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Difference between revisions of "2007 AMC 12B Problems/Problem 18"

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<math>27(c-b)</math>
 
<math>27(c-b)</math>
  
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The difference between two consecutive squares is always an odd number. The consecutive squares with common difference <math>27</math> are <math>13^2=169</math> and <math>14^2=196</math>. One third of the way between them is <math>178</math> and two thirds of the way is <math>187</math>
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The difference between two consecutive squares is always an odd number, therefore <math>c-b</math> is odd. We will show that <math>c-b</math> must be 1. Otherwise we would be looking for two consecutive squares that are at least 81 apart. But already the equation <math>(x+1)^2-x^2 = 27\cdot 3</math> solves to <math>x=40</math>, and <math>40^2</math> has more than three digits.
  
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This gives <math>a=1</math>, <math>b=7</math>, <math>c=8</math>
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The consecutive squares with common difference <math>27</math> are <math>13^2=169</math> and <math>14^2=196</math>. One third of the way between them is <math>178</math> and two thirds of the way is <math>187</math>.
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This gives <math>a=1</math>, <math>b=7</math>, <math>c=8</math>.
  
 
<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math>
 
<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math>

Revision as of 14:51, 5 January 2009

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

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

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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