



 
Hi Pete. Let ABCD be the four digit number x. That is, x = 1000A + 100B + 10C + D. Then (A)(B)(C)(D) = 6 x 4 x 3 x 2. This is 144, but in prime factors it is 3^{2} x 2^{4}. The fundamental theorem of arithmetic (look that up using our Quick Search if you like!) says that any positive integer greater than 1 is the product of a unique set of prime factors. That means that each of This means 1, 2, 3, 4, 6, 8, 9 are the only possibilities for digits. At this point, you can start counting, ensuring each (factor) is not higher than 9: 3^{2} x 2^{4} = (3)(3x2)(2x2x2)(1) = (3)(3)(2x2x2)(2) and that's all 7 combinations. Now you just need to permute each of these. For example, the first combination yields the digits 9, 8, 2, 1 and so that could be 9821 or 2918 or 1892, etc. Watch out for the doubles as you do the permutations! Write back if you need more help. Cheers,  


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