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I was thinking about the possibility of defining a subset of the naturals containing numbers whose digits (appropriately grouped) are its own prime factorization. For example, take 32. It would be a number of this kind if 32 = (3)(2). (Of course it's not.) Are there any members of this set? Well, every prime number is trivially a member, 7 = 7, 11 = 11. The interesting question is, does this set have any composite members? I haven't been able to think of one yet.
I quickly checked the numbers less than 100 and there aren't any examples there. The only single digit numbers in this set are the primes. Consider the double digit numbers, 10 through 19. Let A be a digit, 0 through 9, if 1A = (1)(A), then 1A = A. Since no two digit number is equivalent to a one digit number, this isn't possible. (Note, 1A does not mean 1*A.) What about the numbers 20-29? If 2A = (2)(A), then the most (2)(A) could be is 14, because the most A could be is 7. But we know 2A is in the 20's so this is not possible. Any number in the 30's, 50's, and 70's will have the same problem. Any number 4A can only be grouped (4A) because 4 is not prime. So only the boring numbers (primes) in the 40's make it into the set. Same is true for the 60's, 80's, and 90's. What can we conclude? There is no composite example less than 100.
What about 3 digit numbers? We can start to make arguments by thinking of the grouping possibilities. If ABC is a three digit number, then we could group (A)(B)(C), (AB)(C), (A)(BC). Your mission: find a composite number in this set, or prove that there aren't any.
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kingfal