https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Flyhawkeye&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T15:13:06ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_21&diff=1011262017 AMC 10B Problems/Problem 212019-02-01T00:35:49Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>==Problem==<br />
In <math>\triangle ABC</math>, <math>AB=6</math>, <math>AC=8</math>, <math>BC=10</math>, and <math>D</math> is the midpoint of <math>\overline{BC}</math>. What is the sum of the radii of the circles inscribed in <math>\triangle ADB</math> and <math>\triangle ADC</math>?<br />
<br />
<math>\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3</math><br />
<br />
==Solution==<br />
We note that by the converse of the Pythagorean Theorem, <math>\triangle ABC</math> is a right triangle with a right angle at <math>A</math>. Therefore, <math>AD = BD = CD = 5</math>, and <math>[ADB] = [ADC] = 12</math>. Since <math>A = rs</math>, the inradius of <math>\triangle ADB</math> is <math>\frac{12}{(5+5+6)/2} = \frac 32</math>, and the inradius of <math>\triangle ADC</math> is <math>\frac{12}{(5+5+8)/2} = \frac 43</math>. Adding the two together, we have <math>\boxed{\textbf{(D) } \frac{17}6}</math>.<br />
<br />
==Solution(More explanation)==<br />
We have <math>\triangle ABC</math> a right triangle by dividing each side lengths by <math>2</math> to create a well known <math>3-4-5</math> triangle. We also can know that the median of a right triangle must be equal to half the hypotenuse. Using this property, we have <math>BD=5</math>, <math>CD=5</math>, and <math>AD=5</math>. Now, we can use the Heron's formula to get the area of <math>\triangle ABD</math> as <math>\sqrt{8(2)(3)(3)}=\sqrt{144}=12</math>. Afterward, we can apply this formula again on <math>\triangle ADC</math> to get the area as <math>\sqrt{9(4)(4)(1)}=\sqrt{144}=12</math>. Notice we want the inradius. We can use another property, which is <math>A=rs</math>. This states that <math>\text{Area}=\text{Radius}(\text{Semiperimeter})</math>. (This can be proved by connecting the center of the inscribed circles to the vertices and we can notice the inradius is just the heights of each of the three triangles divided) Finally, we can derive the radii of each inscribed circle. Plugging the semiperimeter and area into the formula, we have <math>12=8r</math> and <math>12=9r</math> for <math>\triangle ABD</math> and <math>\triangle ADC</math>, respectively. Simplifying, we have the radii lengths as <math>r=\frac{3}{2}</math> and <math>\frac{4}{3}</math>. We want the sum, so we have <math>\frac{17}{6}</math>, or <math>\boxed{\text{(D)}}</math> ~Solution by twinbrian<br />
<br />
==See Also==<br />
{{AMC10 box|year=2017|ab=B|num-b=20|num-a=22}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_14&diff=1011252017 AMC 10B Problems/Problem 142019-02-01T00:27:31Z<p>Flyhawkeye: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
An integer <math>N</math> is selected at random in the range <math>1\leq N \leq 2020</math> . What is the probability that the remainder when <math>N^{16}</math> is divided by <math>5</math> is <math>1</math>?<br />
<br />
<math>\textbf{(A)}\ \frac{1}{5}\qquad\textbf{(B)}\ \frac{2}{5}\qquad\textbf{(C)}\ \frac{3}{5}\qquad\textbf{(D)}\ \frac{4}{5}\qquad\textbf{(E)}\ 1</math><br />
<br />
==Solution 1==<br />
By Fermat's Little Theorem, <math>N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}</math> when N is relatively prime to 5. Hence, this happens with probability <math>\boxed{\textbf{(D) } \frac 45}</math>.<br />
<br />
==Solution 2==<br />
Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits <math>0-9</math> .<br />
The pattern for <math>0</math> is <math>0</math>, no matter what power, so <math>0</math> doesn't work. Likewise, the pattern for <math>5</math> is always <math>5</math>. Doing the same for the rest of the digits, we find that the units digits of <math>1^{16}</math>, <math>2^{16}</math> ,<math>3^{16}</math>, <math>4^{16}</math> ,<math>6^{16}</math>, <math>7^{16}</math> ,<math>8^{16}</math> and <math>9^{16}</math> all have the remainder of <math>1</math> when divided by <math>5</math>, so <math>\boxed{\textbf{(D) } \frac 45}</math>.<br />
<br />
==Solution 3 (Casework)==<br />
We can use modular arithmetic for each residue of <math>n \pmod 5</math><br />
<br />
<br />
<br />
If <math>n \equiv 0 \pmod 5</math>, then <math>n^{16} \equiv 0^{16} \equiv 0 \pmod 5</math><br />
<br />
<br />
If <math>n \equiv 1 \pmod 5</math>, then <math>n^{16} \equiv 1^{16} \equiv 1 \pmod 5</math><br />
<br />
<br />
If <math>n \equiv 2 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5</math><br />
<br />
<br />
If <math>n \equiv 3 \pmod 5</math>, then <math>n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5</math><br />
<br />
<br />
If <math>n \equiv 4 \pmod 5</math>, then <math>n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 5</math><br />
<br />
<br />
<br />
In <math>4</math> out of the <math>5</math> cases, the result was <math>1 \pmod 5</math>, and since each case occurs equally as <math>2020 \equiv 0 \pmod 5</math>, the answer is <math>\boxed{\textbf{(D) }\frac{4}{5}}</math><br />
<br />
<br />
{{AMC10 box|year=2017|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AMC&diff=100630Mock AMC2019-01-19T18:20:22Z<p>Flyhawkeye: /* Mock AMC 10 */ Added a Mock AMC 10</p>
<hr />
<div>A '''Mock AMC''' is a contest intended to mimic an actual [[AMC]] (American Mathematics Competitions 8, 10, or 12) exam. A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member and then administered for any of the other community members to take. Sometimes, the administrator may ask other people to sign up to write problems for the contest.<br />
<br />
Mock AMC's are usually very popular in the months leading up to the actual [[AMC]] competition. There is no guarantee that community members will make Mock AMC's in any given year, but it's usually a good bet that someone will.<br />
<br />
Feel free to remove mock tests which are not of good quality, or recommend others.<br />
<br />
Ongoing mock AMCs (as well as other mock contests in general) can now be found in the [https://artofproblemsolving.com/community/c594864_aops_mock_contests AoPS Mock Contests forum].<br />
<br />
== Tips for Writing a Mock AMC ==<br />
Anyone can write a Mock AMC and administer it. If you are interested in writing one, here are some tips:<br />
<br />
* Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.<br />
* Look at famous theorems and formulas and see if there's any way you can make a good problem out of them.<br />
* If you're running out of creative juice and decide to pull problems from contests, try using problems from obscure contests first, if possible. This way, even the more experienced test takers will hopefully find problems that they do not already know how to do.<br />
* Pair up with another user on AoPS and write it together. Two minds are much better than one. With just one person, the problems might be biased toward one subject, but with two people, the chances of this happening are smaller.<br />
<br />
== Past Mock AMCs ==<br />
<br />
Listed below are the [higher-quality] Mock AMCs which have been hosted over AoPS in the past. Feel free to add, remove, or recommend.<br />
<br />
Note that the "level" column represents the originally intended difficulty. In other words, if a person makes a mock [[AMC 12]], the level would be "12", even if the problems themselves are much easier. Recommended mock tests are starred.<br />
<br />
=== Mock AMC 12 ===<br />
<br />
{| class="wikitable" style="text-align:center;width:100%"<br />
|-<br />
|<br />
! scope="col" | '''Author'''<br />
! scope="col" | '''Year'''<br />
! scope="col" | '''Initial Discussion'''<br />
! scope="col" | '''Problems'''<br />
! scope="col" width=80 | '''Answers'''<br />
! scope="col" | '''Results/Discussion'''<br />
|-<br />
! scope="row" | '''Mock AMC #1'''<br />
| mathfanatic<br />
| 2003<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9321 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9353 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9573 6-10]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9574 11-15]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9575 16-20]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9576 21-25]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9365 Results]<br />
|-<br />
! scope="row" | '''Mock AMC #2'''<br />
| mathfanatic<br />
| 2004<br />
| [http://www.artofproblemsolving.com/community/c5h10497p67837 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Solutions]<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC A'''<br />
| JSRosen3<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14138 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14361 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=102483#p102483 Answers] [http://www.artofproblemsolving.com/community/c5h14516 Solutions]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14489 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC B'''<br />
