https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=AlphaMath1&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T00:16:16ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=376982011 AIME I Problems/Problem 12011-03-20T21:06:47Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem 1 ==<br />
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is <math>k\%</math> acid. From jar C, <math>\frac{m}{n}</math> liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.<br />
<br />
== Solution 1==<br />
There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transfering the solutions from jar C, there will be<br />
<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br><br />
<br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right</math> of acid in Jar B.<br />
<br>Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.<br><br />
<cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath><br />
<cmath>\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}</cmath><br />
Add the equations to get<br />
<cmath>\frac{42}{5}+\frac{k}{50}=10</cmath> Solving gives <math>k=80</math>.<br />
<br>If we substitute back in the original equation we get <math>\frac{m}{n}=\frac{2}{3}</math> so <math>3m=2n</math>. Since <math>m</math> and <math>n</math> are relatively prime, <math>m=2</math> and <math>n=3</math>. Thus <math>k+m+n=80+2+3=\boxed{085}</math>.<br />
<br />
== Solution 2 ==<br />
One might cleverly change the content of both Jars. <br />
<br />
Since the end result of both Jars are <math>50\%</math> acid, we can turn Jar A into a 1 gallon liquid with <math>50\%-4(5\%) = 30\%</math> acid <br />
<br />
and Jar B into 1 gallon liquid with <math>50\%-5(2\%) =40\%</math> acid.<br />
<br />
Now, since Jar A and Jar B contain the same amount of liquid, twice as much liquid will be pour into Jar A than Jar B, so <math>\dfrac{2}{3}</math> of Jar C will be pour into Jar A.<br />
<br />
Thus, <math>m=2</math> and <math>n=3</math>.<br />
<br />
<math>\dfrac{30\% + \frac{2}{3} \cdot k\%}{\frac{5}{3}} = 50\%</math><br />
<br />
Solving for <math>k</math> yields <math>k=80</math><br />
<br />
So the answer is <math>80+2+3 = \boxed{85}</math></div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=376672011 AIME I Problems/Problem 12011-03-19T18:15:38Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem 1 ==<br />
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is <math>k\%</math> acid. From jar C, <math>\frac{m}{n}</math> liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.<br />
<br />
== Solution ==<br />
There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transfering the solutions from jar C, there will be<br />
<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br><br />
<br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)=\frac{12}{5}+\frac{k}{100}-\frac{mk}{100n}\right</math> of acid in Jar B.<br />
<br>Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.<br><br />
<cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath><br />
<cmath>\frac{24}{5}-\frac{km}{50n}+\frac{k}{50}=6-\frac{m}{n}</cmath><br />
Add the equations to get<br />
<cmath>\frac{42}{5}+\frac{k}{50}=10</cmath> Solving gives <math>k=80</math>.<br />
<br>If we substitute back in the original equation we get <math>\frac{m}{n}=\frac{2}{3}</math> so <math>3m=2n</math>. Since <math>m</math> and <math>n</math> are relatively prime, <math>m=2</math> and <math>n=3</math>. Thus <math>k+m+n=80+2+3=\boxed{085}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=376662011 AIME I Problems/Problem 12011-03-19T17:59:54Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem 1 ==<br />
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C contains one liter of a solution that is <math>k\%</math> acid. From jar C, <math>\frac{m}{n}</math> liters of the solution is added to jar A, and the remainder of the solution in jar C is added to jar B. At the end both jar A and jar B contain solutions that are 50% acid. Given that <math>m</math> and <math>n</math> are relatively prime positive integers, find <math>k + m + n</math>.<br />
<br />
== Solution ==<br />
There are <math>\frac{45}{100}(4)=\frac{9}{5}</math> L of acid in Jar A. There are <math>\frac{48}{100}(5)=\frac{12}{5}</math> L of acid in Jar B. And there are <math>\frac{k}{100}</math> L of acid in Jar C. After transfering the solutions from jar C, there will be<br />
<br> <math>4+\frac{m}{n}</math> L of solution in Jar A and <math>\frac{9}{5}+\frac{k}{100}\cdot\frac{m}{n}</math> L of acid in Jar A.<br><br />
<br> <math>6-\frac{m}{n}</math> L of solution in Jar B and <math>\frac{12}{5}+\frac{k}{100}\cdot \left(1-\frac{m}{n}\right)</math> of acid in Jar B.<br />
Since the solutions are 50% acid, we can multiply the amount of acid for each jar by 2, then equate them to the amount of solution.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=376652011 AIME I Problems/Problem 112011-03-19T17:51:14Z<p>AlphaMath1: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
<br />
== Solution ==<br />
Note that the cycle of remainders of <math>2^n</math> will start after <math>2^2</math> because remainders of <math>1, 2, 4</math> will not be possible after (the numbers following will always be congruent to 0 modulo 8). Now we have to find the order. Note that <math>2^{100}\equiv 1\mod 125</math>. The order is <math>100</math> starting with remainder <math>8</math>. All that is left is find <math>S</math> in mod <math>1000</math> after some computation.<br />
<cmath>S=2^0+2^1+2^2+2^3+2^4...+2^{102}\equiv 2^{103}-1\equiv 8-1\equiv \boxed{007}\mod 1000</cmath></div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=376642011 AIME I Problems/Problem 112011-03-19T17:06:42Z<p>AlphaMath1: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
<br />
== Solution 1==<br />
Note that the cycle of remainders of <math>2^n</math> will start after <math>2^2</math> because remainders of <math>1, 2, 4</math> will not be possible after (the numbers following will always be congruent to 0 modulo 8). Now we have to find the order. Note that <math>2^{100}\equiv 1\mod 125</math>. The order is <math>100</math> starting with remainder <math>8</math>. All that is left is find <math>S</math> in mod <math>1000</math> after some computation.<br />
