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

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==Problem 18==
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==Problem==
 
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 of a positive integer to the square of the next larger integer. The integer <math>acb</math> lies two thirds of the way between the same two squares. What is <math>a+b+c</math>?
 
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 of a positive integer to the square of the next larger integer. The integer <math>acb</math> lies two thirds of the way between the same two squares. What is <math>a+b+c</math>?
  
 
<math>\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21</math>
 
<math>\mathrm{(A)}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21</math>
  
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==Solution==
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==Solution 1==
 
The difference between <math>acb</math> and <math>abc</math> is given by
 
The difference between <math>acb</math> and <math>abc</math> is given by
  
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==Solution 2==
 
==Solution 2==
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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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One-third the distance from <math>x^2</math> to <math>(x+1)^2</math> is <math>\frac{(x+1)^2 - x^2}{3} = \frac{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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Since <math>\frac{2x+1}{3}</math> must be an integer, and therefore <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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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. Similarly if x is greater than 28, the squares are too large.)
  
 
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>
 
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>
  
 
- JN5537
 
- JN5537
 +
- edited by numerophile
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 +
==Solution 3==
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Let <math>k</math> be the lesser of the two integers. Then the squares of the integers are <math>k^2</math> and <math>k^2+2k+1</math>, and the distance between them is <math>2k+1</math>. Let this be equivalent to <math>3d</math>, so that the one-third of the distance between the squares is equivalent to <math>d</math>. The numbers <math>abc</math> and <math>acb</math> are one-third and two-thirds of the way between <math>k^2</math> and <math>(k+1)^2</math>. Therefore, the distance between these two numbers is also one-third the distance between the squares, or <math>d</math>. Setting these equal to each other, we have
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<math>\frac{2k+1}{3} = 9(c-b)</math>
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<math>\Rightarrow 2k+1 = 27(c-b)</math>.
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Notice that since <math>c</math> and <math>b</math> are digits, their difference is at most <math>9</math> and at least <math>0</math>. Also notice that since <math>acb</math> is greater than <math>abc</math>, <math>c > b</math>. Representing this as an inequality, we have
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<math>27 \le 27(c-b) \le 243</math>.
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Substituting <math>2k+1</math>, we have
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<math>27 \le 2k+1 \le 243</math>
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<math>\Rightarrow 13 \le k \le 121</math>.
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However, we know that <math>abc</math> is a <math>3</math>-digit number, and since <math>k^2</math> is less than <math>abc</math>, <math>k^2</math> must be at most <math>961</math>, or <math>31^2</math>. Therefore <math>k \le 31</math>. Plugging this back into our inequality, we have
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<math>13 \le k \le 31</math>
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<math>\Rightarrow 27 \le 2k+1 \le63</math>
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<math>\Rightarrow 27 \le 27(c-b) \le 63</math>
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<math>\Rightarrow 1 \le (c-b) \le \frac{7}{3}</math>.
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But (c-b) must be an integer, so now we have
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<math>1 \le (c-b) \le 2</math>
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<math>\Rightarrow 27 \le 27(c-b) \le 54</math>
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<math>\Rightarrow 27 \le 2k+1 \le 54</math>
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<math>\Rightarrow 13 \le k \le\frac{53}{2}</math>
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<math>k</math> is also an integer, so now we have
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<math>\Rightarrow 13 \le k \le 26</math>
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<math>\Rightarrow 27 \le 2k+1 \le 53</math>
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<math>\Rightarrow 27 \le 27(c-b) \le 53</math>
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<math>\Rightarrow 1 \le (c-b) \le \frac{53}{27}</math>.
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Once again, <math>(c-b)</math> must be an integer, so we have
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<math>1 \le (c-b) \le 1</math>
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<math>\Rightarrow (c-b) = 1</math>
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<math>\Rightarrow 27(c-b) = 27</math>
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<math>\Rightarrow 2k+1 = 27</math>
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<math>\Rightarrow k = 13</math>
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The two squares are <math>13^2</math> and <math>14^2</math>, or <math>169</math> and <math>196</math>. A third of the distance between them is <math>9</math>, and <math>169 +9 = 178</math>. <math>1 + 7 + 8 = 16 \Rightarrow \boxed{\text{C}}</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}}

Latest revision as of 01:38, 2 November 2024

Problem

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 $\frac{(x+1)^2 - x^2}{3} = \frac{2x+1}{3}$.

Since $\frac{2x+1}{3}$ must be an integer, and therefore $2x+1$ 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. Similarly if x is greater than 28, the squares are too large.)

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 - edited by numerophile

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 (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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