https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bad+math&feedformat=atomAoPS Wiki - User contributions [en]2022-12-06T10:37:37ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_5&diff=1037352000 AIME I Problems/Problem 52019-02-22T16:18:46Z<p>Bad math: Added a new solution</p>
<hr />
<div>== Problem ==<br />
Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is <math>25.</math> One marble is taken out of each box randomly. The [[probability]] that both marbles are black is <math>27/50,</math> and the probability that both marbles are white is <math>m/n,</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. What is <math>m + n</math>?<br />
<br />
== Solution 1==<br />
If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate <math>m/n</math>. The [[Principle of Inclusion-Exclusion]] still requires us to find the individual probability of each box.<br />
<br />
Let <math>a, b</math> represent the number of marbles in each box, and [[without loss of generality]] let <math>a>b</math>. Then, <math>a + b = 25</math>, and since the <math>ab</math> may be reduced to form <math>50</math> on the denominator of <math>\frac{27}{50}</math>, <math>50|ab</math>. It follows that <math>5|a,b</math>, so there are 2 pairs of <math>a</math> and <math>b: (20,5),(15,10)</math>.<br />
<br />
*'''Case 1''': Then the product of the number of black marbles in each box is <math>54</math>, so the only combination that works is <math>18</math> black in first box, and <math>3</math> black in second. Then, <math>P(\text{both white}) = \frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25},</math> so <math>m + n = 26</math>.<br />
<br />
*'''Case 2''': The only combination that works is 9 black in both. Thus, <math>P(\text{both white}) = \frac{1}{10}\cdot \frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>.<br />
<br />
Thus, <math>m + n = \boxed{026}</math>.<br />
<br />
== Solution 2==<br />
Let <math>w_1, w_2, b_1,</math> and <math>b_2</math> represent the white and black marbles in boxes 1 and 2.<br />
<br />
Since there are <math>25</math> marbles in the box:<br />
<br />
<math>w_1 + w_2 + b_1 + b_2 = 25</math><br />
<br />
From the fact that there is a <math>\frac{27}{50}</math> chance of drawing one black marble from each box:<br />
<br />
<math>\frac{b_1 \cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \frac{27}{50} = \frac{54}{100} = \frac{81}{150}</math><br />
<br />
Thinking of the numerator and denominator separately, if <math>\frac{27}{50}</math> was not a reduced fraction when calculating out the probability, then <math>b_1 \cdot b_2 = 27</math>. Since <math>b_1 < 25</math>, this forces the variables to be <math>3</math> and <math>9</math> in some permutation. Without loss of generality, let <math>b_1 = 3</math> and <math>b_2 = 9</math>.<br />
<br />
The denominator becomes:<br />
<math>(3 + w_1)(9 + w_2) = 50</math><br />
<br />
Since there have been <math>12</math> black marbles used, there must be <math>13</math> white marbles. Substituting that in:<br />
<br />
<math>(3 + w_1)(9 + (13 - w_1)) = 50</math><br />
<br />
<math>(3 + w_1)(22 - w_1) = 50</math><br />
<br />
Since the factors of <math>50</math> that are greater than <math>3</math> are <math>5, 10, 25,</math> and <math>50</math>, the quantity <math>3 + w_1</math> must equal one of those. However, since <math>w_1 < 13</math>, testing <math>2</math> and <math>7</math> for <math>w_1</math> does not give a correct product. Thus, <math>\frac{27}{50}</math> must be a reduced form of the actual fraction.<br />
<br />
First assume that the fraction was reduced from <math>\frac{54}{100}</math>, yielding the equations <math>b_1\cdot b_2 = 54</math> and <math>(b_1 + w_1)(b_2 + w_2) = 100</math>.<br />
Factoring <math>b_1 \cdot b_2 = 54</math> and saying WLOG that <math>b_1 < b_2 < 25</math> gives <math>(b_1, b_2) = (3, 18)</math> or <math> (6, 9)</math>. Trying the first pair and setting the denominator equal to 100 gives:<br />
<math>(3 + w_1)(18 + w_2) = 100</math><br />
<br />
<br />
Since <math>w_1 + w_2 = 4</math>, the pairs <math>(w_1, w_2) = (1, 3), (2,2),</math> and <math>(3,1)</math> can be tried, since each box must contain at least one white marble. Plugging in <math>w_1 = w_2 = 2</math> gives the true equation <math>(3 + 2)(18 + 2) =100</math>, so the number of marbles are <math>(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)</math><br />
<br />
Thus, the chance of drawing 2 white marbles is <math>\frac{w_1 \cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \frac{4}{100} = \frac{1}{25}</math> in lowest terms, and the answer to the problem is <math>1 + 25 = \boxed{026}.</math><br />
<br />
<br />
<br />
For completeness, the fraction <math>\frac{81}{150}</math> may be tested. <math>150</math> is the highest necessary denominator that needs to be tested, since the maximum the denominator <math>(w_1+ b_1)(w_2 + b_2)</math> can be when the sum of all integer variables is <math>25</math> is when the variables are <math>6, 6, 6, </math> and <math>7</math>, in some permutation, which gives <math>154</math>. If <math>b_1 \cdot b_2 = 81</math>, this forces <math>b_1 = b_2 = 9</math>, since all variables must be integers under <math>25</math>. The denominator becomes <math>(9 + w_1)(9 + w_2) = 150</math>, and since there are now <math>25 - 18 = 7</math> white marbles total, the denominator becomes <math>(9 + w_1)(16 - w_1) = 150</math>. Testing <math>w_1 = 1</math> gives a solution, and thus <math>w_2 = 6</math>. The complete solution for this case is <math>(w_1, w_2, b_1, b_2) = (1, 6, 9, 9)</math>. Although the distribution and colors of the marbles is different from the last case, the probability of drawing two white marbles is <math>\frac{6 \cdot 1}{ 150}</math>, which still simplifies to <math>\frac {1}{25}</math>.<br />
