$1 \cdot 12$ Consider a 100-digit number consisting only of the digits 1 and 2. You are allowed to pick any 10 consecutive digits and swap the positions of the first 5 and the last 5 digits. If one 100-digit number can be obtained from another by performing this operation several times, then these two numbers are called congruent. How many 100-digit numbers, consisting only of the digits 1 and 2, can be selected such that no two of them are congruent?
[Solution] Since each operation involves moving the first 5 numbers of the 10 numbers taken out 5 positions backward, and the last 5 numbers 5 positions forward, during the operation process, no matter how many times the operation is performed, the difference between a number's position and its initial position is always a multiple of 5. In other words, when the 100 numbers are divided into the following 5 groups:
$S_{j}=\{5 k+j \mid k=0,1,2, \cdots, 19\}, j=1,2,3,4,5$, during the operation process, each number's position can only change within the group, and by appropriately arranging the operation sequence, each number can be moved to any other position within the group. Therefore, under the congruence relation, the relative positions of the 20 digits within each group are irrelevant, i.e., no matter how they are swapped, they are always congruent, and only the number of 1s within the group matters. Thus, as long as the number of 1s in one of the 5 groups is different, the two numbers are definitely not congruent.

For each group of numbers, the number of 1s can be $0,1,2, \cdots, 20$. Hence, by the multiplication principle, the number of mutually incongruent 100-digit numbers composed of the digits 1 and 2 is at most $21^{5}$.