$15 \cdot 3$ ordered pairs of non-negative integers $(m, n)$, if the sum $m+n$ does not require carrying (in decimal), then it is called "simple". Find the number of all simple ordered pairs of non-negative integers whose sum is 1492.
(5th American Invitational Mathematics Examination, 1987)
[Solution] Since there is no carry-over in the summation, there are three ways to add up to 2 in the units place:
$$
0+2, \quad 1+1, \quad 2+0 \text {. }
$$

There are ten ways to add up to 9 in the tens place:
$$
\begin{array}{l}
0+9, \quad 1+8, \quad 2+7, \quad 3+6, \quad 4+5, \\
9+0, \quad 8+1, \quad 7+2, \quad 6+3, \quad 5+4 . \\
\end{array}
$$

There are five ways to add up to 4 in the hundreds place:
$$
0+4, \quad 1+3, \quad 2+2, \quad 3+1, \quad 4+0 .
$$

There are two ways to add up to 1 in the thousands place:
$$
0+1, \quad 1+0 .
$$

Therefore, the total number of simple non-negative integer ordered pairs whose sum is 1492 is
$$
2 \cdot 5 \cdot 10 \cdot 3=300
$$