<br />
==Solution 3==<br />
<br />
We know that <math>\frac{27}{50} = \frac{b_1}{t_1} \cdot \frac{b_2}{t_2}</math>, where <math>b_1</math> and <math>b_2</math> are the number of black marbles in the first and the second box respectively, and <math>t_1</math> and <math>t_2</math> is the total number of marbles in the first and the second boxes respectively. So, <math>t_1 + t_2 = 25</math>. Then, we can realize that <math>\frac{27}{50} = \frac{9}{10} \cdot \frac{3}{5} = \frac{9}{10} \cdot \frac{9}{15}</math>, which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes <math>\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}</math>. The answer is <math>1 + 25 = \boxed{026}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=4|num-a=6}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Bad mathhttps://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_22&diff=1031812018 AMC 10A Problems/Problem 222019-02-18T00:12:49Z<p>Bad math: /* Solution 4 (Fastest) */ 15 is not a valid answer choice, so this method is correct.</p>
<hr />
<div>Let <math>a, b, c,</math> and <math>d</math> be positive integers such that <math>\gcd(a, b)=24</math>, <math>\gcd(b, c)=36</math>, <math>\gcd(c, d)=54</math>, and <math>70<\gcd(d, a)<100</math>. Which of the following must be a divisor of <math>a</math>?<br />
<br />
<math>\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}</math><br />
<br />
== Solution 1 ==<br />
<br />
The gcd information tells us that 24 divides <math>a</math>, both 24 and 36 divide <math>b</math>, both 36 and 54 divide <math>c</math>, and 54 divides <math>d</math>. Note that we have the prime factorizations:<br />
<cmath>\begin{align*}<br />
24 &= 2^3\cdot 3,\\<br />
36 &= 2^2\cdot 3^2,\\<br />
54 &= 2\cdot 3^3.<br />
\end{align*}</cmath><br />
<br />
Hence we have<br />
<cmath>\begin{align*}<br />
a &= 2^3\cdot 3\cdot w\\<br />
b &= 2^3\cdot 3^2\cdot x\\<br />
c &= 2^2\cdot 3^3\cdot y\\<br />
d &= 2\cdot 3^3\cdot z<br />
\end{align*}</cmath><br />
for some positive integers <math>w,x,y,z</math>. Now if 3 divdes <math>w</math>, then <math>\gcd(a,b)</math> would be at least <math>2^3\cdot 3^2</math> which is too large, hence 3 does not divide <math>w</math>. Similarly, if 2 divides <math>z</math>, then <math>\gcd(c,d)</math> would be at least <math>2^2\cdot 3^3</math> which is too large, so 2 does not divide <math>z</math>. Therefore,<br />
<cmath>\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)</cmath><br />
where neither 2 nor 3 divide <math>\gcd(w,z)</math>. In other words, <math>\gcd(w,z)</math> is divisible only by primes that are at least 5. The only possible value of <math>\gcd(a,d)</math> between 70 and 100 and which fits this criterion is <math>78=2\cdot3\cdot13</math>, so the answer is <math>\boxed{\textbf{(D) }13}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
We can say that <math>a</math> and <math>b</math> 'have' <math>2^3 * 3</math>, that <math>b</math> and <math>c</math> have <math>2^2 * 3^2</math>, and that <math>c</math> and <math>d</math> have <math>3^3 * 2</math>. Combining <math>1</math> and <math>2</math> yields <math>b</math> has (at a minimum) <math>2^3 * 3^2</math>, and thus <math>a</math> has <math>2^3 * 3</math> (and no more powers of <math>3</math> because otherwise <math>gcd(a,b)</math> would be different). In addition, <math>c</math> has <math>3^3 * 2^2</math>, and thus <math>d</math> has <math>3^3 * 2</math> (similar to <math>a</math>, we see that <math>d</math> cannot have any other powers of <math>2</math>). We now assume the simplest scenario, where <math>a = 2^3 * 3</math> and <math>d = 3^3 * 2</math>. According to this base case, we have <math>gcd(a, d) = 2 * 3 = 6</math>. We want an extra factor between the two such that this number is between <math>70</math> and <math>100</math>, and this new factor cannot be divisible by <math>2</math> or <math>3</math>. Checking through, we see that <math>6 * 13</math> is the only one that works. Therefore the answer is <math>\boxed{\textbf{(D) } 13}</math><br />
<br />
Solution by JohnHankock<br />
<br />
== Solution 2.1 ==<br />
Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to <math>gcd(a, d)</math> because doing so would later the <math>gcd</math> of <math>(a, b)</math> and <math>(c, d)</math>. This is why: <br />
<br />
The <math>gcd(a, b)</math> is <math>2^3 * 3</math> and the <math>gcd</math> of <math>(c, d)</math> is <math>2 * 3^3</math>. However, the <math>gcd</math> of <math>(b, c) = 2^2 * 3^2</math> (meaning both are divisible by 36). Therefore, <math>a</math> is only divisible by <math>3^1</math> (and no higher power of 3), while <math>d</math> is divisible by only <math>2^1</math> (and no higher power of 2). <br />
<br />
Thus, the <math>gcd</math> of <math>(a, d)</math> can be expressed in the form <math>2 * 3 * k</math> for which <math>k</math> is a number not divisible by <math>2</math> or <math>3</math>. The only answer choice that satisfies this (and the other condition) is <math>\boxed{\textbf{(D) } 13}</math>.<br />
<br />
==Solution 3 (Better notation)==<br />
<br />
First off, note that <math>24</math>, <math>36</math>, and <math>54</math> are all of the form <math>2^x\times3^y</math>. The prime factorizations are <math>2^3\times 3^1</math>, <math>2^2\times 3^2</math> and <math>2^1\times 3^3</math>, respectively. Now, let <math>a_2</math> and <math>a_3</math> be the number of times <math>2</math> and <math>3</math> go into <math>a</math>,respectively. Define <math>b_2</math>, <math>b_3</math>, <math>c_2</math>, and <math>c_3</math> similiarly. Now, translate the <math>lcm</math>s into the following: <br />
