# Find all three digit numbers which are divisible by groups of its digits [closed]

How can I find all three-digit numbers which:

• Do not contain a $$0$$ digit
• Have different digits
• Are divisible by below described groups of its own digits

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

For example:

number = $$132$$

It has only non-zero digits
It has different digits
And it should be divisible by $$13$$, $$12$$, and $$32$$. (omitting one digit)

Thanks a lot in advance for helping me finding these!

## closed as off-topic by Robert Z, José Carlos Santos, Cesareo, Servaes, Brian BorchersJan 20 at 23:02

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• What have you tried? – Robert Z Jan 20 at 9:24

It's actually never possible to find such numbers since for a three digit number $$[abc]$$ $$10a+b \mid 100a+10b+c\iff \frac{100a+10b+c}{10a+b}\in \mathbb Z$$ However $$\frac{100a+10b+c}{10a+b}=\frac{10·(10a+b)+c}{10a+b}=10+\frac{c}{10a+b}\notin \mathbb Z$$ Which is the desired contradiction since
$$a,b,c\in \{n\in\mathbb N: 1≤n≤9\}$$ Therefore $$c<10a+b$$and hence $$10a+b\nmid c$$
A number $$abc$$ formed by the non-zero digits $$a,b,c$$ can never be divisible by $$ab$$ formed by $$a$$ and $$b$$ because if we divide by this number, the residue is $$c$$ which is non-zero and smaller than the number $$ab$$.
Any three digit number $$100a + 10b + c$$, can be expressed as $$10(10a + b) + c$$ $$\Rightarrow 10(10a + b) + c \mod 10a+b = c, c \ne 0$$ This contradicts the last condition