Source: Original, though I wouldn't be surprised to find these printed somewhere else.
>From Philippe Fondanaiche Paris France Good evening, Considering the general equation Y x k = Z, we have the following properties: 1) Whatever Y, k can take only the values 4 and 9. Indeed, taking into consideration the first digit A and the last one N in Y, we have : A = last digit of Nxk and N = Axk + r (carry over < k) So A = last digit of Axk^2 + rxk (a) It is easy to check that for k = 2,3,5,6,7,8 there is no couple (A,r) satisfying the identity (a). For k = 4, there is only one couple A = 2 and N = 8 and for k = 9 there is also a single couple A = 1 and N = 9. 2) For k = 4, the lowest value of Y is 2 178 and for k = 9, the lowest one is 1 089 with the relations 2 178 x 4 = 8 712 and 1 089 x 9 = 9 801. In the set of the 4 digit numbers there is no other solution (easy to demonstrate). 3) Starting from these lowest values of Y, we build new solutions with p digit numbers Y by introducing p-4 times the number 9 between 21 and 78 or between 10 and 89. For example 21 978 x 4 = 87 912 :this the answer of the question 1 with 5 different digits for Y. Other example : 1 099 989 x 9 = 9 899 901. So there is an infinite number of solutions such as 219.....978 or 109......989. 4) By juxtaposing 2 or more identical solutions above mentioned, we create a set of new solutions which can be extended up to the infinite. For example 2 197 821 978 x 4 = 8 791 287 912 or 10 891 089 x 9 = 98 019 801 Best regards.