# The smallest three digit number that is equal to the sum of its digits plus twice the product of its digits

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?

• Do you know how to mathematically formulate the first question? Commented Apr 17, 2013 at 6:15
• the first question i know but the second i'm tired looking-for it it's so many solutions Commented Apr 17, 2013 at 6:23

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, $397$.