<cmath>S=2^0+2^1+2^2+2^3+2^4...+2^{102}\equiv 2^{103}-1\equiv 8-1\equiv \boxed{007}\mod 1000</cmath></div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=376632011 AIME I Problems/Problem 112011-03-19T17:05:15Z<p>AlphaMath1: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
<br />
== Solution 1==<br />
Note that the cycle of remainders of <math>2^n</math> will start after <math>2^2</math> because remainders of <math>1, 2, 4</math> will not be possible after (the numbers following will always be congruent to 0 modulo 8). Now we have to find the order. Note that <math>2^{100}\equiv 1\mod 125</math>. The order is <math>100</math> starting with remainder <math>8</math>. All that is left is find <math>S</math> in mod <math>1000</math> after some computation.<br />
<cmath>S=2^0+2^1+2^2+2^3+2^4...+2^102\equiv 2^103-1\equiv 8-1\equiv \boxed{007}\mod 1000</cmath></div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=376622011 AIME I Problems/Problem 112011-03-19T17:05:02Z<p>AlphaMath1: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.<br />
<br />
== Solution 1==<br />
Note that the cycle of remainders of <math>2^n</math> will start after <math>2^2</math> because remainders of <math>1, 2, 4</math> will not be possible after (the numbers following will always be congruent to 0 modulo 8). Now we have to find the order. Note that <math>2^{100}\equiv 1\mod 125</math>. The order is <math>100</math> starting with remainder <math>8</math>. All that is left is find <math>S</math> in mod <math>1000</math> after some computation.<br />
<math></math>S=2^0+2^1+2^2+2^3+2^4...+2^102\equiv 2^103-1\equiv 8-1\equiv \boxed{007}\mod 1000$.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&diff=376612011 AIME I Problems/Problem 112011-03-19T16:56:24Z<p>AlphaMath1: Created page with '== Problem == Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000<…'</p>
<hr />
<div>== Problem ==<br />
Let <math>R</math> be the set of all possible remainders when a number of the form <math>2^n</math>, <math>n</math> a nonnegative integer, is divided by <math>1000</math>. Let <math>S</math> be the sum of the elements in <math>R</math>. Find the remainder when <math>S</math> is divided by <math>1000</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376602011 AIME I Problems/Problem 92011-03-19T16:54:38Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<cmath>\sqrt{24^3\sin^3 x}=24\cos x</cmath><br />
Square both sides and divide by <math>24^2</math> to get<br />
<cmath>24\sin ^3 x=\cos ^2 x</cmath><br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math><br />
<cmath>24\sin ^3 x=1-\sin ^2 x</cmath><br />
<cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath><br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root, <math>\sin^{-1} \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <br />
<cmath>\sin ^2 x=\frac{1}{9}</cmath><br />
Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376592011 AIME I Problems/Problem 92011-03-19T16:53:52Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<cmath>\sqrt{24^3\sin^3 x}=24\cos x</cmath><br />
Square both sides and divide by <math>24^2</math> to get<br />
<cmath>24\sin ^3 x=\cos ^2 x</cmath><br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math><br />
<cmath>24\sin ^3 x=1-\sin ^2 x</cmath><br />
<cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath><br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\sin^{-1} \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <br />
<cmath>\sin ^2 x=\frac{1}{9}</cmath><br />
Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376582011 AIME I Problems/Problem 92011-03-19T16:53:14Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<cmath>\sqrt{24^3\sin^3 x}=24\cos x</cmath><br />
Square both sides and divide by <math>24^2</math> to get<br />
<cmath>24\sin ^3 x=\cos ^2 x</cmath><br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math><br />
<cmath>24\sin ^3 x=1-\sin ^2 x</cmath><br />
<cmath>24\sin ^3 x+\sin ^2 x - 1=0</cmath><br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <br />
<cmath>\sin ^2 x=\frac{1}{9}</cmath><br />
Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376572011 AIME I Problems/Problem 92011-03-19T16:51:25Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<math>\sqrt{24^3\sin^3 x}=24\cos x</math>.<br />
Square both sides and divide by <math>24^2</math> to get<br />
<math>24\sin ^3 x=\cos ^2 x</math>.<br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math>.<br />
<math>24\sin ^3 x=1-\sin ^2 x</math>.<br />
<math>24\sin ^3 x+\sin ^2 x - 1=0</math>.<br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <math>\sin ^2 x=\frac{1}{9}</math>. Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376562011 AIME I Problems/Problem 92011-03-19T16:50:56Z<p>AlphaMath1: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_{24\sin x} (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<math>\sqrt{24^3\sin^3 x}=24\cos x</math>.<br />
Square both sides and divide by <math>24^2</math> to get<br />
<math>24\sin ^3 x=\cos ^2 x</math><br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math><br />
<math>24\sin ^3 x=1-\sin ^2 x</math><br />
<math>24\sin ^3 x+\sin ^2 x - 1=0</math><br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <math>\sin ^2 x=\frac{1}{9}</math>. Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376552011 AIME I Problems/Problem 92011-03-19T16:50:21Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_(24\sin x) (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<math>\sqrt{24^3\sin^3 x}=24\cos x</math>.<br />
Square both sides and divide by <math>24^2</math> to get<br />
<math>24\sin ^3 x=\cos ^2 x</math><br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math><br />
<math>24\sin ^3 x=1-\sin ^2 x</math><br />
<math>24\sin ^3 x+\sin ^2 x - 1=0</math><br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <math>\sin ^2 x=\frac{1}{9}</math>. Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x</math> to compute our final answer. <math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}</math>.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376542011 AIME I Problems/Problem 92011-03-19T16:50:08Z<p>AlphaMath1: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_(24\sin x) (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<math>\sqrt{24^3\sin^3 x}=24\cos x</math>.<br />
Square both sides and divide by <math>24^2</math> to get<br />