<cmath>\min(a_2,b_2)=3</cmath> <cmath>\min(a_3,b_3)=1</cmath> <cmath>\min(b_2,c_2)=2</cmath> <cmath>\min(b_3,c_3)=2</cmath> <cmath>\min(a_2,c_2)=1</cmath> <cmath>\min(a_3,c_3)=3</cmath> .<br />
<br />
(Unfinished)<br />
~Rowechen Zhong<br />
<br />
==Solution 4 (Fastest)==<br />
Notice that <math>gcd (a,b,c,d)=gcd(gcd(a,b),gcd(b,c),gcd(c,d))=gcd(24,36,54)=6</math>, so <math>gcd(d,a)</math> must be a multiple of <math>6</math>. The only answer choice that gives a value between <math>70</math> and <math>100</math> when multiplied by 6 is <math>\boxed{\textbf{(D) } 13}</math>. - mathleticguyyy<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Intermediate Number Theory Problems]]<br />
{{MAA Notice}}</div>Bad mathhttps://artofproblemsolving.com/wiki/index.php?title=2008_AIME_I_Problems/Problem_1&diff=950212008 AIME I Problems/Problem 12018-06-08T19:10:13Z<p>Bad math: /* Solutions */ Added a third solution.</p>
<hr />
<div>== Problem ==<br />
Of the students attending a school party, <math>60\%</math> of the students are girls, and <math>40\%</math> of the students like to dance. After these students are joined by <math>20</math> more boy students, all of whom like to dance, the party is now <math>58\%</math> girls. How many students now at the party like to dance?<br />
<br />
==Solutions==<br />
===Solution 1===<br />
Say that there were <math>3k</math> girls and <math>2k</math> boys at the party originally. <math>2k</math> like to dance. Then, there are <math>3k</math> girls and <math>2k + 20</math> boys, and <math>2k + 20</math> like to dance.<br />
<br />
Thus, <math>\dfrac{3k}{5k + 20} = \dfrac{29}{50}</math>, solving gives <math>k = 116</math>. Thus, the number of people that like to dance is <math>2k + 20 = \boxed{252}</math>.<br />
<br />
===Solution 2===<br />
Let the number of girls be <math>g</math>. Let the number of total people originally be <math>t</math>.<br />
<br />
We know that <math>\frac{g}{t}=\frac{3}{5}</math> from the problem.<br />
<br />
We also know that <math>\frac{g}{t+20}=\frac{29}{50}</math> from the problem.<br />
<br />
We now have a system and we can solve.<br />
<br />
The first equation becomes:<br />
<br />
<math>3t=5g</math>.<br />
<br />
The second equation becomes: <br />
<br />
<math>50g=29t+580</math><br />
<br />
Now we can sub in <math>30t=50g</math> by multiplying the first equation by <math>10</math>. We can plug this into our second equation.<br />
<br />
<math>30t=29t+580</math><br />
<br />
<math>t=580</math><br />
<br />
We know that there were originally <math>580</math> people. Of those, <math>\frac{2}{5}*580=232</math> like to dance.<br />
<br />
We also know that with these people, <math>20</math> boys joined, all of whom like to dance. We just simply need to add <math>20</math> to get <math>232+20=\boxed{252}</math><br />
<br />
==Solution 3==<br />
Let <math>p</math> denote the total number of people at the party. Then, because we know the proportions of boys to <math>p</math> both before and after 20 boys arrived, we can create the following equation:<br />
<cmath>0.4p+20 = 0.48(p+20)</cmath><br />
Solving for p gives us <math>p=580</math>, so the solution is <math>0.4p+20 = \boxed{252}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=I|before=First Question|num-a=2}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
{{MAA Notice}}</div>Bad mathhttps://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_1&diff=947442012 AIME II Problems/Problem 12018-05-26T16:22:16Z<p>Bad math: Added one more solution.</p>
<hr />
<div>== Problem 1 ==<br />
Find the number of ordered pairs of positive integer solutions <math>(m, n)</math> to the equation <math>20m + 12n = 2012</math>.<br />
<br />
== Solution ==<br />
<br />
==Solution 1==<br />
<br />
Solving for <math>m</math> gives us <math>m = \frac{503-3n}{5},</math> so in order for <math>m</math> to be an integer, we must have <math>3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.</math> The smallest possible value of <math>n</math> is obviously <math>1,</math> and the greatest is <math>\frac{503 - 5}{3} = 166,</math> so the total number of solutions is <math>\frac{166-1}{5}+1 = \boxed{034}</math><br />
<br />
==Solution 2==<br />
<br />
Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Solving for <math>n</math> gives <math>n \equiv 1 \pmod 5</math>. The solutions are the numbers <math>n <br />
= 1, 6, 11, ... , 166</math>. There are <math>\boxed{034}</math> solutions.<br />
<br />
==Solution 3==<br />
<br />
Because the x-intercept of the equation is <math>\frac{2012}{20}</math>, and the y-intercept is <math>\frac{2012}{12}</math>, the slope is <math>\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}</math>. Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: <math>(100,1), (97,6), (94,11)...</math> Because the solutions are only positive, we can generate only 33 more solutions, so in total we have <math>33+1=\boxed{034}</math> solutions.<br />
<br />
== See Also ==<br />
{{AIME box|year=2012|n=II|before=First Problem|num-a=2}}<br />
{{MAA Notice}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Bad mathhttps://artofproblemsolving.com/wiki/index.php?title=Mathematical_Olympiad_Summer_Program&diff=93844Mathematical Olympiad Summer Program2018-04-08T20:37:00Z<p>Bad math: /* History and Culture */</p>
<hr />
<div>The '''Mathematical Olympiad Program''' (abbreviated '''MOP''') is a 3-week intensive problem solving camp held at the Carnegie Mellon University to help high school students prepare for math olympiads, especially the [[International Mathematical Olympiad]]. While the program is free to participants, invitations are limited to the top finishers on the [[USAMO]].<br />
