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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!

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

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  • 2
    $\begingroup$ What have you tried? $\endgroup$ – Robert Z Jan 20 at 9:24
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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 $$10a+b\nmid c$$

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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 $ab$.

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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
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