| beta<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14735 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14764 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14894 Answers] [http://www.artofproblemsolving.com/community/c5h14884 Solutions]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=105741#p105741 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC C'''<br />
| JGeneson<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15001 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15134 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC D'''<br />
| joml88<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16886 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17888 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC E'''<br />
| Silverfalcon<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21997 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22141 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC F'''<br />
| joml88<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22049 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23163 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC G'''<br />
| Lucky707<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24355 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24974 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25087 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h25087p157983 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC H'''<br />
| Silverfalcon<br />
| 2005<br />
| [http://www.artofproblemsolving.com/community/c5h24437p154514 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h24437p154514 Problems]<br />
| throughout thread<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC I'''<br />
| white_horse_king88<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21280 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25181 Problems]<br />
| n/a<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC J'''<br />
| Silverfalcon<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47625 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48129 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=304246#p304246 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48132 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC K'''<br />
| amirhtlusa<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49958 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50515 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC L'''<br />
| amirhtlusa<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61330 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63041 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC M'''<br />
| Silverfalcon<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63542 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78982 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC N'''<br />
| chess64<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98894 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99307 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Results] / [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99566 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC O'''<br />
| mustafa<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121312 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122126 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC P'''<br />
| Anirudh<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=709655#p709655 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716240#p716240 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716262#p716262 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC Q'''<br />
| calc rulz<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125194 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125886 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC R'''<br />
| rnwang2<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126107 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=715597#p715597 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC S'''<br />
| mysmartmouth<br />
| 2007<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127221 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128689 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC T'''<br />
| paladin8<br />
| 2007<br />
| [http://www.artofproblemsolving.com/community/c5h127979p726003 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127979 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC U'''<br />
| Silverfalcon<br />
| 2008<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184067 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185233 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185236 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC V'''<br />
| gfour84<br />
| 2009<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298452 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1634175#p1634175 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637429#p1637429 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637006#p1637006 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302030 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 12'''<br />
| MathTwo<br />
| 2010<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319184 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1759276#p1759276 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778080#p1778080 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328510 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 2/11'''<br />
| Caelestor<br />
| 2011<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2183293#p2183293 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 1/12'''<br />
| Lord.of.AMC<br />
| 2012<br />
| [http://www.artofproblemsolving.com/community/c5h456321p2563731 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2605537#p2605537 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 12'''<br />
| Diehard<br />
| 2012<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459149 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2580327#p2580327 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 12 2012'''<br />
| python123<br />
| 2012<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=456256 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2576205#p2576205 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h456256p2580610 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h456256p2580610 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC A'''<br />
| Binomial-theorem<br />
| 2012<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=473867 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2692082#p2692082 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2717578#p2717578 Solutions]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2707954#p2707954 Results]<br />
|-<br />
! scope = "row" | '''Almost 2016 Mock AMC 11.5'''<br />
| whatshisbucket<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Problems]<br />
| [http://artofproblemsolving.com/community/c209194_almost_2016_mock_amc_11.5 Solutions]<br />
| [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Results / Discussion]<br />
|-<br />
! scope="row" | '''hnkevin42 Mock AMC 12'''<br />
| hnkevin42<br />
| 2016<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5913294 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Results/Discussion]<br />
|-<br />
! scope="row" | '''Last-Minute Mock AMC 12'''<br />
| CountofMC<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Problems]<br />
| [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Answers]<br />
| [https://artofproblemsolving.com/community/c4t334444f4_lastminute_mock_amc_12 Results/Discussion]<br />
|}<br />
<br />
=== Mock AMC 10 ===<br />
<br />
{| class="wikitable" style="text-align:center;width:100%"<br />
|-<br />
|<br />
! scope="col" | '''Author'''<br />
! scope="col" | '''Year'''<br />
! scope="col" | '''Initial Discussion'''<br />
! scope="col" | '''Problems'''<br />
! scope="col" | '''Answers'''<br />
! scope="col" | '''Results/Discussion'''<br />
|-<br />
! scope="row" | '''Mock AMC 10'''<br />
| #H34N1<br />
| 2008<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212730 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214081 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213727 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213732 Results/Discussion]<br />
|-<br />
! scope="row" | '''agent's Mock AMC 10'''<br />
| agentcx<br />
| 2009<br />
| [http://www.artofproblemsolving.com/community/c5h331823p1775678 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1781208#p1781208 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/community/c5h331823p1782769 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 10 Set'''<br />
| AwesomeToad<br />
| 2010<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311120 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1762829#p1762829 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778740#p1778740 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 10/12'''<br />
