Number Plate Puzzle

This is more of an intuitive approach. Since the numbers represented by $ab$ and $cd$ are uniformly distributed over the range $00$ to $99$ this problem is in essence equivalent to that of finding the expected distance between two randomly positioned points on a line segment of a given length $L$ . This is known to be $L/3$ , and so the desired solution in this case is $99/3 = \boxed{33}$ .

EDIT: O.k., now for a proper solution.

The sum of absolute differences between any number $ab$ and all the other numbers equals

$\displaystyle\sum_{k=0}^{ab}(k) + \displaystyle\sum_{k=0}^{99 - ab}(k)$ .

(I included $k=0$ so as to cover the case when $ab = 0$ .) The average absolute difference associated with this number will then be this sum divided by $99$ . We then need to sum all of these averages over the range for $ab$ and then divide this sum by $100$ , as there are $100$ numbers in this range. Exploiting some inherent symmetries, the sum of average absolute differences can be simplified to

$(1/99) * 2 * \displaystyle\sum_{n=0}^{99} \displaystyle\sum_{k=0}^{n}(k) =$

$(2/99) * \displaystyle\sum_{n=0}^{99}(n * (100 - n))$ =

$(2/99) * (100 * \displaystyle\sum_{n=0}^{99}(n) - \displaystyle\sum_{n=0}^{99}(n^{2}))$ .

Using the appropriate summation formulae, this comes out to

$(2/99) * ((100 * 99 * 100 / 2) - (99 * 100 * 199 / 6)) = 10000 - (19900 / 3) = 10100 / 3$ .

Finally, we just need to divide this value by 100 to find the overall average and then take the integer part as our solution. Doing this, we obtain the solution $\lfloor{101 / 3}\rfloor = \boxed{33}$ .