# 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!

• Have you found any? If there are none, there might be a simple argument to see that. If there are some, then I expect a simple search is the easiest way to go. – lulu Jan 20 at 12:48
• How do you get from being divisible by $78$, $75$, and $85$ to a need for being divisible by $78+75+85=238$? – Barry Cipra Jan 20 at 12:55
• @lulu, agreed. I think it'd be OK to relax the nonzero and nonrepeating restrictions (except for repeated $0$s), find all *those* solutions, and show they all have $0$s and/or repeated digits, e.g. $110$. – Barry Cipra Jan 20 at 13:03
• Note: in the header you ask that you number be divisible by the sum of the three two digit numbers, but in the body you just mention divisibility by each of the three digit numbers ("example: 785 should be divisible by 78, 75, and 85.") You should clarify the question. – lulu Jan 20 at 13:29
• @Peter I guess, though I don't understand why the OP doesn't just clarify the point. – lulu Jan 20 at 13:54

The following PARI/GP program finds the $$4$$ solutions , if we only demand that the number is divisible by the sum of the three numbers :
? for(a=1,9,for(b=0,9,for(c=0,9,if(Mod(100*a+10*b+c,20*a+11*b+2*c)==0,if(a*b*c<>

• If we allow zeros, $108$ and $150$ are further solutions. – Peter Jan 20 at 13:53