Question

Let $a,$ $b,$ and $c$ be digits with $a\ne0.$ The three-digit integer $\overline{abc}$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $\overline{acb}$ lies two thirds of the way between the same two squares. What is $a+b+c?$

$10$

$13$

$16$

$18$

$21$

Answer: C

Concepts:perfect square place value divisibility

Difficulty rating: 1930

Solution:

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.

2007 AMC 12B Problem 18

Answer: C

Difficulty rating: 1930

Solution:

Problem 18 in Other Years