<math>24\sin ^3 x=\cos ^2 x</math><br />
Rewrite <math>\cos ^2 x</math> as <math>1-\sin ^2 x</math><br />
<math>24\sin ^3 x=1-\sin ^2 x</math><br />
<math>24\sin ^3 x+\sin ^2 x - 1=0</math><br />
Testing values using the rational root theorem gives <math>\sin x=\frac{1}{3}</math> as a root. <math>\Arcsin \frac{1}{3}</math> does fall in the first quadrant so it satisfies the interval. Thus <math>\sin ^2 x=\frac{1}{9}</math>. Using the Pythagorean Identity gives us <math>\cos ^2 x=\frac{8}{9}</math>. Then we use the definition of <math>\cot ^2 x to compute our final answer. </math>24\cot ^2 x=24\frac{\cos ^2 x}{\sin ^2 x}=24\left(\frac{\frac{8}{9}}{\frac{1}{9}}\right)=24(8)=\boxed{192}$.</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376532011 AIME I Problems/Problem 92011-03-19T16:45:14Z<p>AlphaMath1: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_(24\sin x) (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>.<br />
<br />
== Solution ==<br />
We can rewrite the given expression as<br />
<math>\sqrt{24^3\sin^3 x}=24\cos x</math>.<br />
Square both sides and divide by <math>24^2</math> to get<br />
<math>24\sin ^3 x=\cos ^2 x</math></div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_9&diff=376522011 AIME I Problems/Problem 92011-03-19T16:43:27Z<p>AlphaMath1: Created page with '== Problem == Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_(24\sin x) (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math>'</p>
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<div>== Problem ==<br />
Suppose <math>x</math> is in the interval <math>[0, \pi/2]</math> and <math>\log_(24\sin x) (24\cos x)=\frac{3}{2}</math>. Find <math>24\cot^2 x</math></div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_1&diff=367692011 AMC 10A Problems/Problem 12011-02-10T11:45:46Z<p>AlphaMath1: /* Problem 1 */</p>
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<div>== Problem 1 ==<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
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<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
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== Solution ==</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems/Problem_1&diff=367682011 AMC 10A Problems/Problem 12011-02-10T11:44:23Z<p>AlphaMath1: Created page with '== Problem 1 == A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math>…'</p>
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<div>== Problem 1 ==<br />
A cell phone plan costs <math>\</math><math>20</math> each month, plus <math>5</math>¢ per text message sent, plus 10¢ for each minute used over <math>30</math> hours. In January Michelle sent <math>100</math> text messages and talked for <math>30.5</math> hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00 \qquad\textbf{(B)}\ </math> <math>24.50 \qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
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[[2011 AMC 10A Problems/Problem 1|Solution]]</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=367192011 AMC 10A Problems2011-02-10T02:53:45Z<p>AlphaMath1: /* Problem 1 */</p>
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<div>== Problem 1 ==<br />
A cell phone plan costs <math>\</math>20 each month, plus 5¢ per text message sent, plus 10¢ for each minute used over <math>\</math>30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?<br />
<br />
<math> \textbf{(A)}\ </math> <math>24.00\qquad\textbf{(B)}\ </math> <math>24.50\qquad\textbf{(C)}\ </math> <math>25.50\qquad\textbf{(D)}\ </math> <math>28.00\qquad\textbf{(E)}\ </math> <math>30.00 </math><br />
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[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
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== Problem 2 ==<br />
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[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
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== Problem 3 ==<br />
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[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
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== Problem 4 ==<br />
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[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
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== Problem 5 ==<br />
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[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
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== Problem 6 ==<br />
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[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
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== Problem 7 ==<br />
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[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
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== Problem 8 ==<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
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== Problem 9 ==<br />
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[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
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== Problem 10 ==<br />
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[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
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== Problem 11 ==<br />
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[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
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== Problem 12 ==<br />
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[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
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== Problem 13 ==<br />
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[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
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== Problem 14 ==<br />
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[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
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== Problem 15 ==<br />
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[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
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== Problem 16 ==<br />
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[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
