2007 AMC 12B Problem 18

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

Let the two squares be

N^2

and

(N+1)^2;

their difference is

2N+1

Subtracting the two positions gives

\overline{acb}-\overline{abc}=9(c-b)=\tfrac{2N+1}{3}

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.

Problem 18 in other years