2007 AMC 12B Problem 18

Let the smaller square be $N^2,$ so the larger is $(N+1)^2$ and the gap is $2N+1.$ Then $$\overline{abc}=N^2+\dfrac{2N+1}{3},$$ $$\overline{acb}=N^2+\dfrac{2(2N+1)}{3}.$$ Subtracting, $\overline{acb}-\overline{abc}=9(c-b)$ $=\dfrac{2N+1}{3},$ so $27(c-b)=2N+1.$ Since $\overline{acb}$ is farther along the interval, $c-b$ is positive; and because the right side is odd, $c-b$ is odd. If $c-b\ge3,$ then $N\ge40$ and $N^2$ is not three digits. So $c-b=1,$ giving $N=13.$ The points one third and two thirds of the way from $13^2=169$ to $14^2=196$ are $178$ and $187,$ so $a+b+c=1+7+8=16.$ Thus, the correct answer is C.