A number has three distinct digits. The sum of the digits equals the product of the digits. How many three-digit numbers satisfy this condition?
I believe this might be all the permutations of the divisors of the perfect number, "6"
Note 1 + 2 + 3 = 6 = 1* 2 *3 so...the three digit numbers are
123, 132, 213, 231 312, 321
Mmm I can't think of any others either but I would like to see a proof or a concrete demonstration that there are no others.
Here's a proof that only the three digits 1, 2 and 3 are involved: