2007 AMC 12B Problems/Problem 18

Problem 18

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 $169 +9 = 178$. $1 + 7 + 8 = 16 \Rightarrow \boxed{\text{C}}$.

See Also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions

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