Thursday, 22 December 2016

combinatorics - Find all 3-digit numbers divisible by a sum of groups of its digits

How to find all three-digit number which are divisible by a sum of specific digit groups explained below?



The original number should have only non-zero and non-repeating digits.



example:
301 has a zero digit - cannot be used
331 does not have different digits - cannot be used



And the number should be divisible by two-digit group of its own digits, which are made by omitting one of the number's digits.



example:
785 should be divisible by 78, 75, and 85.




I have come just to this:



If the number is made of digits a,b,c like this [abc], the number should be divisible by
(10a+b)+(10b+c)+(10a+c)=20a+11b+2c



But I am not sure how to find all of the suitable numbers.



Thanks a lot for your time!

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