<br />
== Purpose ==<br />
One purpose of MOP is to select and train the US team for the [[International Mathematical Olympiad]]. This is done at the start of MOP via a [[team selection test]] (TST). The results of the USAMO and the TST are weighted equally when selecting the US IMO team.<br />
<br />
The other important purpose of MOP is to train younger students in Olympiad-level problem solving and broaden their mathematical horizons.<br />
<br />
== Information ==<br />
MOP is currently held at the Carnegie Mellon University. While the dates vary from year to year, MOP is generally held in the last three weeks of June. <br />
<br />
Invitations are extended to the top non-Canadian finishers on USAMO. Students receiving invitations can be divided into four groups:<br />
<br />
USAMO winners: The Americans among the top 12 finishers on USAMO are invited to MOP regardless of their age. Additionally, they are invited to take the Team Selection Test and are viewed as potential members of the American IMO team for that year.<br />
<br />
Top non-senior USAMO finishers: In addition to the winners, the next 18 or so non-senior non-Canadian finishers are invited to attend MOP. This group is viewed as potential IMO team members for future years, although in extreme circumstances (including 2006) IMO team members for that year have been drawn from this pool.<br />
<br />
Top 30 freshmen and sophomores: The top 30 freshmen on USAMO and USAJMO are invited to attend MOP with the goal of providing them with a foundation in Olympiad-level mathematics.<br />
<br />
In 2008, another group was added. The girls who will be representing the United States at the China Girls Math Olympiad will attend MOP to prepare for that contest. This is group is colloquially known as Pink MOP.<br />
<br />
== Structure of the Program ==<br />
<br />
MOP is divided into three groups that roughly correspond with the first three kinds of invitations. Black MOP consists of that year's USAMO winners and contains the IMO team members and alternates. Blue MOP is for the second group of invitees and mostly consists of students who just completed their junior or sophomore year of high school, although in exceptional cases some 7th and 8th graders have participated. Finally, Red MOP consists of all the freshmen who were invited to participate, as well as Girls' Math Olympiad participants. Students and instructors have discretion in selecting which group they're part of and may choose to transfer part way through the program; this generally involves members of Black dropping down to Blue or occasionally members of Red promoting themselves to Blue. The three groups take classes and practice tests separately, are given different levels of material to practice with, and to a certain extent are distinct socially.<br />
<br />
Each Weekday consists of three instructional sessions: 8:30 AM - 10:00 AM, 10:15 PM - 11:45 PM, and 1:15 PM - 2:45 PM. Classes usually consist of a lecture followed by a problem set. Solutions are often presented by students with the supervision of an instructor.<br />
<br />
Timed and graded olympiad style tests are an integral part of MOP. Every few days, a 4-hour, 4-question test is administered in place of the afternoon lecture, and is graded with comments within 2-3 days.<br />
<br />
Team tests also occur weekly. Students are divided into teams of five, in 2008 consisting of one or two blue MOPpers each, and work on a set of thirty problems for approximately half a week. On the day of the contest, the teams present solutions to problems which have not yet been presented, in arbitrary order. The fun starts when all of the easy problems have been taken, and teams resort to certain creative methods in order to solve a problem.<br />
<br />
The combination of these makes MOP an extraordinarily intense experience. One participant at 2007 MOP calculated that by the end of the second week members of Blue MOP had already spent more time in a classroom than most calculus classes do in a year, and by the end of the third week participants had spent 170 hours over 19 days either in class or taking practice test for an average of roughly 9 hours a day of math- and that's before time spent doing problem sets and working on the team contest outside of class is included.<br />
<br />
== History and Culture ==<br />
MOP was created in 1974 as a training camp for the first United States IMO team. <br />
<br />
At the time that MOP was established the official name was simply "Mathematical Olympiad Program", which was the source of the original abbreviation "MOP". At some point, however, the official name was changed to "Mathematical Olympiad Summer Program" and the official abbreviation became "MOSP". Despite this change, participants and alumni almost universally continued to refer to the program as "MOP". Although some administrators continued to use "MOSP" in official documents, students used "MOP" in every setting. One former participant testifies, "Any lost souls using the other appellation are looked upon with pity and regret." Finally, the administration relented in 2017, officially renaming the camp as "MOP".<br />
<br />
Previous locations for MOP have included [[IMSA]], [[Rutgers University]], West Point (US Military Academy), and the US Naval Academy.<br />
<br />
MOP is not only a training camp but also a competition in and of itself. In addition to the regularly administered practice olympiads and the weekly team contest, returning students write and administer the [[ELMO]] (an amorphous acronym) and the USEMO (the USEless Math Olympiad).<br />
<br />
