Problem:
Find the number of ordered pairs of positive integers such that and neither nor has a zero digit.
Solution:
Count the number of such ordered pairs with . If is a one-digit number, then 's digits are , and . There are choices for in this case. In the case where is a two-digit number, represent the digits of as and . Then 's digits are , and . Because and , there are choices for and choices for , and so there are choices for . In the case where is a three-digit number, represent the digits of as , and . Then 's digits are , and . Because , or , there are choices for . Thus the number of pairs with is . Because each such pair can be reversed to give another allowable pair with , there are pairs
Count the number of forbidden pairs, that is, pairs in which or has a zero digit. If or has units digit , then both do, and the given equation reduces to , where and . Thus, in this case, there are forbidden pairs. In the case where neither nor has units digit , then exactly one of them must be of the form , where neither nor is . There are such values of and such values for for a total of forbidden pairs in this case. Therefore the total number of forbidden pairs is , and there are of the requested pairs.
The problems on this page are the property of the MAA's American Mathematics Competitions