Talk:2-Digit Numbers divisible by both Product and Sum of Digits

"It remains to demonstrate that these are the only ones."
I just did a computer search and those are the only ones. How should I say this? --kc_kennylau (talk) 11:48, 13 May 2017 (EDT)


 * This sounds like the Four Colour Theorem situation, i.e. proof by cases, where the number of cases is greater than anyone is willing to deal with or is useful for learning. Instead of writing down other 87 possibilities, a better way would be to narrow down the set in steps. --Julius (talk) 12:04, 13 May 2017 (EDT)


 * Say, the product condition can be checked like this: let $ a \in \{ 1, ..., 9 \}$, $ b \in \{ 0, ...,9 \} $, $ n \in \N $ such that $ 10a + b = abn $. Then $ n = \frac{ 10 }{ b } + \frac{ 1 }{ a } $ and to keep $ n $ natural only some combinations of a and b will work, e.g. $ a = 1 \implies b = \{ 2, 5 \} $, $ a = 2 \implies b = 4$, and so on. Then the sum property could be directly check upon a much smaller set. As for the rest of computer proofs, such results should most probably be mentioned, but not used as a direct proof for a proposition. --Julius (talk) 12:22, 13 May 2017 (EDT)


 * For a set as small as the 2-digit numbers, it could be considered completely appropriate to make one proof an exhaustive one which literally goes through every single 2-digit number and checks them all.


 * You can eliminate a whole bunch of them by doing mundane stuff like: pointing out that a 2-digit number starting with an even digit and ending with an odd one won't be divisible by the product of its digits, because an even number does not divide an odd one, and then pointing out that a 2-digit number with 2 odd digits won't be divisible by the sum of its digits for the same reason. This leaves only the even numbers.


 * This is a theorem which may yield multiple proofs, as it's high-school accessible (middle-school accessible, even) and all sorts of bright tricks may be brought into play. --prime mover (talk) 15:09, 13 May 2017 (EDT)