https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Osmannal4&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T12:00:28ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_16&diff=1748702017 AMC 10A Problems/Problem 162022-06-14T17:28:31Z<p>Osmannal4: /* Problem */</p>
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<div>==Problem==<br />
There are <math>10</math> horses, named Horse <math>1</math>, Horse <math>2</math>, . . . , Horse <math>10</math>. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse <math>k</math> runs one lap in exactly <math>k</math> minutes. At time <math>0</math> all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time <math>S > 0</math>, in minutes, at which all <math>10</math> horses will again simultaneously be at the starting point is <math>S=2520</math>. Let <math>T > 0</math> be the least time, in minutes, such that at least <math>5</math> of the horses are again at the starting point. What is the sum of the digits of <math>T?</math><br />
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<math>\textbf{(A) }2 \qquad \textbf{(B) }3 \qquad \textbf{(C) }4 \qquad \textbf{(D) }5 \qquad \textbf{(E) }6</math><br />
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==Solution 1==<br />
If we have horses, <math>a_1, a_2, \ldots, a_n</math>, then any number that is a multiple of all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that <math>\text{LCM}(1,2,3,2\cdot2,2\cdot3) = 12</math>. Finally, <math>1+2 = \boxed{\ 3}</math>.<br />
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==Solution 2==<br />
We are trying to find the smallest number that has <math>5</math> one-digit divisors. Therefore we try to find the LCM for smaller digits, such as <math>1</math>,<math> 2</math>, <math>3</math>, or <math>4</math>. We quickly consider <math>12</math> since it is the smallest number that is the LCM of <math>1</math>, <math>2</math>, <math>3</math> and <math>4</math>. Since <math>12</math> has <math>5</math> single-digit divisors, namely <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>6</math>, our answer is <math>1+2 = \boxed{\textbf{(B)}\ 3}</math><br />
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==Solution 3 (Speedy Guess and Check)==<br />
First, for 5 horses to simultaneously pass the starting line after <math>T</math> seconds, <math>T</math> must be divisible by the amount of seconds it takes each of the 5 horses to pass the starting line, meaning all of the horses must be divisors of <math>T</math>, and therefore meaning <math>T</math> must have at least <math>5</math> <math>1</math>-digit divisors. Since we want to minimize <math>T</math>, we will start by guessing the lowest natural number, <math>1</math>. <math>1</math> has only <math>1</math> factor, so it does not work, we now repeat the process for the numbers between <math>2</math> and <math>12</math> (This should not take more than a minute) to get that <math>12</math> is the first number to have <math>7</math> or more single-digit divisors (<math>1, 2, 3, 4, 6</math>). The sum of the digits of <math>12</math> is <math>1+2 = \boxed{\textbf{(B)}\ 3}</math>, which is our answer.<br />
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Note that this solution is rather fast with this problem because the numbers given in the question are low, this may not always be the case, however, when given higher numbers.<br />
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==Solution 4 (Bash)==<br />
By inspection, <math>(1, 2, 3, 4, 6)</math> yields the lowest answer of <math>12</math> and the sum of the digits is <math>1+2 \Longrightarrow \boxed{\textbf{(B)}\ 3}</math><br />
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~JH. L<br />
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==Video Solution==<br />
https://youtu.be/umr2Aj9ViOA<br />
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==See Also==<br />
{{AMC10 box|year=2017|ab=A|num-b=15|num-a=17}}<br />
{{AMC12 box|year=2017|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
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[[Category:Introductory Number Theory Problems]]</div>Osmannal4https://artofproblemsolving.com/wiki/index.php?title=2019_AIME_II_Problems/Problem_10&diff=1133942019 AIME II Problems/Problem 102019-12-24T20:22:02Z<p>Osmannal4: /* Solution 2 */</p>
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<div>==Problem 10==<br />
There is a unique angle <math>\theta</math> between <math>0^{\circ}</math> and <math>90^{\circ}</math> such that for nonnegative integers <math>n</math>, the value of <math>\tan{\left(2^{n}\theta\right)}</math> is positive when <math>n</math> is a multiple of <math>3</math>, and negative otherwise. The degree measure of <math>\theta</math> is <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime integers. Find <math>p+q</math>.<br />
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==Solution 1==<br />
Note that if <math>\tan \theta</math> is positive, then <math>\theta</math> is in the first or third quadrant, so <math>0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}</math>. Also notice that the only way <math>\tan{\left(2^{n}\theta\right)}</math> can be positive for all <math>n</math> that are multiples of <math>3</math> is when <math>2^0\theta, 2^3\theta, 2^6\theta</math>, etc. are all the same value <math>\pmod{180^{\circ}}</math>. This happens if <math>8\theta = \theta \pmod{180^{\circ}}</math>, so <math>7\theta = 0^{\circ} \pmod{180^{\circ}}</math>. Therefore, the only possible values of theta between <math>0^{\circ}</math> and <math>90^{\circ}</math> are <math>\frac{180}{7}^{\circ}</math>, <math>\frac{360}{7}^{\circ}</math>, and <math>\frac{540}{7}^{\circ}</math>. However <math>\frac{180}{7}^{\circ}</math> does not work since <math>\tan{2 \cdot \frac{180}{7}^{\circ}}</math> is positive, and <math>\frac{360}{7}^{\circ}</math> does not work because <math>\tan{4 \cdot \frac{360}{7}^{\circ}}</math> is positive. Thus, <math>\theta = \frac{540}{7}^{\circ}</math>. <math>540 + 7 = \boxed{547}</math>.<br />
