Find the number of multiples of 3 which have six digits, none of which is greater than 5 .
The first digit can be any number from 1 to 5 , making 5 possibilities. Each of the succeeding digits, from the tenthousands digit to the tens digit, can be any of the six digits from 0 to 5 . Finally, we claim that there are exactly two possibilities for the last digit. Given the first five digits, if we append the digits 0,1 , and 2 in turn, we get three consecutive integers, exactly one of which is a multiple of 3 . The same happens when we add the digits 3,4 , and 5 .

Thus the total number of multiples of 3 is $5 \cdot 6 \cdot 6 \cdot 6 \cdot 6 \cdot 2=12960$.