| djmathman<br />
| 2013<br />
| [http://www.artofproblemsolving.com/community/c5h556673s1_first_mock_amcs_of_the_20132014_season Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h556673p3294854 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h556673p3324794 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h556673p3324794 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 10 2014-2015*'''<br />
| AlcumusGuy<br />
| 2014<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=618080&hilit=AlcumusGuy%27s+mock+amc+10 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=618080&hilit=AlcumusGuy%27s+mock+amc+10 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h618080p3710795 Answers/Results]<br />
| [http://artofproblemsolving.com/community/c56018 Problem Discussion]<br />
|-<br />
! scope="row" | '''Kelvin the Frog v2015'''<br />
| BOGTRO<br />
| 2015<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=623373&hilit=Kelvin+the+frog Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=623373&hilit=Kelvin+the+frog Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 10/12'''<br />
| joey8189681<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h624714p3741743 Initial Discussion]<br />
| [https://www.dropbox.com/s/1c8hhl6yt2awpix/Mock%20AMC%2010.pdf?dl=0 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h624714p3754593 Answers]<br />
| [http://www.artofproblemsolving.com/community/q2h626092p3756574 Results]<br />
|-<br />
! scope="row" | '''May Mock AMC 10 Contest'''<br />
| azmath333<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Answers] [http://www.artofproblemsolving.com/community/c5h1082684p4825662 Solutions]<br />
| [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Results / Discussion] <br />
|-<br />
! scope="row" | '''2015 Mock AMC 10*'''<br />
| droid347<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p5051006 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p5045295 Results / Discussion]<br />
|-<br />
! scope="row" | '''July Mock AMC 10 Contest'''<br />
| akshaygowrish<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1104125_july_mock_amc_10_contest Initial Discussion]<br />
| [https://docs.google.com/document/d/12OXd-mmS_T7SyMcqw1hfvykNlcj-4TR4mMXlRJ09uBg/edit?usp=sharing Problems]<br />
| [https://docs.google.com/document/d/1-Q_w5XNKcRaffvpCLSdoJ51TpRpcDAgYLmV9OlczBCs/edit?usp=sharing Answer Key]<br />
| [http://artofproblemsolving.com/community/c5h1104125p5154283 Results / Discussion]<br />
|-<br />
! scope = "row" |'''August Mock AMC 10'''<br />
| azmath333<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1115462_august_mock_amc_10 Initial Discussion]<br />
| [http://latex.artofproblemsolving.com/miscpdf/augustmockamc10.pdf Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Results] / [http://www.artofproblemsolving.com/community/c121880_august_mock_amc_10_discussion_forum Discussion]<br />
|-<br />
! scope = "row" | '''September Mock AMC 10'''<br />
| cpma213<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1137447p5320386 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1137447p5320386 Problems]<br />
| n/a<br />
| [http://artofproblemsolving.com/community/c5h1137447p5479954 Results / Discussion]<br />
|-<br />
! scope = "row" | '''December 2015 Mock AMC 1^3*(3+7)'''<br />
| mathisawesome2169<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1173465p5647504 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1173465p5647504 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/community/c5h1173465p5707101 Results / Discussion]<br />
|-<br />
! scope = "row" | '''New Year's Mock AMC10*'''<br />
| checkmatetang<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1171366p5627439 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1171366p5627439 Problems]<br />
| [http://artofproblemsolving.com/community/c5h1171366p5711384 Answers]<br />
| [http://artofproblemsolving.com/community/c5h1171366p5711384 Results] / [http://artofproblemsolving.com/community/c202656_new_years_mock_amc10_discussion Discussion]<br />
|-<br />
! scope = "row" | '''2015-2016 Mock AMC 10*'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1168398p5595105 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1168398p5595105 Problems]<br />
| [http://artofproblemsolving.com/community/c147536h1182005_mock_amc10_wrapup Answers]<br />
| [http://artofproblemsolving.com/community/c5h1168398p5595105 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 10 2015-2016*'''<br />
| AlcumusGuy<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1177413p5687990 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1177413_mock_amc_10_20152016_released Problems]<br />
| [http://artofproblemsolving.com/community/c212844_mock_amc_10_20152016 Solutions]<br />
| [http://artofproblemsolving.com/community/c5h1177413p5769839 Results / Discussion]<br />
|-<br />
! scope = "row" | '''January 2016 Mock AMC 10*'''<br />
| atmchallenge<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Results/Discussion]<br />
|-<br />
! scope = "row" | '''2015-2016 Mock AMC 10* #2'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1188467 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1188467 Problems]<br />
| [http://artofproblemsolving.com/community/c147536h1175930 Solutions]<br />
| [http://artofproblemsolving.com/community/c5h1188467 Results / Discussion]<br />
|-<br />
<br />
! scope="row" | '''MeepyMeepMeep Mock AMC 10'''<br />
| MeepyMeepMeep, speck<br />
| 2016<br />
| [http://artofproblemsolving.com/community/c5h1195432p5852002 Initial Discussion]<br />
| [https://www.dropbox.com/s/uw0herhuqk8zbp5/Mock%20AMC%2010.pdf?dl=0 Problems]<br />
| n/a<br />
| n/a<br />
|- <br />
<br />
!scope="row" | eisirrational's Mock AMC 10<br />
| eisirrational, illogical_21, whatshisbucket<br />
| 2017<br />
| [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Problems]<br />
| [http://artofproblemsolving.com/community/c418591h1392426p7811150 Answer Key]<br />
| [https://artofproblemsolving.com/community/c418591_mock_amc_10_discussion Discussion]<br />
|-<br />
<br />
!scope="row" | Summer Mock AMC 10<br />
| Rowechen, OmicronGamma, FedeX333X, KenV, kvedula2004, ItzVineeth<br />
| 2017<br />
| [https://artofproblemsolving.com/community/c5h1471834_summer_mock_amc_series Initial Discussion]<br />
| [https://drive.google.com/file/d/0B3M-fxa6QG0_ODNpSnJZcEt3djQ/view Problems]<br />
| [http://latex.artofproblemsolving.com/miscpdf/hehdrwpi.pdf?t=1500270651030 Answers]<br />
| [https://artofproblemsolving.com/community/c481237_2017_summer_mock_amc_10_discussion_forum Statistics / Discussion]<br />
|-<br />
<br />
!scope="row" | scrabbler94's Mock AMC 10<br />
| scrabbler94<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Problems]<br />
| [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Answers]<br />
| [http://artofproblemsolving.com/community/c5h1569848p9704902 Results] / [https://artofproblemsolving.com/community/c593716_scrabbler94s_mock_amc_10_discussion Discussion]<br />
|-<br />
<br />
!scope="row" | 2018 AMC 10C (A Mock AMC 10)<br />
| QIDb602<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Problems]<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Answers]<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Results] / [https://artofproblemsolving.com/community/c630141 Discussion]<br />
|-<br />
<br />
! scope="row" | '''2018 Mock AMC 10'''<br />
| blue8931<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1606066 Initial Discussion]<br />
| [https://artofproblemsolving.com/downloads/printable_post_collections/628064 Problems]<br />
| [https://artofproblemsolving.com/community/c628116h1632436_official_answer_key Answer Key]<br />
| [https://artofproblemsolving.com/community/c628116_2018_mock_amc_10_discussion_forum Results / Discussion]<br />
|-<br />
<br />
!scope="row" | 2018 Memorial Day Mock AMC 10<br />
| QIDb602<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Initial Discussion]<br />
| [https://drive.google.com/file/d/1paAHspy5PH1J9fMoNYTb7cIEH3Di72K5/view Problems]<br />
| [https://drive.google.com/file/d/1tipZuHE11jmmOfnSDnhkwKnvIOnbsbaL/view?usp=sharing Answers]<br />
| [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Results] / [https://artofproblemsolving.com/community/c671484 Discussion]<br />
|-<br />
<br />
!scope="row" | Autumn Mock AMC 10<br />
| Krypton36, AlastorMoody, alphaone001, dchenmathcounts, InternetPerson10, kootrapali, mathdragon2000, MathGeek2018, Stormersyle<br />
| 2018<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11032970 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11032970 Problems]<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11331788 Answers]<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11332566 Results] / [https://artofproblemsolving.com/community/c761744_autumn_mock_amc_10_discussion_forum_d Discussion]<br />
|-<br />
<br />
!scope="row" | 2018 December Mock AMC 10<br />
| mathchampion1, kcbhatraju, kootrapali, chocolatelover111<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Problems]<br />