Popular pastimes at MOP include chess, card games, Mafia (which was banned after a police incident in 2007), Starcraft (which was explicitly banned in 2009) and Ultimate Frisbee.<br />
<br />
== Links ==<br />
*[http://www.unl.edu/amc/a-activities/a6-MOP/MOP.shtml AMC MOP page]</div>Bad mathhttps://artofproblemsolving.com/wiki/index.php?title=ELMO&diff=88002ELMO2017-10-28T19:59:12Z<p>Bad math: /* Details */ Contest names updated, facts fixed.</p>
<hr />
<div>== Summary ==<br />
The '''ELMO''' is an annual math olympiad that happens at [[MOP]]. Its initials have stood for many things over time, originating from a backronym originally called the "Experimental Lincoln Math Olympiad". The black MOPpers and returning MOPpers write and organize the test, and lead the teams of competitors, which consist of rookie MOPpers. The team leaders present their teams' solutions to the graders, imitating the grading process of the [[IMO]]. The ELMO began in 1999.<br />
<br />
== Details ==<br />
The ELMO is a student-written contest similar in format to the [[IMO]] (that is, 6 problems over two days with 4.5 hours per day); it is a tradition for all new students not in Black (and occasionally a couple returning ones) to take the ELMO during a weekend of [[MOP]].<br />
<br />
Every year, the ELMO committee, consisting of most of the returning MOPpers as well as the Black group, create a new name for the ELMO based on these initials. Names in past years have included "Eric Larsen Math Olympiad"(2008?), " Entirely Legitimate (Junior) Math Olympiad"(2009), and "Exceedingly Luck-Based Math Olympiad" (2010), Ex-experimental Math Olympiad, Easy Little Math Olympiad, Extremely Last-Minute Olympiad, e^log Math Olympiad, End Letter Missing, Exceedingly Loquacious Math Olympiad, English Language Master's Open (2011), Everyone Lives at Most Once (2013), Ego Loss May Occur (2014), Ex-Lincoln Math Olympiad (2015), and The vEry badLy naMed cOntest (2017). <br />
<br />
Part of this finite simple group (usually in Black) make up the grading committee, and previous MOPpers submit problems to be included on the test. (The head organizer for the ELMO is called the Supreme Grand Ayatollah, a tradition started By [[David B. Rush]] in 2009.) All returning students then vote on the [[problems]] that the grading committee decides to put on the shortlist, to determine what 6 problems will be on the test. Problems are voted on in pairs, rather than individually; for example, someone would say, "I want G1/N1 to be problems 1 and 4, all in favor?" rather than "G1 should be problem 1". They also vote on what a rubric for grading these problems will be, especially how many points would be gained by making partial progress on the problem and how much lost for certain common mistakes (often this rubric only includes the most common solution; unique ideas are determined on a case-by-case basis). The sheer amount of ideas that need to be voted down often cause these coordination meetings to devolve into [[chaos]], shouting matches, and thrown insults.<br />
<br />
The people not on grading committee (this usually includes all Blue and Green returnees, and occasionally a Black or two) act as "team leaders" for their "country" (each country gets 1 or 2 leaders), and each picks students to serve on their team (where is a variable depending on number of willing team leaders and number of returning MOPpers). Picking order is supposedly random but has often consisted of team leaders with higher test scores picking first in the first round, with every other round reversed so it seems fair. Throughout all this, the ELMO room is kept securely locked so that no one not involved in ELMO coordination can see any potential test problems, and more importantly, team picking order. The team leaders are responsible for voting on problems they feel their team would be strong at, reading solutions of their team members, and attending coordination sessions to argue with the graders about how many points their team members' solutions are worth. After ELMO is finished, "medals" are awarded, with medal distribution percentages similar to those at IMO; i.e. 10% receive Gold, etc. Approximately 70%(?) of ELMO participants receive some sort of award or medal.<br />
<br />
Shortlist problems are publicly released after ELMO on their forum page.<br />
<br />
== Results ==<br />
The winning team in 2010 was "Kanto", consisting of Youkow Homma, Thomas Swayze, Archit Kulkarni, and Gil Goldschlager, with team leaders Brian Wai and Thomas Lu. Due to the experimental nature of Green Group, this particular year had many co-captaining teams. The team "Yang Dominion", led by Dai and Patrick Yang, placed second. The winning individual score was 33, and the winning team score was 80.<br />
<br />
The individual top 5 in 2010 were:<br />
<br />
1. Thomas Swayze<br />
2. Bowei Liu<br />
3. Eric Schneider<br />
Silver:<br />
4. Bobby Shen<br />
5. Reed LaFleche<br />
<br />
== Resources ==<br />
<br />
* [http://web.archive.org/web/20030805142824/www.people.fas.harvard.edu/~gcarroll/math/mop99/elmo99.tex The first ELMO, in .tex format] [[1999 ELMO|Added to the wiki]]</div>Bad mathhttps://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki:FAQ&diff=87214AoPS Wiki:FAQ2017-08-27T01:45:50Z<p>Bad math: /* Acronym Correction */</p>
<hr />
<div>{{shortcut|[[A:FAQ]]}}<br />
<br />
This is a community created list of Frequently Asked Questions about Art of Problem Solving. If you have a request to edit or add a question on this page, please make it [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=416&t=414129 here].<br />
<br />
== General==<br />
<br />
<br />
