https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ayyou365&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T14:06:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_19&diff=1400042021 AMC 10A Problems/Problem 192020-12-20T03:59:06Z<p>Ayyou365: </p>
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<div>==Problem==<br />
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We don't know that yet lol</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_19&diff=1400032021 AMC 10A Problems/Problem 192020-12-20T03:58:04Z<p>Ayyou365: </p>
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<div>We don't know that yet lol</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_19&diff=1400022021 AMC 10A Problems/Problem 192020-12-20T03:56:19Z<p>Ayyou365: Created page with "There's supposed to be questions here?"</p>
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<div>There's supposed to be questions here?</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_18&diff=1400012021 AMC 10A Problems/Problem 182020-12-20T03:50:51Z<p>Ayyou365: </p>
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<div>No question for you</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_18&diff=1400002021 AMC 10A Problems/Problem 182020-12-20T03:50:20Z<p>Ayyou365: </p>
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<div>No question for you lol</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_10A_Problems/Problem_18&diff=1399992021 AMC 10A Problems/Problem 182020-12-20T03:49:51Z<p>Ayyou365: Created page with "Lmao no question for you"</p>
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<div>Lmao no question for you</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10B_Problems/Problem_25&diff=1124142014 AMC 10B Problems/Problem 252019-11-30T21:07:11Z<p>Ayyou365: /* Video Solution */</p>
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<div>==Problem==<br />
In a small pond there are eleven lily pads in a row labeled <math>0</math> through <math>10</math>. A frog is sitting on pad <math>1</math>. When the frog is on pad <math>N</math>, <math>0<N<10</math>, it will jump to pad <math>N-1</math> with probability <math>\frac{N}{10}</math> and to pad <math>N+1</math> with probability <math>1-\frac{N}{10}</math>. Each jump is independent of the previous jumps. If the frog reaches pad <math>0</math> it will be eaten by a patiently waiting snake. If the frog reaches pad <math>10</math> it will exit the pond, never to return. What is the probability that the frog will escape before being eaten by the snake?<br />
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<math> \textbf {(A) } \frac{32}{79} \qquad \textbf {(B) } \frac{161}{384} \qquad \textbf {(C) } \frac{63}{146} \qquad \textbf {(D) } \frac{7}{16} \qquad \textbf {(E) } \frac{1}{2} </math><br />
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==Solution==<br />
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Notice that the probabilities are symmetrical around the fifth lily pad. If the frog is on the fifth lily pad, there is a <math>\frac{1}{2}</math> chance that it escapes and a <math>\frac{1}{2}</math> that it gets eaten. Now, let <math>P_k</math> represent the probability that the frog escapes if it is currently on pad <math>k</math>. We get the following system of <math>5</math> equations:<br />
<cmath>P_1=\frac{9}{10}\cdot P_2</cmath><br />
<cmath>P_2=\frac{2}{10}\cdot P_1 + \frac{8}{10}\cdot P_3</cmath><br />
<cmath>P_3=\frac{3}{10}\cdot P_2 + \frac{7}{10}\cdot P_4</cmath><br />
<cmath>P_4=\frac{4}{10}\cdot P_3 + \frac{6}{10}\cdot P_5</cmath><br />
<cmath>P_5=\frac{5}{10}</cmath><br />
We want to find <math>P_1</math>, since the frog starts at pad <math>1</math>. Solving the above system (really long process) yields <math>P_1=\frac{63}{146}</math>, so the answer is <math>\boxed{(C)}</math>.<br />
==Video Solution==<br />
https://m.youtube.com/watch?v=dQw4w9WgXcQ<br />
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==See Also==<br />
{{AMC10 box|year=2014|ab=B|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Ayyou365https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_8_Problems/Problem_24&diff=973602010 AMC 8 Problems/Problem 242018-08-23T21:19:08Z<p>Ayyou365: /* Solution 1 */</p>
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<div>==Problem==<br />
What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?<br />
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<math> \textbf{(A)}\ 2^{24}<10^8<5^{12}\\<br />
\textbf{(B)}\ 2^{24}<5^{12}<10^8 \\<br />
\textbf{(C)}\ 5^{12}<2^{24}<10^8 \\<br />
\textbf{(D)}\ 10^8<5^{12}<2^{24} \\<br />
\textbf{(E)}\ 10^8<2^{24}<5^{12} </math><br />
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==Solution 1==<br />
Use brute force.<br />
<math>10^8=100,000,000</math>, <br />
<math>5^{12}=244,140,625</math>, and<br />
<math>2^{24}=16,777,216</math>.<br />
Therefore, <math>\boxed{\text{(A)}2^{24}<10^8<5^{12}}</math> is the answer.<br />
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== Solution 2==<br />
Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.<br />
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== Solution 3==<br />
First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.<br />
<math>10^8</math> is fine as is.<br />
We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.<br />
We can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>.<br />
We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.<br />
Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\textbf{(A)}\ 2^{24}<10^8<5^{12}</math>.<br />
Solution by coolak<br />
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==See Also==<br />
{{AMC8 box|year=2010|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Ayyou365