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Difference between revisions of "2012 AMC 10A Problems/Problem 17"

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==Problem==
 
==Problem==
  
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Let <math>a</math> and <math>b</math> be relatively prime integers with <math>a>b>0</math> and <math>\frac{a^3-b^3}{(a-b)^3}</math> = <math>\frac{73}{3}</math>. What is <math>a-b</math>?
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Let <math>a</math> and <math>b</math> be relatively prime positive integers with <math>a>b>0</math> and <math>\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}</math>. What is <math>a-b</math>?
  
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math>
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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5</math>
  
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== Solution 1 (Meta) ==
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==Solution 1 (Quick Insight)==
  
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Since <math>a</math> and <math>b</math> are both integers, <math>a^3-b^3</math> and <math>(a-b)^3</math> are both integers as well. Then, for the given fraction to simplify to <math>\frac{73}{3}</math>, the denominator <math>(a-b)^3</math> must be a multiple of <math>3.</math> Thus, <math>a-b</math> is a multiple of <math>3</math>. Looking at the answer choices, the only multiple of <math>3</math> is <math>\boxed{\textbf{(C)}\ 3}</math>.
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Since <math>a</math> and <math>b</math> are relatively prime, <math>a^3-b^3</math> and <math>(a-b)^3</math> are both integers as well. Then, for the given fraction to simplify to <math>\frac{73}{3}</math>, the denominator <math>(a-b)^3</math> must be a multiple of <math>3.</math> Thus, <math>a-b</math> is a multiple of <math>3</math>. Looking at the answer choices, the only multiple of <math>3</math> is <math>\boxed{\textbf{(C)}\ 3}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==
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Therefore <math>a=10</math> and <math>b=7</math> is a solution.
 
Therefore <math>a=10</math> and <math>b=7</math> is a solution.
 
So <math>a-b=\boxed{\textbf{(C)}\ 3}</math>
 
So <math>a-b=\boxed{\textbf{(C)}\ 3}</math>
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'''Note:'''
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From <math>9ab=70(a-b)^2</math>, the Euclidean Algorithm gives <math>\gcd(a-b,a)=\gcd(a-b,b)=1</math>. Thus <math>(a-b)^2</math> is relatively prime to <math>ab</math>, and clearly <math>9</math> and <math>70</math> are coprime as well. The solution must therefore be <math>(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}</math> and <math>ab=70</math>.
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== Solution 4 ==
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Slightly expanding, we have that
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<math>\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}</math>.
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Canceling the <math>(a-b)</math>, cross multiplying, and simplifying, we obtain that
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<math>0=70a^2-149ab+70b^2</math>.
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Dividing everything by <math>b^2</math>, we get that
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<math>0=70\left(\frac{a}{b}\right)^2-149\left(\frac{a}{b} \right)+70</math>.
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Applying the quadratic formula....and following the restriction that <math>a>b>0</math>....
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<math>\frac{a}{b}=\frac{10}{7}</math>.
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Hence, <math>7a=10b</math>.
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Since they are relatively prime, <math>a=10</math>, <math>b=7</math>.
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<math>10-7=\boxed{\textbf{(C)}\ 3}</math>.
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==Solution 5==
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Note that the denominator, when simplified, gets <math>3.</math> We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly <math>\boxed{\textbf{(C)}\ 3}</math> ~mathboy282
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==Solution 6==
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Let us rewrite the expression as <math>\frac{(a-b)^2 + 3ab}{(a-b)^2}</math>. Now letting <math>x = a - b</math>, we simplify the expression to <math>\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}</math>. Cross multiplying and doing a bit of simplification, we obtain that <math>ab = \frac{70x^2}{9}</math>. Since <math>a</math> and <math>b</math> are both integers, we know that <math>\frac{70x^2}{9}</math> has to be an integer. Experimenting with values of <math>x</math>, we get that <math>x = 3</math> which means <math>ab = 70</math>. We could prime factor from here to figure out possible values of <math>a</math> and <math>b</math>, but it is quite obvious that <math>a = 10</math> and <math>b=7</math>, so our desired answer is  <math>\boxed{\textbf{(C)}\ 3}</math> ~triggod
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==Solution 7==
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Since the two numbers are integers, and both <math>(a-b)^3</math> and <math>a^3-b^3</math> would yield integers, for the denominator to have a factor of 3, <math>(a-b)</math> must have a factor of 3. Only choice <math>\boxed{\textbf{(C)}}</math> has a factor of 3. ~hh99754539
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Note: isn't this solution (7) identical to solution #1??
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==Solution 8==
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We expand difference of cubes and cancel <math>a-b</math> from the numerator and denominator and see that <math>\dfrac{a^2-ab+b^2}{(a-b)^2}=\dfrac{73}3.</math> Obviously, we can not equate the numerator and denominator quite yet since that would imply that <math>a-b</math> is irrational (<math>\sqrt3</math>). We try the easiest thing to make <math>\dfrac{73}3</math>'s denominator a square: simply multiply by <math>\dfrac33,</math> giving <math>\dfrac{219}9.</math> Setting the denominators to be equal, we see that <math>(a-b)^2=9\implies a-b=3.</math>
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~ Technodoggo
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== Video Solution by OmegaLearn==
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https://youtu.be/ZWqHxc0i7ro?t=417
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~ pi_is_3.14
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==Video Solution==
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https://youtu.be/8SXVrlH71jk
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~savannahsolver
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:24, 14 June 2024

Problem

Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$. What is $a-b$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$

Solution 1 (Quick Insight)

Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$, the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$. Looking at the answer choices, the only multiple of $3$ is $\boxed{\textbf{(C)}\ 3}$.

