2007 AMC 12B Problems/Problem 18

Problem

Let $a$ , $b$ , and $c$ be digits with $a\ne 0$ . The three-digit integer $abc$ lies one third of the way from the square of a positive integer to the square of the next larger integer. The integer $acb$ lies two thirds of the way between the same two squares. What is $a+b+c$ ?

$\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21$

Solution 1

The difference between $acb$ and $abc$ is given by

$(100a + 10c + b) - (100a + 10b + c) = 9(c-b)$

The difference between the two squares is three times this amount or

$27(c-b)$

The difference between two consecutive squares is always an odd number, therefore $c-b$ is odd. We will show that $c-b$ must be 1. Otherwise we would be looking for two consecutive squares that are at least 81 apart. But already the equation $(x+1)^2-x^2 = 27\cdot 3$ solves to $x=40$ , and $40^2$ has more than three digits.

The consecutive squares with common difference $27$ are $13^2=169$ and $14^2=196$ . One third of the way between them is $178$ and two thirds of the way is $187$ .

This gives $a=1$ , $b=7$ , $c=8$ .

$a+b+c = 16 \Rightarrow \mathrm{(C)}$

Solution 2

One-third the distance from $x^2$ to $(x+1)^2$ is $\frac{2x^2 + (x+1)^2}{3} = \frac{3x^2+2x+1}{3}$ .

Since this must be an integer, $3x^2+2x+1$ is divisible by $3$ . Since $3x^2$ is always divisible by $3$ , $2x+1$ must be divisible by $3$ .

Therefore, x must be $10, 13, 16, 19, 22, 25,$ or $28$ . (1, 4, and 7 don't work because their squares are too small)

Guessing and checking, we find that $x=13$ works, so the integer $abc$ is one-third of the way from $169$ to $196$ , which is $178$ . $1+7+8 = 16.$

- JN5537

Solution 3

Let $k$ be the lesser of the two integers. Then the squares of the integers are $k^2$ and $k^2+2k+1$ , and the distance between them is $2k+1$ . Let this be equivalent to $3d$ , so that the one-third of the distance between the squares is equivalent to $d$ . The numbers $abc$ and $acb$ are one-third and two-thirds of the way between $k^2$ and $(k+1)^2$ . Therefore, the distance between these two numbers is also one-third the distance between the squares, or $d$ . Setting these equal to each other, we have

$\frac{2k+1}{3} = 9(c-b)$

$\Rightarrow 2k+1 = 27(c-b)$ .

Notice that since $c$ and $b$ are digits, their difference is at most $9$ and at least $0$ . Also notice that since $acb$ is greater than $abc$ , $c > b$ . Representing this as an inequality, we have

$27 \le 27(c-b) \le 243$ .

Substituting $2k+1$ , we have

$27 \le 2k+1 \le 243$

$\Rightarrow 13 \le k \le 121$ .

However, we know that $abc$ is a $3$ -digit number, and since $k^2$ is less than $abc$ , $k^2$ must be at most $961$ , or $31^2$ . Therefore $k \le 31$ . Plugging this back into our inequality, we have

$13 \le k \le 31$

$\Rightarrow 27 \le 2k+1 \le63$

$\Rightarrow 27 \le 27(c-b) \le 63$

$\Rightarrow 1 \le (c-b) \le \frac{7}{3}$ .

But (c-b) must be an integer, so now we have

$1 \le (c-b) \le 2$

$\Rightarrow 27 \le 27(c-b) \le 54$

$\Rightarrow 27 \le 2k+1 \le 54$

$\Rightarrow 13 \le k \le\frac{53}{2}$

$k$ is also an integer, so now we have

$\Rightarrow 13 \le k \le 26$

$\Rightarrow 27 \le 2k+1 \le 53$

$\Rightarrow 27 \le 27(c-b) \le 53$

$\Rightarrow 1 \le (c-b) \le \frac{53}{27}$ .

Once again, $(c-b)$ must be an integer, so we have

$1 \le (c-b) \le 1$

$\Rightarrow (c-b) = 1$

$\Rightarrow 27(c-b) = 27$

$\Rightarrow 2k+1 = 27$

$\Rightarrow k = 13$

The two squares are $13^2$ and $14^2$ , or $169$ and $196$ . A third of the distance between them is $9$ , and $169 +9 = 178$ . $1 + 7 + 8 = 16 \Rightarrow \boxed{\text{C}}$ .

2007 AMC 12B (Problems • Answer Key • Resources)
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2007 AMC 12B Problems/Problem 18

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also