Find all three digit numbers which are divisible by groups of its digits

Question

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!

Answer 1

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 $$10a+b\nmid c$$

Answer 2

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 $ab$.

Answer 3

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

• Are divisible by below-described groups of its own digits