| [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Answers]<br />
| [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Results] / [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Discussion]<br />
|-<br />
<br />
|}<br />
<br />
=== Mock AMC 8 ===<br />
<br />
{| class="wikitable" style="text-align:center;width:100%"<br />
|-<br />
|<br />
! scope="col" | '''Author'''<br />
! scope="col" | '''Year'''<br />
! scope="col" | '''Initial Discussion'''<br />
! scope="col" | '''Problems'''<br />
! scope="col" width=80 | '''Answers'''<br />
! scope="col" | '''Results/Discussion'''<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| mathfanatic<br />
| 2004<br />
| [http://www.artofproblemsolving.com/community/c5h15878p111575 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15878 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114070#p114070 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h15878p114404 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| 13375P34K43V312<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107507 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108455 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=630802#p630802 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h108455p623522 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| ahaanomegas<br />
| 2012<br />
| [http://www.artofproblemsolving.com/community/c5h459346p2577870 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2598111#p2598111 Problems]<br />
| n/a<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| utahjazz<br />
| 2012<br />
| [http://www.artofproblemsolving.com/community/c5h485041p2717878 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728186#p2728186 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2731391#p2731391 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| iNomOnCountdown<br />
| 2014<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&hilit=iNomonCountdown%27s+mock+amc+8 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&hilit=iNomonCountdown%27s+mock+amc+8 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&start=0 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&start=0 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8'''<br />
| Tan<br />
| 2014<br />
| [http://www.artofproblemsolving.com/community/c5h613297p3648226 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h613297p3648227 Problems]<br />
| [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Answers]<br />
| [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Results / Discussion]<br />
|-<br />
! scope = "row" | '''2015 Hard Mock AMC 8*'''<br />
| Not_a_Username, 8invalid8<br />
| 2015<br />
| [http://artofproblemsolving.com/community/q1h1128391p5227860 Initial Discussion]<br />
| [http://latex.artofproblemsolving.com/miscpdf/kashimyo.pdf Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c3h1152332p5457736 Initial Discussion]<br />
| [http://artofproblemsolving.com/downloads/printable_post_collections/163667 Problems]<br />
| [http://artofproblemsolving.com/community/category-admin/165328 Forum]<br />
| [http://artofproblemsolving.com/community/c165328h1155882p5484131 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8 #2'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c3h1167425p5586244 Initial Discussion]<br />
| [http://artofproblemsolving.com/downloads/printable_post_collections/165838 Problems]<br />
| [http://artofproblemsolving.com/community/c147536_the_mock_amc_8_forum Forum]<br />
| [http://artofproblemsolving.com/community/c3h1167425p5586244 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8 2015'''<br />
| Alberty44<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c194276_discusssion Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1168822p5674003 Problems]<br />
| n/a<br />
| [http://artofproblemsolving.com/community/c194276h1175928p5673996 Results/Discussion]<br />
|-<br />
! scope = "row" | '''Christmas AMC 8'''<br />
| Mudkipswims42<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1177489p5688588 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1177489p5688588 Problems]<br />
| [http://www.artofproblemsolving.com/community/c229455_christmas_amc8_discussion_forum Forum]<br />
| [http://artofproblemsolving.com/community/c5h1177489p5880200 Results/Discussion]<br />
|- <br />
<br />
! scope = "row" | '''Mock AMC 8!'''<br />
| eisirrational, pretzel, AOPS12142015, Th3Numb3rThr33, e_power_pi_times_i<br />
| 2017<br />
| [https://artofproblemsolving.com/community/c5h1526187_mock_amc_8 Initial Discussion]<br />
| [http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi84LzFkNzUyMWYwZTViOTQ1MzRmZmU3ODE0NmI2MzIzMGUxNjA2MTcwLnBkZg==&rn=TW9jayBBTUMgOCB2Ny5wZGY= Problems]<br />
| [http://artofproblemsolving.com/community/c3h1545004p9368333 Answer Key/Solutions]<br />
| [https://artofproblemsolving.com/community/c561318_mock_amc_8_2017_discussion Discussion]<br />
|-<br />
<br />
! scope = "row" | '''Mock Summ(er)ation AMC 8!'''<br />
| mathchamp1, kevinmathz, Gali, mathdragon2000, reddragon644, ShreyJ, Quadrastic, Mathnerd1223334444<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1672811_2018_summeration_mock_amc_8 Initial Discussion]<br />
| [https://math-adventures.weebly.com/uploads/1/1/8/8/118819793/amc_8_problems.pdf Problems]<br />
| n/a<br />
| n/a<br />
|-<br />
<br />
! scope = "row" | '''popcorn1's AMC 8 2018'''<br />
| popcorn1<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1715072_popcorn1s_amc_8_2018 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c594864h1715072p11074872 Problems]<br />
| Answer Key / Solutions - not released<br />
| Discussion - not released<br />
|-<br />
<br />
! scope = "row" | ''' Stormersyle's Mock AMC 8<br />
| Stormersyle <br />
| 2018<br />
| [http://artofproblemsolving.com/community/c594864h1722790p11146826 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c594864h1722790p11146826 Problems]<br />
| [https://latex.artofproblemsolving.com/miscpdf/qgdafjlp.pdf?t=1544494957792 Answer Key/Solutions]<br />
| Discussion - not released<br />
|-<br />
<br />
! scope = "row" | '''mathchampion1's Christmas Mock AMC 8'''<br />
| mathchampion1<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Problems]<br />
| Answer Key / Solutions - not released<br />
| Discussion - not released<br />
|-}<br />
<br />
== See also ==<br />
* [[American Mathematics Competitions]]<br />
* [[Math books]]<br />
* [[Mathematics competitions]]<br />
* [[Mock AIME]]<br />
* [[Mock MathCounts]]<br />
* [[Mock USAMO]]<br />
* [[Mock USAJMO]]<br />
* [[Resources for mathematics competitions]]<br />
* [[AoPS Past Contests]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=Mock_AMC&diff=100629Mock AMC2019-01-19T18:18:47Z<p>Flyhawkeye: /* Mock AMC 12 */ Removed a mock AMC 10</p>
<hr />
<div>A '''Mock AMC''' is a contest intended to mimic an actual [[AMC]] (American Mathematics Competitions 8, 10, or 12) exam. A number of '''Mock AMC''' competitions have been hosted on the [[Art of Problem Solving]] message boards. They are generally made by one community member and then administered for any of the other community members to take. Sometimes, the administrator may ask other people to sign up to write problems for the contest.<br />
<br />
Mock AMC's are usually very popular in the months leading up to the actual [[AMC]] competition. There is no guarantee that community members will make Mock AMC's in any given year, but it's usually a good bet that someone will.<br />
<br />
Feel free to remove mock tests which are not of good quality, or recommend others.<br />
<br />
Ongoing mock AMCs (as well as other mock contests in general) can now be found in the [https://artofproblemsolving.com/community/c594864_aops_mock_contests AoPS Mock Contests forum].<br />
<br />
== Tips for Writing a Mock AMC ==<br />
Anyone can write a Mock AMC and administer it. If you are interested in writing one, here are some tips:<br />
<br />
* Look at past [[AMC]]/[[AHSME]] tests to get a feel for what kind of problems you should write and what difficulty level they should be.<br />
* Look at famous theorems and formulas and see if there's any way you can make a good problem out of them.<br />
* If you're running out of creative juice and decide to pull problems from contests, try using problems from obscure contests first, if possible. This way, even the more experienced test takers will hopefully find problems that they do not already know how to do.<br />
* Pair up with another user on AoPS and write it together. Two minds are much better than one. With just one person, the problems might be biased toward one subject, but with two people, the chances of this happening are smaller.<br />
<br />
== Past Mock AMCs ==<br />
<br />
Listed below are the [higher-quality] Mock AMCs which have been hosted over AoPS in the past. Feel free to add, remove, or recommend.<br />
<br />
Note that the "level" column represents the originally intended difficulty. In other words, if a person makes a mock [[AMC 12]], the level would be "12", even if the problems themselves are much easier. Recommended mock tests are starred.<br />
