https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Triggod&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-16T09:02:48Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142863 2012 AMC 10A Problems/Problem 17 2021-01-20T14:33:08Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> <br /> ==Solution 6==<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab}{(a-b)^2}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible values of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~triggod<br /> <br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142862 2012 AMC 10A Problems/Problem 17 2021-01-20T14:30:10Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> <br /> ==Solution 6==<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab}{(a-b)^2}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~triggod<br /> <br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142861 2012 AMC 10A Problems/Problem 17 2021-01-20T14:29:04Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> <br /> ==Solution 6==<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab)}{a-b}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~triggod<br /> <br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142856 2012 AMC 10A Problems/Problem 17 2021-01-20T08:55:24Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> <br /> ==Solution 6==<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab)}{a-b}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~triggod<br /> <br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142855 2012 AMC 10A Problems/Problem 17 2021-01-20T08:54:58Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> <br /> ==Solution 6==<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab)}{a-b}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142854 2012 AMC 10A Problems/Problem 17 2021-01-20T08:53:39Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> <br /> ==Solution 6==<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab)}{a-b}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;3&lt;/math&gt;<br /> ~triggod<br /> <br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142853 2012 AMC 10A Problems/Problem 17 2021-01-20T08:52:36Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> Solution 6<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab)}{a-b}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;3&lt;/math&gt;<br /> ~triggod<br /> <br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod https://artofproblemsolving.com/wiki/index.php?title=2012_AMC_10A_Problems/Problem_17&diff=142852 2012 AMC 10A Problems/Problem 17 2021-01-20T08:50:55Z <p>Triggod: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; be relatively prime positive integers with &lt;math&gt;a&gt;b&gt;0&lt;/math&gt; and &lt;math&gt;\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}.&lt;/math&gt; What is &lt;math&gt;a-b?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are relatively prime, &lt;math&gt;a^3-b^3&lt;/math&gt; and &lt;math&gt;(a-b)^3&lt;/math&gt; are both integers as well. Then, for the given fraction to simplify to &lt;math&gt;\frac{73}{3}&lt;/math&gt;, the denominator &lt;math&gt;(a-b)^3&lt;/math&gt; must be a multiple of &lt;math&gt;3.&lt;/math&gt; Thus, &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;. Looking at the answer choices, the only multiple of &lt;math&gt;3&lt;/math&gt; is &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Using difference of cubes in the numerator and cancelling out one &lt;math&gt;(a-b)&lt;/math&gt; in the numerator and denominator gives &lt;math&gt;\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}&lt;/math&gt;.<br /> <br /> Set &lt;math&gt;x = a^2 + b^2&lt;/math&gt;, and &lt;math&gt;y = ab&lt;/math&gt;. Then &lt;math&gt;\frac{x + y}{x - 2y} = \frac{73}{3}&lt;/math&gt;. Cross multiplying gives &lt;math&gt;3x + 3y = 73x - 146y&lt;/math&gt;, and simplifying gives &lt;math&gt;\frac{x}{y} = \frac{149}{70}&lt;/math&gt;. Since &lt;math&gt;149&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are relatively prime, we let &lt;math&gt;x = 149&lt;/math&gt; and &lt;math&gt;y = 70&lt;/math&gt;, giving &lt;math&gt;a^2 + b^2 = 149&lt;/math&gt; and &lt;math&gt;ab = 70&lt;/math&gt;. Since &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;, the only solution is &lt;math&gt;(a,b) = (10, 7)&lt;/math&gt;, which can be seen upon squaring and summing the various factor pairs of &lt;math&gt;70&lt;/math&gt;.<br /> <br /> Thus, &lt;math&gt;a - b = \boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> '''Remarks:'''<br /> <br /> An alternate method of solving the system of equations involves solving the second equation for &lt;math&gt;a&lt;/math&gt;, by plugging it into the first equation, and solving the resulting quartic equation with a substitution of &lt;math&gt;u = b^2&lt;/math&gt;. The four solutions correspond to &lt;math&gt;(\pm10, \pm7), (\pm7, \pm10).