==== Can I change my user name? ====<br />
<br />
:As indicated during the time of your registration, you are unable to change your username. However, if you have a strong reason for doing so or if your username contains your real name, you can PM an admin and they will check your account to determine if your username can be changed. Should your username change request be approved, any edits to the AoPSWiki, Reaper results, or any cached content on Google or other search engines will not be changed.<br />
<br />
==== Something looks weird (e.g., blurry, missing line, etc.) ====<br />
<br />
: This is likely due to your browser zoom level. Please make sure your zoom level is at 100% for correct rendering of the web page.<br />
<br />
==== Can I make more than one account?==== <br />
<br />
:Short answer: No.<br />
<br />
:Long answer: Multiple accounts (multis) are banned on AoPS. Having more than one account leads to issues of not remembering on what account you did what. Using multiple accounts to "game" the system, (e.g. increase rating for posts or in online games) will lead to bans on all accounts associated to you. If you have already made additional accounts, please choose one account and stop using the others.<br />
<br />
====What software does Art of Problem Solving use to run the website?====<br />
<br />
:* Search: Solr<br />
:* Wiki: MediaWiki<br />
:* Asymptote and Latex are generated through their respective binary packages<br />
:* Videos: YouTube<br />
<br />
:All other parts of the website are custom built.<br />
<br />
====Can you make an AoPS App?====<br />
<br />
:There are currently no plans to build an app. Any app built would work almost identically to the website, and not be any faster, so there is little value in building an app.<br />
<br />
== Forums ==<br />
<br />
====How do I create a forum?====<br />
<br />
:To create an AoPS forum, an user must be on the AoPS community for at least 2 weeks. To create a forum, hover over the community tab, then click "My AoPS." You should now see your avatar, and a list of your friends. Now, click on the "My Forums" tab. There, you would be able to see which forum you moderate or administrate, as well as the private forums you can access. Click on the "+" button at the top right. This should lead you to a forum creating page.<br />
<br />
==== How do I format my post, e.g. bold text, add URLs, etc.? ====<br />
<br />
:AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found [[BBCode:Tutorial|here]].<br />
<br />
==== How do I hide content in the forums? ====<br />
:Wrap the content you want to hide in [hide] tags.<br />
[hide]Content[/hide]<br />
:If you want to customize the label, instead of saying "Click here to reveal hidden text", you can do something like:<br />
[hide=Label to display]Content[/hide]<br />
<br />
==== I got the message "You can not post at this time" when trying to post, why? ====<br />
<br />
:New users are not allowed to post messages with URLs and various other things. Once you have five posts you can post normally.<br />
<br />
====I got the message "Too many messages." when trying to send a private message, why?====<br />
:To prevent PM spam abuse, users with less than five forum posts are limited to four private messages within a forty-eight hour period.<br />
<br />
==== If I make more posts, it means I'm a better user, right? ====<br />
<br />
:Absolutely not. Post quality is far more important than post quantity. Users making a lot of senseless posts are often considered worse users, or spammers.<br />
<br />
==== I have made some posts but my post count did not increase. Why? ====<br />
<br />
:When you post in some of the forums, such as the Test Forum, Mafia Forum, Fun Factory, and most user created forums, the post does not count towards your overall post count.<br />
<br />
==== The time the post was posted seems wrong ====<br />
<br />
:Posts may say something weird for when posted, "such as 5 minutes ago" or some time in the future when you just posted. This is due to your system clock being incorrect. Please update your system clock to the correct time. Many operating systems have an option to keep the clock accurate automatically. Mac users may wish to check [http://www.macinstruct.com/node/92 Synchronize your Mac's Clock With A Time Server]. You can also check [http://www.time.gov the US offical time.]<br />
<br />
==== How does AoPS select moderators? ====<br />
<br />
:When a new moderator is needed in the forums, AoPS administrators first check if any current moderators could serve as a moderator of the forum which needs a moderator. Should none be found, AoPS administrators and/or other moderators scour the forum looking for productive users. They may also ask for suggestions from other moderators or trusted users on the site. Once they have pinpointed a possible candidate based on their long term usage of the site, productive posts in the forum, and having no recent behavioral issues, that user is asked if he or she would like to moderate the forum. <br />
<br />
:Less active forums often have no moderator. Inappropriate posts should be reported by users and administrators will take appropriate action.<br />
<br />
:AoPS receives MANY requests to be a moderator. As they receive so many, it is possible that you won't get a response should you request to be one. Also, AoPS very rarely makes someone a mod for asking to be one, so '''please do not ask'''.<br />
<br />
==== I believe a post needs corrective action. What should I do? ====<br />
<br />
