# Find least positive three digit integer equal to sum of its digits plus twice the product of digits?

Find least positive three digit integer equal to sum of its digits plus twice the product of digits ?
My Attempt
let $a,b,c$ be three digits.According to given condition $abc=a+b+c+2*a*b*c$. This leads to $9(11a+b)=2*a*b*c$ Thus one digit is divisible by 9 or two digits divisible by three.$(11a+b)$ is even implies digit $a$ and $b$ is of same parity. I am stuck here and can't proceed further. Any help is appreciated. Thanks in advance.

## 1 Answer

A good start. Now I would say we need the digits large to make this work because you have $99a$ on the left and $2a\cdot b\cdot c$ on the right, so $b\cdot c \gt 49$. As you want a factor $9$ and making $c$ large doesn't make $abc$ too much larger, I would try $c=9$. That gives $11a+b=2a\cdot b, b=\frac {11a}{2a-1}$. As $2a-1$ is coprime with $a$, it must divide into $11$, which gives $a=6, b=6$ for a candidate number of $669.$ All that is left is to prove there is none smaller. Given $b \cdot c \gt 49$, and assuming $c \neq 9$ the only other choices are $(7,8), (8,7), (9,6), (9,7), (9,8)$. The first two would force $a=9$ for the divisibility and that is larger than $669$ so we don't have to check them.

• I have eventually reached this solution , after posting this problem , but then came the surprise. Answer is 397. So 669 is a solution , but not the least one. – rugi Jul 27 '17 at 6:02
• The commonality of $b=9$ in the other cases makes it easy to check. I had to go to bed last night. – Ross Millikan Jul 27 '17 at 13:54