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== Problem 17 ==<br />
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[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
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== Problem 18 ==<br />
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[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
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== Problem 19 ==<br />
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[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
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== Problem 20 ==<br />
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[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
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== Problem 21 ==<br />
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[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
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== Problem 22 ==<br />
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[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
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== Problem 23 ==<br />
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[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
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== Problem 24 ==<br />
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[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
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== Problem 25 ==<br />
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[[2011 AMC 10A Problems/Problem 25|Solution]]</div>AlphaMath1https://artofproblemsolving.com/wiki/index.php?title=2011_AMC_10A_Problems&diff=367132011 AMC 10A Problems2011-02-10T02:41:50Z<p>AlphaMath1: /* Problem 8 */</p>
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<div>== Problem 1 ==<br />
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[[2011 AMC 10A Problems/Problem 1|Solution]]<br />
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== Problem 2 ==<br />
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[[2011 AMC 10A Problems/Problem 2|Solution]]<br />
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== Problem 3 ==<br />
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[[2011 AMC 10A Problems/Problem 3|Solution]]<br />
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== Problem 4 ==<br />
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[[2011 AMC 10A Problems/Problem 4|Solution]]<br />
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== Problem 5 ==<br />
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[[2011 AMC 10A Problems/Problem 5|Solution]]<br />
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== Problem 6 ==<br />
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[[2011 AMC 10A Problems/Problem 6|Solution]]<br />
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== Problem 7 ==<br />
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[[2011 AMC 10A Problems/Problem 7|Solution]]<br />
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== Problem 8 ==<br />
[[2011 AMC 10A Problems/Problem 8|Solution]]<br />
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== Problem 9 ==<br />
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[[2011 AMC 10A Problems/Problem 9|Solution]]<br />
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== Problem 10 ==<br />
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[[2011 AMC 10A Problems/Problem 10|Solution]]<br />
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== Problem 11 ==<br />
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[[2011 AMC 10A Problems/Problem 11|Solution]]<br />
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== Problem 12 ==<br />
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[[2011 AMC 10A Problems/Problem 12|Solution]]<br />
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== Problem 13 ==<br />
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[[2011 AMC 10A Problems/Problem 13|Solution]]<br />
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== Problem 14 ==<br />
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[[2011 AMC 10A Problems/Problem 14|Solution]]<br />
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== Problem 15 ==<br />
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[[2011 AMC 10A Problems/Problem 15|Solution]]<br />
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== Problem 16 ==<br />
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[[2011 AMC 10A Problems/Problem 16|Solution]]<br />
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== Problem 17 ==<br />
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[[2011 AMC 10A Problems/Problem 17|Solution]]<br />
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== Problem 18 ==<br />
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[[2011 AMC 10A Problems/Problem 18|Solution]]<br />
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== Problem 19 ==<br />
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[[2011 AMC 10A Problems/Problem 19|Solution]]<br />
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== Problem 20 ==<br />
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[[2011 AMC 10A Problems/Problem 20|Solution]]<br />
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== Problem 21 ==<br />
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[[2011 AMC 10A Problems/Problem 21|Solution]]<br />
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== Problem 22 ==<br />
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[[2011 AMC 10A Problems/Problem 22|Solution]]<br />
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== Problem 23 ==<br />
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[[2011 AMC 10A Problems/Problem 23|Solution]]<br />
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== Problem 24 ==<br />
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[[2011 AMC 10A Problems/Problem 24|Solution]]<br />
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== Problem 25 ==<br />
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[[2011 AMC 10A Problems/Problem 25|Solution]]</div>AlphaMath1