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

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

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

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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2007 AMC 12B Problems/Problem 18

Problem 18

Solution

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