There appeared in "Nouvelles Annales de Mathématiques" the following
puzzle as a modification of one of my "Canterbury Puzzles." Arrange the
nine digits in three groups of two, three, and four digits, so that the
first two numbers when multiplied together make the third. Thus, 12 ×
483 = 5,796. I now also propose to include the cases where there are
one, four, and four digits, such as 4 × 1,738 = 6,952. Can you find all
the possible solutions in both cases?