AIME 1987 · 第 1 题

Solution

Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective digit in $1492$ (the values of $n$ are then fixed). Thus, the number of ordered pairs will be $(1 + 1)(4 + 1)(9 + 1)(2 + 1) = 2\cdot 5\cdot 10\cdot 3 = \boxed{300}$.

If you do not understand the above solution, consider this. For every positive integer $m$, there is only one whole number $n$ that you can add to it to obtain the required sum. Also, the total number of non-negative integers that are smaller than or equal to an integer $a$ is $(a + 1)$ because there are $(a - 1)$ positive integers that are less than it, in addition to $0$ and itself.