An [ordered pair](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(m,n)$ of [non-negative](https://artofproblemsolving.com/wiki/index.php/Non-negative) [integers](https://artofproblemsolving.com/wiki/index.php/Integer) is called "simple" if the [addition](https://artofproblemsolving.com/wiki/index.php/Addition) $m+n$ in base $10$ requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to $1492$.
Since no carrying over is allowed, the range of possible values of any digit of $m$ is from $0$ to the respective [digit](https://artofproblemsolving.com/wiki/index.php/Digit) in $1492$ (the values of $n$ are then fixed). Thus, the number of [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) 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](https://artofproblemsolving.com/wiki/index.php/Positive_integer) $m$, there is only one [whole number](https://artofproblemsolving.com/wiki/index.php/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.