- For each natural number $n$ of $3$ decimal digits (thus with the first non-zero digit), we consider the number $n_0$ obtained by eliminating its possible digit equal to zero. For example, if $n = 205$ then $n_0 = 25$. Determine the number of three-digit integers $n$ for which $n_0$ is a divisor of $n$ other than $n$. I find a solution but my method of resolution is very large. Help me!!
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asked Apr 1, 2015 at 8:20
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The interesting ones are the ones of the form $n=100a+c$ with $a,c\in\{1,...,9\}$. Then $$ 10a+c\mid 100a+c\iff 10a+c\mid 90a $$ so for some integer $k$ we must have $$ 90a=k(10a+c)\iff c=\frac{90a-10ak}{k} $$ and to ensure $1\leq c\leq 9$ we must require $$ \frac{90a}{10a+9}\leq k\leq\frac{90a}{10a+1} $$ furthermore, $k$ has to divide $90a$ for $c$ to be an integer. It is then easy to check that $$ \begin{align} a&=1&&k\in\{5,6,7,8\}&5,6&\mid 90a\\ a&=2&&k\in\{7,8\}\\ a&=3&&k\in\{7,8\}\\ a&=4&&k\in\{8\}&8&\mid90a\\ a&=5&&k\in\{8\}\\ a&=6&&k\in\{8\}\\ a&=7&&k\in\{8\}\\ a&=8&&k\in\emptyset\\ a&=9&&k\in\emptyset \end{align} $$ so we are left with $(a,k,c)\in\{(1,5,8),(1,6,5),(4,8,5)\}$ so only $$ \begin{align} n&=108\\ n&=105\\ n&=405 \end{align} $$ of the form $100a+c$ with $a,c\in\{1,2,...,9\}$ will work - and they do! $$ \begin{align} 108/18&=6\\ 105/15&=7\\ 405/45&=9 \end{align} $$
The less interesting cases are $a\mid 100a$ and naturally $100a+10b=10(10a+b)$ so $10a+b\mid100a+10b$.
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$\begingroup$ @DomenicoVuono: It is slightly more elegant than trial and error for the $81$ cases of $a,c\in\{1,2,...,9\}$, but maybe not much faster. This has to do with the very limited character of the question "space" in the first place :) $\endgroup$– StringApr 1, 2015 at 9:34