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, $12 \times 483 = 5,796.$ I now also propose to include the cases where there are one, four, and four digits, such as $4 \times 1,738 = 6,952.$ Can you find all the possible solutions in both cases?
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