A Not-so 2018 Problem

$\overline{abc}=100a+10b+c$

$100a+10b+c=2a^3+b^2+8c$

$(100a-2a^3)+(10b-b^2) =7c$

$2a(50-a^2)+b(10-b)=7c$

It is obvious that $a, b, c$ are positive integers with $1\leq a \leq9$ and $0 \leq b, c, \leq 9$

We consider the endpoints at $a=1$ and $a=9$

When $a=1$ ,

$2a(50-a^2)=98$

$b(10-b)\geq0$

$7c\geq0+98$

$c\geq14$ , which is impossible.

When $a=9$ ,

$2a(50-a^2)=-558<0$

Note that $b(10-b)=-(b-5)^2+25\leq25$

$2a(50-a^2)+b(10-b)<0$

$7c<0$ ,which is impossible.

The same argument holds for $a=8$ .

When $a=7$ ,

$2a(50-a^2)=14$

$\Rightarrow 14+b(10-b)=7c$

It is obvious that $b(10-b)$ is divisible by 7

If so, $b=0$ or $b=3$ or $b=7$

When $b=0$ , $c=2$

When $b=3$ , $c=5$

When $b=7$ , $c=5$

When $a=6$ ,

$2a(50-a^2)=168$

$2a(50-a^2) +b(10-b)\geq 168+0$

$7c\geq168$

$c \geq 24$ , which is impossible.

The same argument holds for $a=2,3,4,5$

We conclude that there are only $3$ possible sets of solutions.

The $3$ three-digit integers are $702,735,775$ .

Their sum $=702+735+775=2212$