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Difference between revisions of "2007 AMC 12B Problems/Problem 18"
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<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math>
 
<math>a+b+c = 16 \Rightarrow \mathrm{(C)}</math>
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==Solution==
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One-third the distance from <math>x^2</math> to <math>(x+1)^2</math> is <math>\frac{2x^2 + (x+1)^2}{3} = \frac{3x^2+2x+1}{3}</math>.
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Since this must be an integer, <math>3x^2+2x+1</math> is divisible by <math>3</math>. Since <math>3x^2</math> is always divisible by <math>3</math>, <math>2x+1</math> must be divisible by <math>3</math>.
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Therefore, x must be <math>10, 13, 16, 19, 22, 25, </math> or <math>28</math>. (1, 4, and 7 don't work because their squares are too small)
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Guessing and checking, we find that <math>x=13</math> works, so the integer <math>abc</math> is one-third of the way from <math>169</math> to <math>196</math>, which is <math>178</math>. <math>1+7+8 = 16.</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2007|ab=B|num-b=17|num-a=19}}
 
{{AMC12 box|year=2007|ab=B|num-b=17|num-a=19}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:22, 16 June 2020

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

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

Solution

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 $178$. $1+7+8 = 16.$

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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