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==Solution 2==<br />
As in the previous solution, we note that <math>\tan \theta</math> is positive when <math>\theta</math> is in the first or third quadrant. In order for <math>\tan\left(2^n\theta\right)</math> to be positive for all <math>n</math> divisible by <math>3</math>, we must have <math>\theta</math>, <math>2^3\theta</math>, <math>2^6\theta</math>, etc to lie in the first or third quadrants. We already know that <math>\theta\in(0,90)</math>. We can keep track of the range of <math>2^n\theta</math> for each <math>n</math> by considering the portion in the desired quadrants, which gives <br />
<cmath>n=1 \implies (90,180)</cmath><br />
<cmath>n=2\implies (270,360)</cmath><br />
<cmath>n=3 \implies (180,270)</cmath><br />
<cmath>n=4 \implies (90,180)</cmath><br />
<cmath>n=5\implies(270,360)</cmath><br />
<cmath>n=6 \implies (180,270)</cmath><br />
<cmath>\cdots</cmath><br />
at which point we realize a pattern emerging. Specifically, the intervals repeat every <math>3</math> after <math>n=1</math>. We can use these repeating intervals to determine the desired value of <math>\theta</math> since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.<br />
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Initially, the lower bound is <math>0</math> (at <math>n=0</math>), then increases to <math>\frac{90}{2}=45</math> at <math>n=1</math>. This then becomes <math>45+\frac{45}{2}</math> at <math>n=2</math>, <math>45+\frac{45}{2}</math> at <math>n=3</math>, <math>45+\frac{45}{2}+\frac{45}{2^3}</math> at <math>n=4</math>,<math>45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}</math> at <math>n=5</math>. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as <math>n</math> approaches infinity, the lower bound converges to <br />
<cmath>\sum_{k=0}^{\infty}\left(45+\frac{45}{2}\right)\cdot \left(\frac{1}{8}\right)^k=\frac{45+\frac{45}{2}}{1-\frac{1}{8}}=\frac{\frac{135}{2}}{\frac{7}{8}}=\frac{540}{7}\implies p+q=540+7=\boxed{547}</cmath>-ktong<br />
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==Solution 3==<br />
Since <math>tan\left(\theta\right) > 0</math>, <math>0 < \theta < 90</math>. Since <math>tan\left(2\theta\right) < 0</math>, <math>\theta</math> has to be in the second half of the interval (0, 90) ie (45, 90). Since <math>tan\left(4\theta\right) < 0</math>, <math>\theta</math> has to be in the second half of that interval ie (67.5, 90). And since <math>tan\left(8\theta\right) > 0</math>, <math>\theta</math> has to be in the first half of (67.5, 90). Inductively, the pattern repeats: <math>\theta</math> is in the first half of the second half of the second half of the first half of the second half of the second half... of the interval (0, 90). Consider the binary representation of numbers in the interval (0, 1). Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... . Similarly, numbers in the second half of the second half start with 0.11... etc. So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be <math>0.11011011011..._2 = \frac{6}{7}_{10}</math>. So we want the number which is 6/7 of the way through the interval (0, 90) so <math>\theta = \frac{6}{7}\cdot 90 = \frac{540}{7}</math> and <math>p+q = 540 + 7 = \boxed{547}</math><br />
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==Solution 4==<br />
With some simple arithmetic and guess and check, we can set the lower bound and upper bounds for the "first round of <math>3</math> powers of two", which are <math>\frac{540}{8} = 67.5</math> and <math>\frac{630}{8} = 78.75</math>. Going on to the "second round of <math>3</math> powers of two, we set the new lower and upper bounds as <math>\frac{360 \times 12.5}{64} = 70.3125</math> and <math>\frac{360 \times 13.75}{64} = 77.34375</math> using some guess and check and bashing. Now, it is obvious that the bounds for the "zeroth round of <math>3</math> powers of two" are <math>0</math> and <math>90</math>, and notice that <math>90 - 78.75 = 11.25</math> and <math>78.75 - 77.34375 = 1.40625</math> and <math>\frac{11.25}{1.40625} = 8</math>.<br />
This is obviously a geometric series, so setting <math>11.25</math> as <math>u</math>, we obtain <math>90 - (u + \frac{u}{8} + \frac{u}{64} + ...)</math> = <math>90 - \frac{u}{1-\frac{1}{8}}</math> = <math>\frac{u}{\frac{7}{8}}</math> = <math>\frac{45}{4} \times \frac{8}{7}</math> which simplifies to <math>\frac{90}{7}</math>. We can now finally subtract <math>\frac{90}{7}</math> from <math>\frac{630}{7}</math> and then we get <math>\frac{540}{7}</math> as the unique angle, so <math>\boxed{547}</math> is our answer.<br />
-fidgetboss_4000<br />
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==See Also==<br />
{{AIME box|year=2019|n=II|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Osmannal4https://artofproblemsolving.com/wiki/index.php?title=1978_IMO_Problems/Problem_1&diff=1108921978 IMO Problems/Problem 12019-11-04T22:48:47Z<p>Osmannal4: /* Solution */</p>
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<div>==Problem==<br />
<math>m</math> and <math>n</math> are positive integers with <math>m < n</math>. The last three decimal digits of <math>1978^m</math> are the same as the last three decimal digits of <math>1978^n</math>. Find <math>m</math> and <math>n</math> such that <math>m + n</math> has the least possible value.<br />
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==Solution==<br />
{{Solution}}<br />
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Solution is available here:<br />
https://www.youtube.com/watch?v=SRl4Wnd60os</div>Osmannal4