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

(→‎Solution 1(The not so good solution))
Line 35: Line 35:
 
So <math>a-b=\boxed{\textbf{(C)}\ 3}</math>
 
So <math>a-b=\boxed{\textbf{(C)}\ 3}</math>
  
 +
== Solution 4 ==
 +
 +
Slightly expanding, we have that
 +
<math>\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}</math>.
 +
Canceling the <math>(a-b)</math>, cross multiplying, and simplifying, we obtain that
 +
<math>0=70a^2-149ab+70b^2</math>.
 +
Dividing everything by <math>b^2</math>, we get that
 +
<math>0=70(\frac{a}{b})^2-149(\frac{a}{b})+70</math>.
 +
Applying the quadratic formula....and following the restriction that <math>a>b>0</math>, we get that
 +
<math>\frac{a}{b}=\frac{10}{7}</math>.
 +
Hence, <math>7a=10b</math>.
 +
Since they are relatively prime, <math>a=10</math>, <math>b=7</math>.
 +
<math>10-7=\boxed{\textbf{(C)}\ 3}</math>.
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:57, 4 February 2018

Problem

Let $a$ and $b$ be relatively prime integers with $a>b>0$ and $\frac{a^3-b^3}{(a-b)^3}$ = $\frac{73}{3}$. 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 (The not so good solution)

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

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 $0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$. Applying the quadratic formula....and following the restriction that $a>b>0$, we get that $\frac{a}{b}=\frac{10}{7}$. Hence, $7a=10b$. Since they are relatively prime, $a=10$, $b=7$. $10-7=\boxed{\textbf{(C)}\ 3}$.

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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