Final answer:
The maximum size of a set containing 2-digit sequences where each element is at least 3 nudges away from every other is 9. This is determined by dividing the number of options for each digit place (10) by 3, leading to 3 choices per digit place, resulting in 3 x 3 possible sequences.
Step-by-step explanation:
The problem posed is essentially one of maximizing a set of 2-digit sequences where each sequence is at least 3 nudges away from every other. A nudge is defined as increasing or decreasing a digit by 1, or swapping a 0 with a 9 and vice versa. Understanding that there are 100 possible 2-digit sequences (including those starting with 0, like 09), we must ensure that no sequence in the set $s can be transformed into any other with fewer than three nudges.
For the maximum possible set, we have to take into account that the nudges can be in either tens or ones place. To be three nudges apart, the numbers must differ by at least 3 in either the tens or the ones place. For example, the sequence 00 could not be in the same set as 01, 02, or 10, but could be with 03 or 30. By systematically applying this rule, we can find an arrangement that fulfills the condition.
Considering all available sequences, one efficient approach to build the set is to start with 00 and include every third number (03, 06, 09, etc.) for the tens place and do the same for the ones place. This ensures each selected number is three changes away from any other in the set.
Using this method, we can construct a set of 10 sequences for each digit place, resulting in 10 x 10 = 100 possible sequences. However, since each sequence must be at least 3 nudges away from any other, we must divide this total by 3 in each dimension (tens and ones), leading to 10/3 options rounded down for both, which gives us 3 x 3 = 9 sequences in the maximal set $s.