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

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 $ab$.


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 $$10a+b\nmid c$$

Tags:

Combinatorics

Decimal Expansion

Divisibility

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