&lt;/math&gt; <br /> <br /> Also, we can solve for &lt;math&gt;a-b&lt;/math&gt; directly instead of solving for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;a^2-2ab+b^2=149-2(70)=9 \implies a-b=3.&lt;/math&gt;<br /> <br /> Note that if you double &lt;math&gt;x&lt;/math&gt; and double &lt;math&gt;y&lt;/math&gt;, you will get different (but not relatively prime) values for &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; that satisfy the original equation.<br /> <br /> == Solution 3 ==<br /> <br /> The first step is the same as above which gives &lt;math&gt;\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Then we can subtract &lt;math&gt;3ab&lt;/math&gt; and then add &lt;math&gt;3ab&lt;/math&gt; to get &lt;math&gt;\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}&lt;/math&gt;, which gives &lt;math&gt;1+\frac{3ab}{(a-b)^2}=\frac{73}{3}&lt;/math&gt;. &lt;math&gt;\frac{3ab}{(a-b)^2}=\frac{70}{3}&lt;/math&gt;.<br /> Cross multiply &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;. Since &lt;math&gt;a&gt;b&lt;/math&gt;, take the square root. &lt;math&gt;a-b=3\sqrt{\frac{ab}{70}}&lt;/math&gt;.<br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and relatively prime, &lt;math&gt;\sqrt{\frac{ab}{70}}&lt;/math&gt; is an integer. &lt;math&gt;ab&lt;/math&gt; is a multiple of &lt;math&gt;70&lt;/math&gt;, so &lt;math&gt;a-b&lt;/math&gt; is a multiple of &lt;math&gt;3&lt;/math&gt;.<br /> Therefore &lt;math&gt;a=10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt; is a solution.<br /> So &lt;math&gt;a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;<br /> <br /> '''Note:'''<br /> <br /> From &lt;math&gt;9ab=70(a-b)^2&lt;/math&gt;, the Euclidean Algorithm gives &lt;math&gt;\gcd(a-b,a)=\gcd(a-b,b)=1&lt;/math&gt;. Thus &lt;math&gt;(a-b)^2&lt;/math&gt; is relatively prime to &lt;math&gt;ab&lt;/math&gt;, and clearly &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;70&lt;/math&gt; are coprime as well. The solution must therefore be &lt;math&gt;(a-b)^2=9 \rightarrow a-b=\boxed{\textbf{(C)}\ 3}&lt;/math&gt; and &lt;math&gt;ab=70&lt;/math&gt;.<br /> <br /> == Solution 4 ==<br /> <br /> Slightly expanding, we have that <br /> &lt;math&gt;\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}&lt;/math&gt;.<br /> <br /> Canceling the &lt;math&gt;(a-b)&lt;/math&gt;, cross multiplying, and simplifying, we obtain that<br /> <br /> &lt;math&gt;0=70a^2-149ab+70b^2&lt;/math&gt;.<br /> Dividing everything by &lt;math&gt;b^2&lt;/math&gt;, we get that<br /> <br /> &lt;math&gt;0=70(\frac{a}{b})^2-149(\frac{a}{b})+70&lt;/math&gt;.<br /> <br /> Applying the quadratic formula....and following the restriction that &lt;math&gt;a&gt;b&gt;0&lt;/math&gt;....<br /> <br /> &lt;math&gt;\frac{a}{b}=\frac{10}{7}&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;7a=10b&lt;/math&gt;.<br /> <br /> Since they are relatively prime, &lt;math&gt;a=10&lt;/math&gt;, &lt;math&gt;b=7&lt;/math&gt;.<br /> <br /> &lt;math&gt;10-7=\boxed{\textbf{(C)}\ 3}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> Note that the denominator, when simplified, gets &lt;math&gt;3.&lt;/math&gt; 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 &lt;math&gt;\boxed{\textbf{(C)}\ 3}&lt;/math&gt; ~mathboy282<br /> <br /> Solution 6<br /> Let us rewrite the expression as &lt;math&gt;\frac{(a-b)^2 + 3ab)}{a-b}&lt;/math&gt;. Now letting &lt;math&gt;x = a - b&lt;/math&gt;, we simplify the expression to &lt;math&gt;\frac{70x^2 + 3ab}{x} = \frac{73}{3}&lt;/math&gt;. Cross multiplying and doing a bit of simplification, we obtain that &lt;math&gt;ab = \frac{70x^2}{9}&lt;/math&gt;. Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are both integers, we know that &lt;math&gt;\frac{70x^2}{9}&lt;/math&gt; has to be an integer. Experimenting with values of &lt;math&gt;x&lt;/math&gt;, we get that &lt;math&gt;x = 3&lt;/math&gt; which means &lt;math&gt;ab = 70&lt;/math&gt;. We could prime factor from here to figure out possible vlaues of &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, but it is quite obvious that &lt;math&gt;a = 10&lt;/math&gt; and &lt;math&gt;b=7&lt;/math&gt;, so our desired answer is &lt;math&gt;3&lt;/math&gt;<br /> ~triggod<br /> <br /> == Video Solution ==<br /> https://youtu.be/ZWqHxc0i7ro?t=417<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Video Solution==<br /> https://youtu.be/8SXVrlH71jk<br /> <br /> ~savannahsolver<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}<br /> {{MAA Notice}}</div> Triggod