Solution 2

Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}$.

Set $x = a^2 + b^2$, and $y = ab$. Then $\frac{x + y}{x - 2y} = \frac{73}{3}$. Cross multiplying gives $3x + 3y = 73x - 146y$, and simplifying gives $\frac{x}{y} = \frac{149}{70}$. Since $149$ and $70$ are relatively prime, we let $x = 149$ and $y = 70$, giving $a^2 + b^2 = 149$ and $ab = 70$. Since $a>b>0$, the only solution is $(a,b) = (10, 7)$, which can be seen upon squaring and summing the various factor pairs of $70$.

Thus, $a - b = \boxed{\textbf{(C)}\ 3}$.

Remarks:

An alternate method of solving the system of equations involves solving the second equation for $a$, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of $u = b^2$. The four solutions correspond to $(\pm10, \pm7), (\pm7, \pm10).$

Also, we can solve for $a-b$ directly instead of solving for $a$ and $b$: $a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.$

Note that if you double $x$ and double $y$, you will get different (but not relatively prime) values for $a$ and $b$ that satisfy the original equation.

Solution 3

The first step is the same as above which gives $\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$.

Then we can subtract $3ab$ and then add $3ab$ to get $\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}$, which gives $1+\frac{3ab}{(a-b)^2}=\frac{73}{3}$. $\frac{3ab}{(a-b)^2}=\frac{70}{3}$. Cross multiply $9ab=70(a-b)^2$. Since $a>b$, take the square root. $a-b=3\sqrt{\frac{ab}{70}}$. Since $a$ and $b$ are integers and relatively prime, $\sqrt{\frac{ab}{70}}$ is an integer. $ab$ is a multiple of $70$, so $a-b$ is a multiple of $3$. Therefore $a=10$ and $b=7$ is a solution. So $a-b=\boxed{\textbf{(C)}\ 3}$

Note:

From $9ab=70(a-b)^2$, the Euclidean Algorithm gives $\gcd(a-b,a)=\gcd(a-b,b)=1$. Thus $(a-b)^2$ is relatively prime to $ab$, and clearly $9$ and $70$ are coprime as well. The solution must therefore be $(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}$ and $ab=70$.

Solution 4

Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$.

Canceling the $(a-b)$, cross multiplying, and simplifying, we obtain that

$0=70a^2-149ab+70b^2$. Dividing everything by $b^2$, we get that

$0=70\left(\frac{a}{b}\right)^2-149\left(\frac{a}{b} \right)+70$.

Applying the quadratic formula....and following the restriction that $a>b>0$....

$\frac{a}{b}=\frac{10}{7}$.

Hence, $7a=10b$.

Since they are relatively prime, $a=10$, $b=7$.

$10-7=\boxed{\textbf{(C)}\ 3}$.

Solution 5

Note that the denominator, when simplified, gets $3.$ We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly $\boxed{\textbf{(C)}\ 3}$ ~mathboy282


Solution 6

Let us rewrite the expression as $\frac{(a-b)^2 + 3ab}{(a-b)^2}$. Now letting $x = a - b$, we simplify the expression to $\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}$. Cross multiplying and doing a bit of simplification, we obtain that $ab = \frac{70x^2}{9}$. Since $a$ and $b$ are both integers, we know that $\frac{70x^2}{9}$ has to be an integer. Experimenting with values of $x$, we get that $x = 3$ which means $ab = 70$. We could prime factor from here to figure out possible values of $a$ and $b$, but it is quite obvious that $a = 10$ and $b=7$, so our desired answer is $\boxed{\textbf{(C)}\ 3}$ ~triggod

Solution 7

Since the two numbers are integers, and both $(a-b)^3$ and $a^3-b^3$ would yield integers, for the denominator to have a factor of 3, $(a-b)$ must have a factor of 3. Only choice $\boxed{\textbf{(C)}}$ has a factor of 3. ~hh99754539

Note: isn't this solution (7) identical to solution #1??

Solution 8

We expand difference of cubes and cancel $a-b$ from the numerator and denominator and see that $\dfrac{a^2-ab+b^2}{(a-b)^2}=\dfrac{73}3.$ Obviously, we can not equate the numerator and denominator quite yet since that would imply that $a-b$ is irrational ($\sqrt3$). We try the easiest thing to make $\dfrac{73}3$'s denominator a square: simply multiply by $\dfrac33,$ giving $\dfrac{219}9.$ Setting the denominators to be equal, we see that $(a-b)^2=9\implies a-b=3.$

~ Technodoggo

Video Solution by OmegaLearn

https://youtu.be/ZWqHxc0i7ro?t=417

~ pi_is_3.14

Video Solution

https://youtu.be/8SXVrlH71jk

~savannahsolver

See Also

2012 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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All AMC 10 Problems and Solutions

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