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.


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.

  • $\begingroup$ 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. $\endgroup$ – rugi Jul 27 '17 at 6:02
  • $\begingroup$ The commonality of $b=9$ in the other cases makes it easy to check. I had to go to bed last night. $\endgroup$ – Ross Millikan Jul 27 '17 at 13:54

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