Marija likes to play with all sorts of things, numbers among them. This time she asks the following. Given a natural number N, how many natural numbers X with exactly N digits have the product of the digits in even positions equal to the product of the digits in odd positions?
Positions are numbered from left to right. The leading digit, the one with the largest place value, is position 1, so it lies in an odd position. Since X has exactly N digits, the leading digit is not 0.
A product over no digits is defined as 1. So when N is 1 there is no even position and the product of the digits in even positions is 1.