:If you believe a post needs moderative action, you may report it by clicking the "!" icon on the upper-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better. Usually, you shouldn't publicly post such things on a thread itself, which is called "backseat moderation" and is considered rude and unproductive.<br />
<br />
==== How long of a non-commented thread is considered reviving? ====<br />
<br />
:If any post is still on-topic and isn't spammy, it isn't considered reviving. However, everyone has a different period of time that they consider reviving. In general, apply common sense.<br />
<br />
==== How do I post images? ====<br />
<br />
:While AoPS forums have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restorations, etc. We also have limited disk space which causes us to remove attachments from time to time. Therefore, we recommend using a third party image hosting solution. There are many options, but we recommend imgur.com.<br />
<br />
:* Go to imgur.com<br />
:* Click upload images<br />
:* Follow the on screen instructions to upload the image to imgur.<br />
:* After uploading, you'll be presented the uploaded image along with links on the left.<br />
:* Find the link labeled BBCode (message boards & forums)<br />
:* Copy the code, and paste into your post. It will look something like:<br />
<br />
[img]http://i.imgur.com/aBcDeFgH.jpg[/img]<br />
<br />
== Blogs ==<br />
==== How come I can't create a blog? ====<br />
:One needs to have at least 5 posts in order to make a blog.<br />
==== How do I make my blog look nice? ====<br />
:Many AoPSers make their blogs look awesome by applying [[CSS|CSS]], which is a high-level stylesheet language. This can be done by typing CSS code into the CSS box in the Blog Control Panel.<br />
<br />
<br />
== Alcumus ==<br />
<br />
==== How is rating computed? ====<br />
:The rating is more of a prediction of what percentage of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. However, if you are predicted to get most correct, and you miss one or two problems, the rating, or prediction of percentage correct, will go down.<br />
<br />
==== I am stuck on a problem, changing the topic does not change the problem. ====<br />
:Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before moving on.<br />
<br />
==== Why can't I change topics? ====<br />
:Alcumus provides review problems to make sure you still recall information learned from the past. You are not able to skip these problems. You will need to answer the problem before the topic changes to the currently selected topic.<br />
<br />
== Contests ==<br />
==== Where can I find past contest questions and solutions? ====<br />
:In the [http://www.artofproblemsolving.com/Forum/resources.php Contests] section.<br />
<br />
==== How do I get problems onto the contest page? ====<br />
<br />
:Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.<br />
<br />
==== What are the guidelines for posting problems to be added to the contests section? ====<br />
:Refer to the [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=144&t=195579 guidelines in this post].<br />
<br />
==== Why is the wiki missing many contest questions? ====<br />
:Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations. See, for example, [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1391106#p1391106 this post].<br />
<br />
==== What if I find an error on a problem? ====<br />
:Please post an accurate description of the problem in [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=426693 this thread]. If the problem is on the wiki, you can edit it yourself.<br />
<br />
== LaTeX and Asymptote ==<br />
==== What is LaTeX, and how do I use it? ====<br />
<br />
:<math>\LaTeX</math> is a typesetting markup language that is useful to produce properly formatted mathematical and scientific expressions.<br />
<br />
==== How can I download LaTeX to use on the forums? ====<br />
<br />
:There are no downloads necessary; the forums and the wiki render LaTeX commands between dollar signs. <br />
<br />
==== How can I download LaTeX for personal use? ====<br />
:You can download TeXstudio [http://texstudio.sourceforge.net here] or TeXnicCenter [http://www.texniccenter.org here]<br />
<br />
==== Where can I find a list of LaTeX commands? ====<br />
:See [[LaTeX:Symbols|here]].<br />
<br />
==== Where can I test LaTeX commands? ====<br />
<br />
:[[A:SAND|Sandbox]] or [http://www.artofproblemsolving.com/Resources/texer.php TeXeR]. You can also use our [http://artofproblemsolving.com/community/c67_test_forum Test Forum].<br />
<br />
==== Where can I find examples of Asymptote diagrams and code? ====<br />
<br />
:Search this wiki for the <tt><nowiki><asy></nowiki></tt> tag or the Forums for the <tt><nowiki>[asy]</nowiki></tt> tag. See also [[Asymptote:_Useful_commands_and_their_Output|these examples]] and [[Proofs without words|this article]] (click on the images to obtain the code).<br />
<br />
==== How can I draw 3D diagrams? ====<br />
<br />
:See [[Asymptote: 3D graphics]].<br />
<br />
==== What is the cse5 package? ==== <br />
<br />
:See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&t=149650 here]. The package contains a set of shorthand commands that implement the behavior of usual commands, for example <tt>D()</tt> for <tt>draw()</tt> and <tt>dot()</tt>, and so forth.<br />
<br />
==== What is the olympiad package? ====<br />
<br />
:See [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=519&t=165767 here]. The package contains a set of commands useful for drawing diagrams related to [[:Category:Olympiad Geometry Problems|olympiad geometry problems]].<br />
<br />
== AoPSWiki ==<br />
==== Is there a guide for wiki syntax? ====<br />
<br />
:See [http://en.wikipedia.org/wiki/Help:Wiki_markup wiki markup], [[AoPSWiki:Tutorial]], and [[Help:Contents]].<br />
