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How can I do the following?

Find the least three-digits number that is equal to the sum of its digits plus twice the product of its digit?

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    $\begingroup$ Do you know how to mathematically formulate the first question? $\endgroup$ – k1next Apr 17 '13 at 6:15
  • $\begingroup$ the first question i know but the second i'm tired looking-for it it's so many solutions $\endgroup$ – leava_sinus Apr 17 '13 at 6:23
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You are looking for integers $a$, $b$, and $c$, with $0\leq a,b,c\leq 9$ ($c\neq 0$), and

$$ a+10b+100c=a+b+c+2abc $$ Simplifying, that becomes

$$ 9b+99c=2abc $$ We look for the smallest solution, so we want the smallest integer $c$ satisfying $$ c=\frac{9b}{2ab-99} $$ From this, we need $2ab\geq100$ or $ab\geq 50$. So the possible values of $a$ and $b$ are - $(a,b) = (6,9),(7,8),(7,9),(8,7),(8,8),(8,9),(9,6),(9,7),(9,8)$ or $(9,9)$.

For these, respectively, we have $2ab-99 = 9,13,27,13,29,45,9,27,45$ and $63$. Now, we can rule out those with $13$ and $29$, as $9b$ will not be divisible by those numbers. None of the valid values for $b$ are divisible by $5$, which also rules out $45$. We're down to $(a,b)=(6,9),(7,9),(9,6),$ $(9,7)$ and $(9,9)$. Five is easy enough to check. This gives:

$$ (6,9):\quad c=\frac{81}{9}=9\\ (7,9):\quad c=\frac{81}{27}=3\\ (9,6):\quad c=\frac{54}{9}=6\\ (9,7):\quad c=\frac{63}{27}\not\in\mathbb{Z}\\ (9,9):\quad c=\frac{81}{63}\not\in\mathbb{Z}\\ $$ It's quite obvious from here that the smallest is $c=3$, which gives our number, $397$.

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