<br />
=== Mock AMC 12 ===<br />
<br />
{| class="wikitable" style="text-align:center;width:100%"<br />
|-<br />
|<br />
! scope="col" | '''Author'''<br />
! scope="col" | '''Year'''<br />
! scope="col" | '''Initial Discussion'''<br />
! scope="col" | '''Problems'''<br />
! scope="col" width=80 | '''Answers'''<br />
! scope="col" | '''Results/Discussion'''<br />
|-<br />
! scope="row" | '''Mock AMC #1'''<br />
| mathfanatic<br />
| 2003<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9321 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9353 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9573 6-10]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9574 11-15]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9575 16-20]<br />
[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9576 21-25]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9365 Results]<br />
|-<br />
! scope="row" | '''Mock AMC #2'''<br />
| mathfanatic<br />
| 2004<br />
| [http://www.artofproblemsolving.com/community/c5h10497p67837 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10497 Solutions]<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC A'''<br />
| JSRosen3<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14138 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14361 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=102483#p102483 Answers] [http://www.artofproblemsolving.com/community/c5h14516 Solutions]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14489 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC B'''<br />
| beta<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14735 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14764 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14894 Answers] [http://www.artofproblemsolving.com/community/c5h14884 Solutions]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=105741#p105741 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC C'''<br />
| JGeneson<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15001 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15134 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15251 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC D'''<br />
| joml88<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16886 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17888 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17891 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC E'''<br />
| Silverfalcon<br />
| 2004<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21997 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22141 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22344 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC F'''<br />
| joml88<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22049 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23163 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23177 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC G'''<br />
| Lucky707<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24355 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24974 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25087 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h25087p157983 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC H'''<br />
| Silverfalcon<br />
| 2005<br />
| [http://www.artofproblemsolving.com/community/c5h24437p154514 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h24437p154514 Problems]<br />
| throughout thread<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC I'''<br />
| white_horse_king88<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21280 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25181 Problems]<br />
| n/a<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC J'''<br />
| Silverfalcon<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47625 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48129 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=304246#p304246 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=48132 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC K'''<br />
| amirhtlusa<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49958 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50515 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=321102#p321102 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC L'''<br />
| amirhtlusa<br />
| 2005<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61330 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63041 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63258 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC M'''<br />
| Silverfalcon<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63542 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78982 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=79749 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC N'''<br />
| chess64<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98894 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99307 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99344 Results] / [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=99566 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC O'''<br />
| mustafa<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121312 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122126 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126071 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC P'''<br />
| Anirudh<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=709655#p709655 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716240#p716240 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=716262#p716262 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC Q'''<br />
| calc rulz<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125194 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125886 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128102 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC R'''<br />
| rnwang2<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=126107 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=715597#p715597 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127421 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC S'''<br />
| mysmartmouth<br />
| 2007<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127221 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128689 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131403 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC T'''<br />
| paladin8<br />
| 2007<br />
| [http://www.artofproblemsolving.com/community/c5h127979p726003 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=127979 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129159 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC U'''<br />
| Silverfalcon<br />
| 2008<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=184067 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185233 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185236 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185238 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC V'''<br />
| gfour84<br />
| 2009<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=298452 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1634175#p1634175 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637429#p1637429 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1637006#p1637006 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=302030 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 12'''<br />
| MathTwo<br />
| 2010<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=319184 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1759276#p1759276 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778080#p1778080 Results]/[http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328510 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 2/11'''<br />
| Caelestor<br />
| 2011<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2183293#p2183293 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=390907 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 1/12'''<br />
| Lord.of.AMC<br />
| 2012<br />
| [http://www.artofproblemsolving.com/community/c5h456321p2563731 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2605537#p2605537 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2610048#p2610048 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 12'''<br />
| Diehard<br />
| 2012<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459149 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2580327#p2580327 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=459710 Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 12 2012'''<br />
| python123<br />
| 2012<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=456256 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2576205#p2576205 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h456256p2580610 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h456256p2580610 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC A'''<br />