<br />
==== What do I do if I see a mistake in the wiki? ====<br />
<br />
:Edit the page and correct the error! You can edit most pages on the wiki. Click the "edit" button on the right sidebar to edit a page.<br />
<br />
==== Why can't I edit the wiki? ====<br />
<br />
:You must be a registered user on AoPS to edit. To be registered, make sure you give a correct email, and activate your account.<br />
<br />
== Miscellaneous ==<br />
==== Is it possible to join the AoPS Staff? ====<br />
<br />
:Yes. Mr. Rusczyk will sometimes hire a small army of college students to work as interns, graders, and teaching assistants. You must be at least in your second semester of your senior year and be legal to work in the U.S. (at least 16).<br />
<br />
==== What is the minimum age to be an assistant in an Art of Problem Solving class? ====<br />
<br />
:You must have graduated from high school, or at least be in the second term of your senior year.<br />
<br />
==AoPS Acronyms==<br />
*'''AFK'''- Away from keyboard<br />
*'''AoPS'''- Art of Problem Solving, the website you're on right now!<br />
*'''AIME'''- American Invitational Mathematics Examination<br />
*'''AMC'''- American Math Competitions<br />
*'''ATM'''- At the Moment<br />
*'''brb'''- Be right Back<br />
*'''BTW'''- By the way<br />
*'''CEMC''' - Centre for Mathematics and Computing<br />
*'''C&P or C+P' or CP''' - Counting and Probability or Contests and Programs<br />
*'''EBWOP'''- Editing by way of post<br />
*'''FTW'''- For the Win, a game on AoPS<br />
*'''gg'''- Good Game<br />
*'''gj'''- Good Job<br />
*'''glhf'''-Good Luck Have Fun<br />
*'''gtg''' - Got to go<br />
*'''HSM''' - High School Math Forum<br />
*'''ID(R)K'''-I Don't (Really) Know<br />
*'''iff'''-If and only if<br />
*'''IIRC'''- If I recall correctly<br />
*'''IKR'''- I know right?<br />
*'''IMO'''- In my opinion (or International Math Olympiad, depending on context)<br />
*'''JMO'''- United States of America Junior Mathematical Olympiad<br />
*'''lol'''- Laugh Out Loud<br />
*'''MC'''- Mathcounts, a popular math contest for Middle School students.<br />
*'''mod(s)'''- Moderator(s)<br />
*'''MOEMS'''- Math Olympiads for Elementary and Middle Schools<br />
*'''MO(S)P'''- Mathematical Olympiad (Summer) Program<br />
*'''MSM'''- Middle School Math Forum<br />
*'''NT'''- Number Theory<br />
*'''OBC'''- Online by computer<br />
*'''OMG'''- Oh My Gosh<br />
*'''OP'''- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering<br />
*'''QED'''- Quod erat demonstrandum, Latin for "Which was to be proven"; some English mathematicians use it as an acronym for Quite Elegantly Done<br />
*'''QS&A'''- Questions, Suggestions, and Announcements Forum<br />
*'''ro(t)fl''' - Rolling on the floor laughing<br />
*'''smh''' - Shaking my head<br />
*'''sqrt''' - Square root<br />
*'''Sticky'''- A post pinned to the top of a forum - a thing no one reads but really should read<br />
*'''ToS'''- Terms of Service - a thing no one reads but really should read<br />
*'''USA(J)MO'''- USA (Junior) Mathematical Olympiad<br />
*'''V/LA'''- Vacation or Long Absence/Limited Access<br />
*'''WLOG'''- Without loss of generality<br />
*'''wrt'''- With respect to<br />
*'''wtg''' - Way to go<br />
*'''tytia'''- Thank you, that is all<br />
*'''xD'''- Bursting Laugh<br />
<br />
== FTW! ==<br />
<br />
Please see the [//artofproblemsolving.com/ftw/faq For the Win! FAQ] for many common questions.<br />
<br />
== School ==<br />
<br />
==== What if I miss a class? ====<br />
:There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. You can also ask questions on the class message board. Don't worry, though, classroom participation usually isn't weighted heavily.<br />
<br />
==== Is there audio or video in class? ====<br />
:There is generally no audio or video in the class. The classes are generally text and image based, in an interactive chat room environment, which allows students to ask questions at any time during the class. In addition to audio and video limiting interactivity and being less pedagogically effective, the technology isn't quite there yet for all students to be able to adequately receive streaming audio and video.<br />
<br />
==== What if I want to drop out of a class? ====<br />
:For any course with more than 2 classes, students can drop the course any time before the third class begins and receive a full refund. No drops are allowed after the third class has started. To drop the class, go to the My Classes section by clicking the My Classes link at the top-right of the website. Then find the area on the right side of the page that lets you drop the class. A refund will be processed within 10 business days.<br />
<br />
==== For my homework, there is suppose to be a green bar but it's orange, why? ====<br />
<br />
:For the bar to turn green, the writing problem must be attempted. Once it is attempted the bar will turn green.<br />
<br />
==== I need more time for my homework, what should I do? ====<br />
<br />
:There is a "Request Extension" button in the homework tab of your class. This will automatically extend the due date to 2 days after the normal deadline. If you want more time you need to ask for it in the little comment box, stating the reason why you want the extension, and how much time you want. This request will be looked at by the teachers and they will decide if you get the extension or not. Note that you can only use this button 3 times.<br />
:Otherwise, you can send an email to extensions@aops.com with your username, class name and ID, if known, and reason for extension. Someone should get back to you within a couple days.</div>Bad math