46. Find 4 triples of non-negative integers such that each number from 1 to 81 inclusive can be represented as the sum of four numbers - one from each triple.
46. Since the sum of four non-negative numbers must be 1, three of the four addends must be zeros. Therefore, the first triplet can be $1, 2, 3$, and each of the remaining triplets can have one zero. To get a sum of 4, one of the triplets, other than the first, must have the number 3. Let it be the second number in the second triplet. To get a sum of 7, take the third number in the second triplet to be 6. This gives sums equal to the numbers from 1 to 9. To get a sum of 10, take the second number in the third triplet to be 9. Continuing, we get:

$$
\begin{array}{rrr}
1 & 2 & 3 \\
0 & 3 & 6 \\
0 & 9 & 18 \\
0 & 27 & 54
\end{array}
$$