# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Robert Z, José Carlos Santos, Cesareo, Servaes, Brian Borchers
If this question can be reworded to fit the rules in the help center, please edit the question.

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