| Binomial-theorem<br />
| 2012<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=473867 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2692082#p2692082 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2717578#p2717578 Solutions]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2707954#p2707954 Results]<br />
|-<br />
! scope = "row" | '''Almost 2016 Mock AMC 11.5'''<br />
| whatshisbucket<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Problems]<br />
| [http://artofproblemsolving.com/community/c209194_almost_2016_mock_amc_11.5 Solutions]<br />
| [http://www.artofproblemsolving.com/community/c5h1175087p5665743 Results / Discussion]<br />
|-<br />
! scope="row" | '''hnkevin42 Mock AMC 12'''<br />
| hnkevin42<br />
| 2016<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5913294 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1187484p5778754 Results/Discussion]<br />
|-<br />
! scope="row" | '''Last-Minute Mock AMC 12'''<br />
| CountofMC<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Problems]<br />
| [https://artofproblemsolving.com/community/c594864h1582044_lastminute_mock_amc_12 Answers]<br />
| [https://artofproblemsolving.com/community/c4t334444f4_lastminute_mock_amc_12 Results/Discussion]<br />
|}<br />
<br />
=== Mock AMC 10 ===<br />
<br />
{| class="wikitable" style="text-align:center;width:100%"<br />
|-<br />
|<br />
! scope="col" | '''Author'''<br />
! scope="col" | '''Year'''<br />
! scope="col" | '''Initial Discussion'''<br />
! scope="col" | '''Problems'''<br />
! scope="col" | '''Answers'''<br />
! scope="col" | '''Results/Discussion'''<br />
|-<br />
! scope="row" | '''Mock AMC 10'''<br />
| #H34N1<br />
| 2008<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212730 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214081 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213727 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213732 Results/Discussion]<br />
|-<br />
! scope="row" | '''agent's Mock AMC 10'''<br />
| agentcx<br />
| 2009<br />
| [http://www.artofproblemsolving.com/community/c5h331823p1775678 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1781208#p1781208 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/community/c5h331823p1782769 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 10 Set'''<br />
| AwesomeToad<br />
| 2010<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=311120 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1762829#p1762829 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1778740#p1778740 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 10/12'''<br />
| djmathman<br />
| 2013<br />
| [http://www.artofproblemsolving.com/community/c5h556673s1_first_mock_amcs_of_the_20132014_season Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h556673p3294854 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h556673p3324794 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h556673p3324794 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 10 2014-2015*'''<br />
| AlcumusGuy<br />
| 2014<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=618080&hilit=AlcumusGuy%27s+mock+amc+10 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=618080&hilit=AlcumusGuy%27s+mock+amc+10 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h618080p3710795 Answers/Results]<br />
| [http://artofproblemsolving.com/community/c56018 Problem Discussion]<br />
|-<br />
! scope="row" | '''Kelvin the Frog v2015'''<br />
| BOGTRO<br />
| 2015<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=623373&hilit=Kelvin+the+frog Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=133&t=623373&hilit=Kelvin+the+frog Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3731642#p3731642 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 10/12'''<br />
| joey8189681<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h624714p3741743 Initial Discussion]<br />
| [https://www.dropbox.com/s/1c8hhl6yt2awpix/Mock%20AMC%2010.pdf?dl=0 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h624714p3754593 Answers]<br />
| [http://www.artofproblemsolving.com/community/q2h626092p3756574 Results]<br />
|-<br />
! scope="row" | '''May Mock AMC 10 Contest'''<br />
| azmath333<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1082684_mock_amc_1012_contest Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Answers] [http://www.artofproblemsolving.com/community/c5h1082684p4825662 Solutions]<br />
| [http://www.artofproblemsolving.com/community/c5h1082684p4812575 Results / Discussion] <br />
|-<br />
! scope="row" | '''2015 Mock AMC 10*'''<br />
| droid347<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p4893513 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p5051006 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1094505p5045295 Results / Discussion]<br />
|-<br />
! scope="row" | '''July Mock AMC 10 Contest'''<br />
| akshaygowrish<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1104125_july_mock_amc_10_contest Initial Discussion]<br />
| [https://docs.google.com/document/d/12OXd-mmS_T7SyMcqw1hfvykNlcj-4TR4mMXlRJ09uBg/edit?usp=sharing Problems]<br />
| [https://docs.google.com/document/d/1-Q_w5XNKcRaffvpCLSdoJ51TpRpcDAgYLmV9OlczBCs/edit?usp=sharing Answer Key]<br />
| [http://artofproblemsolving.com/community/c5h1104125p5154283 Results / Discussion]<br />
|-<br />
! scope = "row" |'''August Mock AMC 10'''<br />
| azmath333<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1115462_august_mock_amc_10 Initial Discussion]<br />
| [http://latex.artofproblemsolving.com/miscpdf/augustmockamc10.pdf Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1115462p5270430 Results] / [http://www.artofproblemsolving.com/community/c121880_august_mock_amc_10_discussion_forum Discussion]<br />
|-<br />
! scope = "row" | '''September Mock AMC 10'''<br />
| cpma213<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1137447p5320386 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1137447p5320386 Problems]<br />
| n/a<br />
| [http://artofproblemsolving.com/community/c5h1137447p5479954 Results / Discussion]<br />
|-<br />
! scope = "row" | '''New Year's Mock AMC10*'''<br />
| checkmatetang<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1171366p5627439 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1171366p5627439 Problems]<br />
| [http://artofproblemsolving.com/community/c5h1171366p5711384 Answers]<br />
| [http://artofproblemsolving.com/community/c5h1171366p5711384 Results] / [http://artofproblemsolving.com/community/c202656_new_years_mock_amc10_discussion Discussion]<br />
|-<br />
! scope = "row" | '''2015-2016 Mock AMC 10*'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1168398p5595105 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1168398p5595105 Problems]<br />
| [http://artofproblemsolving.com/community/c147536h1182005_mock_amc10_wrapup Answers]<br />
| [http://artofproblemsolving.com/community/c5h1168398p5595105 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 10 2015-2016*'''<br />
| AlcumusGuy<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1177413p5687990 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1177413_mock_amc_10_20152016_released Problems]<br />
| [http://artofproblemsolving.com/community/c212844_mock_amc_10_20152016 Solutions]<br />
| [http://artofproblemsolving.com/community/c5h1177413p5769839 Results / Discussion]<br />
|-<br />
! scope = "row" | '''January 2016 Mock AMC 10*'''<br />
| atmchallenge<br />
| 2015<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5700377 Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1178971p5838468 Results/Discussion]<br />
|-<br />
! scope = "row" | '''2015-2016 Mock AMC 10* #2'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1188467 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1188467 Problems]<br />
| [http://artofproblemsolving.com/community/c147536h1175930 Solutions]<br />
| [http://artofproblemsolving.com/community/c5h1188467 Results / Discussion]<br />
|-<br />
<br />
! scope="row" | '''MeepyMeepMeep Mock AMC 10'''<br />
| MeepyMeepMeep, speck<br />
| 2016<br />
| [http://artofproblemsolving.com/community/c5h1195432p5852002 Initial Discussion]<br />
| [https://www.dropbox.com/s/uw0herhuqk8zbp5/Mock%20AMC%2010.pdf?dl=0 Problems]<br />
| n/a<br />
| n/a<br />
|- <br />
<br />
!scope="row" | eisirrational's Mock AMC 10<br />
| eisirrational, illogical_21, whatshisbucket<br />
| 2017<br />
| [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c5h1372991_eisirrationals_mock_amc_10 Problems]<br />
| [http://artofproblemsolving.com/community/c418591h1392426p7811150 Answer Key]<br />
| [https://artofproblemsolving.com/community/c418591_mock_amc_10_discussion Discussion]<br />
|-<br />
<br />
!scope="row" | Summer Mock AMC 10<br />
| Rowechen, OmicronGamma, FedeX333X, KenV, kvedula2004, ItzVineeth<br />
| 2017<br />
| [https://artofproblemsolving.com/community/c5h1471834_summer_mock_amc_series Initial Discussion]<br />
