$ ordered pairs. These pairs must be pairwise distinct, otherwise the two corresponding lists would agree in their first two places. Since there are at most $n^2$ such ordered pairs, you need $n^2\ge 100$.
This would be possible with $n=10$ if and only if there existed $3$ mutually orthogonal Latin squares of order $10$. Whether these exist is an open problem as of 2005.
You can succeed with $n=11$ using all $121$ lists of the form
$$
(x,y,x+y,x+2y, x+3y)
$$
where $x,y$ range over $\{0,1,\dots,10\}$, and all arithmetic is taken modulo $11$.

In terms of coding theory, you are looking for the smallest number $n$ such that $A_n(5, 4)\ge 100$, where $A_q(\ell, d)$ is the size of the largest code with length $\ell$, distance $d$, and an alphabet of size $q$.
