1. Using the digits $1,9,0,1,2,0,1,9$, form a four-digit number divisible by 3 (each digit can be used as many times as it appears). Which is:

a) the largest such number;

b) the fourth largest such number?
1. a) The largest number that can be formed from these four digits is 9921; since it is divisible by 3, it is precisely the answer we are looking for.

b) If we place the two largest digits on the first two positions from the left, i.e., 99, and since $3 \mid 9+9$, we conclude that the remaining two positions must be filled with two digits whose sum is divisible by 3 (so that the entire number is divisible by 3), which leaves the possibilities 21, 12, and 00. Therefore, 9921, 9912, and 9900 are the largest three four-digit numbers divisible by 3 that can be formed from the given digits, so the fourth largest number must have a digit different from 9 in the second position. The next largest digit we can place in the second position is 2, and if we place the digit 9 in the third position (to make the number as large as possible), the digit 1 must then come in the fourth position (due to divisibility by 3). Therefore, the answer is 9291.