16) What number is obtained by summing all the four-digit numbers with distinct digits, where each digit can only be $1, 2, 3$ or 6?
(A) 79992,
(B) 13332,
(C) 123456
(D) 100000,
(E) 63210.
16. The answer is $(\mathbf{A})$.

Consider the four-digit numbers composed of the digits $1,2,3$ and 6, where the first digit is 1. These are six:

$$
1236,1263,1326,1362,1623,1632 \text {. }
$$

Similarly, there are six numbers where the first digit is 2, six where it is 3, and six where it is 6. This happens for all four digits of the numbers in question. Suppose now we write all the numbers of the type described by the problem in a column and add them. In the units column, we add six times 1, six times 2, six times 3, and six times 6, resulting in 72. We write 2 as the last digit of the sum and add 7 to the sum of the tens column. This sum then becomes $72+7=79$. We write 9 as the tens digit of the sum and carry over 7 to the hundreds column. In this column, the sum is 79 again, and the hundreds digit of the sum is 9 with a carry-over of 7 to the thousands column. In this column, which is the last to consider, the sum is still 79. In conclusion, the sum is 79992.

[Problem proposed by F. Mugelli.]