Q:

Digits with Nudges-Combinatorics

Given a $2$-digit sequence (which can start with $0$), let us define a nudge as the operation of either increasing or decreasing one of the digits by $1$, or changing a $0$ to a $9$ or a $9$ to a $0$. For example, we can nudge $19$ to get $09$, $29$, $18$, or $10$.
Suppose $S$ is a set of $2$-digit sequences such that it takes at least $3$ nudges to transform any element of $S$ into another element of $S$. What is the maximum possible number of elements of $S$?

A:

I'll provide a formula for finding the minimal number of nudges needed to transform one number into another .
Let $\overline{ab}$ and $\overline{cd}$ two numbers . If we want to nudge $\overline{cd}$ and get to $\overline{ab}$ we can simply reverse the operations needed to go from $\overline{ab}$ to $\overline{cd}$.
We should transform $a$ to $c$ . Assume for example that $a<c$ .To get to $c$ (in a minimal number of nudges) we could either go through $a,a+1,a+2,\ldots,c$ or in the reverse way : $a,a-1,\ldots ,0,9,8,\ldots ,c$ . For the first way there are $c-a$ nudges and for the second there are $10-(c-a)$ .
This means to use a minimal number of nudges we will choose $\min(c-a,10-(c-a))$. 
Now use the same argument for transforming $b$ to $d$ .
Putting everything together :
The minimal number of nudges to go from $\overline{ab}$ to $\overline{cd}$ is :
$$\min(\mid c-a \mid,10-\mid c-a \mid )+\min(\mid d-b \mid,10-\mid d-b \mid )$$

