Dtyjlui

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!

It's actually never possible to find such numbers since for a three digit number $[abc]$
$$10a+b mid 100a+10b+ciff frac{100a+10b+c}{10a+b}in mathbb Z$$
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

$$a,b,cin {ninmathbb N: 1≤n≤9}$$ Therefore $$c<10a+b$$and hence $$10a+bnmid c$$

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

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