| [https://drive.google.com/file/d/0B3M-fxa6QG0_ODNpSnJZcEt3djQ/view Problems]<br />
| [http://latex.artofproblemsolving.com/miscpdf/hehdrwpi.pdf?t=1500270651030 Answers]<br />
| [https://artofproblemsolving.com/community/c481237_2017_summer_mock_amc_10_discussion_forum Statistics / Discussion]<br />
|-<br />
<br />
!scope="row" | scrabbler94's Mock AMC 10<br />
| scrabbler94<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Problems]<br />
| [https://artofproblemsolving.com/community/c5h1569848_mock_2018_amc_10_test_scrabbler94 Answers]<br />
| [http://artofproblemsolving.com/community/c5h1569848p9704902 Results] / [https://artofproblemsolving.com/community/c593716_scrabbler94s_mock_amc_10_discussion Discussion]<br />
|-<br />
<br />
!scope="row" | 2018 AMC 10C (A Mock AMC 10)<br />
| QIDb602<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Problems]<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Answers]<br />
| [https://artofproblemsolving.com/community/c594864h1596208_released_2018_amc_10c_a_mock_amc_10 Results] / [https://artofproblemsolving.com/community/c630141 Discussion]<br />
|-<br />
<br />
! scope="row" | '''2018 Mock AMC 10'''<br />
| blue8931<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1606066 Initial Discussion]<br />
| [https://artofproblemsolving.com/downloads/printable_post_collections/628064 Problems]<br />
| [https://artofproblemsolving.com/community/c628116h1632436_official_answer_key Answer Key]<br />
| [https://artofproblemsolving.com/community/c628116_2018_mock_amc_10_discussion_forum Results / Discussion]<br />
|-<br />
<br />
!scope="row" | 2018 Memorial Day Mock AMC 10<br />
| QIDb602<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Initial Discussion]<br />
| [https://drive.google.com/file/d/1paAHspy5PH1J9fMoNYTb7cIEH3Di72K5/view Problems]<br />
| [https://drive.google.com/file/d/1tipZuHE11jmmOfnSDnhkwKnvIOnbsbaL/view?usp=sharing Answers]<br />
| [https://artofproblemsolving.com/community/c594864h1645806_released_2018_memorial_day_mock_amc_10 Results] / [https://artofproblemsolving.com/community/c671484 Discussion]<br />
|-<br />
<br />
!scope="row" | Autumn Mock AMC 10<br />
| Krypton36, AlastorMoody, alphaone001, dchenmathcounts, InternetPerson10, kootrapali, mathdragon2000, MathGeek2018, Stormersyle<br />
| 2018<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11032970 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11032970 Problems]<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11331788 Answers]<br />
| [http://artofproblemsolving.com/community/c594864h1710983p11332566 Results] / [https://artofproblemsolving.com/community/c761744_autumn_mock_amc_10_discussion_forum_d Discussion]<br />
|-<br />
<br />
!scope="row" | 2018 December Mock AMC 10<br />
| mathchampion1, kcbhatraju, kootrapali, chocolatelover111<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1743064_2018_december_mock_amc_10_released Problems]<br />
| [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Answers]<br />
| [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Results] / [https://artofproblemsolving.com/community/c784638_december_mock_amc_10_2018_discussion_forum Discussion]<br />
|-<br />
<br />
|}<br />
<br />
=== Mock AMC 8 ===<br />
<br />
{| class="wikitable" style="text-align:center;width:100%"<br />
|-<br />
|<br />
! scope="col" | '''Author'''<br />
! scope="col" | '''Year'''<br />
! scope="col" | '''Initial Discussion'''<br />
! scope="col" | '''Problems'''<br />
! scope="col" width=80 | '''Answers'''<br />
! scope="col" | '''Results/Discussion'''<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| mathfanatic<br />
| 2004<br />
| [http://www.artofproblemsolving.com/community/c5h15878p111575 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15878 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114070#p114070 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h15878p114404 Results / Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| 13375P34K43V312<br />
| 2006<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=107507 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108455 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=630802#p630802 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h108455p623522 Results/Discussion]<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| ahaanomegas<br />
| 2012<br />
| [http://www.artofproblemsolving.com/community/c5h459346p2577870 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2598111#p2598111 Problems]<br />
| n/a<br />
| n/a<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| utahjazz<br />
| 2012<br />
| [http://www.artofproblemsolving.com/community/c5h485041p2717878 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2728186#p2728186 Problems]<br />
| n/a<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2731391#p2731391 Results]<br />
|-<br />
! scope="row" | '''Mock AMC 8'''<br />
| iNomOnCountdown<br />
| 2014<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&hilit=iNomonCountdown%27s+mock+amc+8 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&hilit=iNomonCountdown%27s+mock+amc+8 Problems]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&start=0 Answers]<br />
| [http://www.artofproblemsolving.com/Forum/viewtopic.php?f=132&t=590451&start=0 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8'''<br />
| Tan<br />
| 2014<br />
| [http://www.artofproblemsolving.com/community/c5h613297p3648226 Initial Discussion]<br />
| [http://www.artofproblemsolving.com/community/c5h613297p3648227 Problems]<br />
| [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Answers]<br />
| [https://docs.google.com/spreadsheets/d/1U1QJ9r4r2xFbRByk9gXOhEVV48ScjoLWZp4IFanbKF0/edit#gid=1475864782 Results / Discussion]<br />
|-<br />
! scope = "row" | '''2015 Hard Mock AMC 8*'''<br />
| Not_a_Username, 8invalid8<br />
| 2015<br />
| [http://artofproblemsolving.com/community/q1h1128391p5227860 Initial Discussion]<br />
| [http://latex.artofproblemsolving.com/miscpdf/kashimyo.pdf Problems]<br />
| [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Answers]<br />
| [http://www.artofproblemsolving.com/community/c5h1128391p5423312 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c3h1152332p5457736 Initial Discussion]<br />
| [http://artofproblemsolving.com/downloads/printable_post_collections/163667 Problems]<br />
| [http://artofproblemsolving.com/community/category-admin/165328 Forum]<br />
| [http://artofproblemsolving.com/community/c165328h1155882p5484131 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8 #2'''<br />
| PersonPsychopath<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c3h1167425p5586244 Initial Discussion]<br />
| [http://artofproblemsolving.com/downloads/printable_post_collections/165838 Problems]<br />
| [http://artofproblemsolving.com/community/c147536_the_mock_amc_8_forum Forum]<br />
| [http://artofproblemsolving.com/community/c3h1167425p5586244 Results / Discussion]<br />
|-<br />
! scope = "row" | '''Mock AMC 8 2015'''<br />
| Alberty44<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c194276_discusssion Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1168822p5674003 Problems]<br />
| n/a<br />
| [http://artofproblemsolving.com/community/c194276h1175928p5673996 Results/Discussion]<br />
|-<br />
! scope = "row" | '''Christmas AMC 8'''<br />
| Mudkipswims42<br />
| 2015<br />
| [http://artofproblemsolving.com/community/c5h1177489p5688588 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c5h1177489p5688588 Problems]<br />
| [http://www.artofproblemsolving.com/community/c229455_christmas_amc8_discussion_forum Forum]<br />
| [http://artofproblemsolving.com/community/c5h1177489p5880200 Results/Discussion]<br />
|- <br />
<br />
! scope = "row" | '''Mock AMC 8!'''<br />
| eisirrational, pretzel, AOPS12142015, Th3Numb3rThr33, e_power_pi_times_i<br />
| 2017<br />
| [https://artofproblemsolving.com/community/c5h1526187_mock_amc_8 Initial Discussion]<br />
| [http://services.artofproblemsolving.com/download.php?id=YXR0YWNobWVudHMvYi84LzFkNzUyMWYwZTViOTQ1MzRmZmU3ODE0NmI2MzIzMGUxNjA2MTcwLnBkZg==&rn=TW9jayBBTUMgOCB2Ny5wZGY= Problems]<br />
| [http://artofproblemsolving.com/community/c3h1545004p9368333 Answer Key/Solutions]<br />
| [https://artofproblemsolving.com/community/c561318_mock_amc_8_2017_discussion Discussion]<br />
|-<br />
<br />
! scope = "row" | '''Mock Summ(er)ation AMC 8!'''<br />
| mathchamp1, kevinmathz, Gali, mathdragon2000, reddragon644, ShreyJ, Quadrastic, Mathnerd1223334444<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1672811_2018_summeration_mock_amc_8 Initial Discussion]<br />
| [https://math-adventures.weebly.com/uploads/1/1/8/8/118819793/amc_8_problems.pdf Problems]<br />
| n/a<br />
| n/a<br />
|-<br />
<br />
! scope = "row" | '''popcorn1's AMC 8 2018'''<br />
| popcorn1<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1715072_popcorn1s_amc_8_2018 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c594864h1715072p11074872 Problems]<br />
| Answer Key / Solutions - not released<br />
