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2007 AMC 12B Problems/Problem 18

Revision as of 22:52, 20 February 2008 by Chickendude (talk | contribs) (New page: ==Problem 18== Let <math>a</math>, <math>b</math>, and <math>c</math> be digits with <math>a\ne 0</math>. The three-digit integer <math>abc</math> lies one third of the way from the square...)
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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 $a+b+c$?

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

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

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

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