| Discussion - not released<br />
|-<br />
<br />
! scope = "row" | ''' Stormersyle's Mock AMC 8<br />
| Stormersyle <br />
| 2018<br />
| [http://artofproblemsolving.com/community/c594864h1722790p11146826 Initial Discussion]<br />
| [http://artofproblemsolving.com/community/c594864h1722790p11146826 Problems]<br />
| [https://latex.artofproblemsolving.com/miscpdf/qgdafjlp.pdf?t=1544494957792 Answer Key/Solutions]<br />
| Discussion - not released<br />
|-<br />
<br />
! scope = "row" | '''mathchampion1's Christmas Mock AMC 8'''<br />
| mathchampion1<br />
| 2018<br />
| [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Initial Discussion]<br />
| [https://artofproblemsolving.com/community/c594864h1752494_released_mock_amc_8_christmas_edition Problems]<br />
| Answer Key / Solutions - not released<br />
| Discussion - not released<br />
|-}<br />
<br />
== See also ==<br />
* [[American Mathematics Competitions]]<br />
* [[Math books]]<br />
* [[Mathematics competitions]]<br />
* [[Mock AIME]]<br />
* [[Mock MathCounts]]<br />
* [[Mock USAMO]]<br />
* [[Mock USAJMO]]<br />
* [[Resources for mathematics competitions]]<br />
* [[AoPS Past Contests]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_22&diff=1002522007 iTest Problems/Problem 222019-01-11T02:05:03Z<p>Flyhawkeye: /* See Also */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the value of <math>c</math> such that the system of equations<br />
<cmath> |x+y|=2007 \\ |x-y|=c</cmath><br />
has exactly two solutions <math>(x,y)</math> in real numbers.<br />
<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12 \quad</math><br />
<br />
<math>\text{(N) } 13 \quad<br />
\text{(O) } 14 \quad<br />
\text{(P) } 15 \quad<br />
\text{(Q) } 16 \quad<br />
\text{(R) } 17 \quad<br />
\text{(S) } 18 \quad<br />
\text{(T) } 223 \quad<br />
\text{(U) } 678 \quad<br />
\text{(V) } 2007 \quad </math><br />
<br />
== Solution ==<br />
<br />
<math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solutions. This occurs at <math>c=\boxed{0\ \text{(A)}}</math>.<br />
<br />
==See Also==<br />
{{iTest box|year=2007|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_22&diff=1002512007 iTest Problems/Problem 222019-01-11T02:04:52Z<p>Flyhawkeye: </p>
<hr />
<div>== Problem ==<br />
<br />
Find the value of <math>c</math> such that the system of equations<br />
<cmath> |x+y|=2007 \\ |x-y|=c</cmath><br />
has exactly two solutions <math>(x,y)</math> in real numbers.<br />
<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12 \quad</math><br />
<br />
<math>\text{(N) } 13 \quad<br />
\text{(O) } 14 \quad<br />
\text{(P) } 15 \quad<br />
\text{(Q) } 16 \quad<br />
\text{(R) } 17 \quad<br />
\text{(S) } 18 \quad<br />
\text{(T) } 223 \quad<br />
\text{(U) } 678 \quad<br />
\text{(V) } 2007 \quad </math><br />
<br />
== Solution ==<br />
<br />
<math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solutions. This occurs at <math>c=\boxed{0\ \text{(A)}}</math>.<br />
<br />
==See Also==<br />
{{iTest box|year=2007|num-b=20|num-a=22}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_22&diff=1002502007 iTest Problems/Problem 222019-01-11T02:04:02Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the value of <math>c</math> such that the system of equations<br />
<cmath> |x+y|=2007 \\ |x-y|=c</cmath><br />
has exactly two solutions <math>(x,y)</math> in real numbers.<br />
<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12 \quad</math><br />
<br />
<math>\text{(N) } 13 \quad<br />
\text{(O) } 14 \quad<br />
\text{(P) } 15 \quad<br />
\text{(Q) } 16 \quad<br />
\text{(R) } 17 \quad<br />
\text{(S) } 18 \quad<br />
\text{(T) } 223 \quad<br />
\text{(U) } 678 \quad<br />
\text{(V) } 2007 \quad </math><br />
<br />
== Solution ==<br />
<br />
<math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solutions. This occurs at <math>c=\boxed{0\ \text{(A)}}</math>.</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_22&diff=1002492007 iTest Problems/Problem 222019-01-11T02:03:45Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the value of <math>c</math> such that the system of equations<br />
<cmath> |x+y|=2007 \\ |x-y|=c</cmath><br />
has exactly two solutions <math>(x,y)</math> in real numbers.<br />
<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12 \quad</math><br />
<br />
<math>\text{(N) } 13 \quad<br />
\text{(O) } 14 \quad<br />
\text{(P) } 15 \quad<br />
\text{(Q) } 16 \quad<br />
\text{(R) } 17 \quad<br />
\text{(S) } 18 \quad<br />
\text{(T) } 223 \quad<br />
\text{(U) } 678 \quad<br />
\text{(V) } 2007 \quad </math><br />
<br />
== Solution ==<br />
<br />
<math>2007|x-y|=c</math> must be one line; otherwise, the system would have four solutions. This occurs at <math>c=\boxed{0\ \textbf{(A)}}</math>.</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_22&diff=1002482007 iTest Problems/Problem 222019-01-11T01:59:39Z<p>Flyhawkeye: /* Problem */</p>
<hr />
<div>== Problem ==<br />
<br />
Find the value of <math>c</math> such that the system of equations<br />
<cmath> |x+y|=2007 \\ |x-y|=c</cmath><br />
has exactly two solutions <math>(x,y)</math> in real numbers.<br />
<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12 \quad</math><br />
<br />
<math>\text{(N) } 13 \quad<br />
\text{(O) } 14 \quad<br />
\text{(P) } 15 \quad<br />
\text{(Q) } 16 \quad<br />
\text{(R) } 17 \quad<br />
\text{(S) } 18 \quad<br />
\text{(T) } 223 \quad<br />
\text{(U) } 678 \quad<br />
\text{(V) } 2007 \quad </math><br />
<br />
== Solution ==</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_14&diff=1002472007 iTest Problems/Problem 142019-01-11T00:33:58Z<p>Flyhawkeye: </p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>\phi(n)</math> be the number of positive integers <math>k< n</math> which are relatively prime to <math>n</math>. For how many distinct values of <math>n</math> is <math>\phi(n)=12</math>?<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12\quad<br />
\text{(N) } 13\quad </math><br />
<br />
== Solution ==<br />
<br />
See https://artofproblemsolving.com/community/q2h598845p3554139.<br />
<br />
==See Also==<br />
{{iTest box|year=2007|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Number Theory Problems]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_14&diff=1002462007 iTest Problems/Problem 142019-01-11T00:33:06Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>\phi(n)</math> be the number of positive integers <math>k< n</math> which are relatively prime to <math>n</math>. For how many distinct values of <math>n</math> is <math>\phi(n)=12</math>?<br />
<br />
<math>\text{(A) } 0 \quad<br />
\text{(B) } 1 \quad<br />
\text{(C) } 2 \quad<br />
\text{(D) } 3 \quad<br />
\text{(E) } 4 \quad<br />
\text{(F) } 5 \quad<br />
\text{(G) } 6 \quad<br />
\text{(H) } 7 \quad<br />
\text{(I) } 8 \quad<br />
\text{(J) } 9 \quad<br />
\text{(K) } 10 \quad<br />
\text{(L) } 11 \quad<br />
\text{(M) } 12\quad<br />
\text{(N) } 13\quad </math><br />
<br />
== Solution ==<br />
<br />
See https://artofproblemsolving.com/community/q2h598845p3554139.</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_iTest_Problems/Problem_9&diff=1002452007 iTest Problems/Problem 92019-01-11T00:24:20Z<p>Flyhawkeye: </p>
<hr />
<div>==Problem==<br />
Suppose that <math>m</math> and <math>n</math> are positive integers such that <math>m < n</math>, the geometric mean of <math>m</math> and <math>n</math> is greater than <math>2007</math>, and the arithmetic mean of <math>m</math> and <math>n</math> is less than <math>2007</math>. How many pairs <math>(m, n)</math> satisfy these conditions?<br />
<br />
<math>\mathrm{(A)}\,0\quad\mathrm{(B)}\,1\quad\mathrm{(C)}\,2\quad\mathrm{(D)}\,3\quad\mathrm{(E)}\,4\quad\mathrm{(F)}\,5\quad\mathrm{(G)}\,6\quad\mathrm{(H)}\,7\quad\mathrm{(I)}\,2007</math><br />
<br />
==Solution==<br />
Since the arithmetic mean is less than 2007 and the geometric mean is greater than 2007, the arithmetic mean must be less than the geometric mean. But by the [[AM-GM inequality]], this is impossible. Therefore no such pairs <math>(m, n)</math> exist, and the answer is <math>0\Rightarrow\boxed{A}</math>.<br />
<br />
==See Also==<br />
{{iTest box|year=2007|num-b=8|num-a=10}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_9&diff=999341983 AIME Problems/Problem 92019-01-01T23:41:20Z<p>Flyhawkeye: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Find the minimum value of <math>\frac{9x^2\sin^2 x + 4}{x\sin x}</math> for <math>0 < x < \pi</math>.<br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Let <math>y=x\sin{x}</math>. We can rewrite the expression as <math>\frac{9y^2+4}{y}=9y+\frac{4}{y}</math>.<br />
<br />
Since <math>x>0</math> and <math>\sin{x}>0</math> because <math>0< x<\pi</math>, we have <math>y>0</math>. So we can apply [[AM-GM]]:<br />
<br />
<cmath>9y+\frac{4}{y}\ge 2\sqrt{9y\cdot\frac{4}{y}}=12</cmath><br />
<br />
The equality holds when <math>9y=\frac{4}{y}\Longleftrightarrow y^2=\frac49\Longleftrightarrow y=\frac23</math>.<br />
<br />
Therefore, the minimum value is <math>\boxed{012}</math>. This is reached when plugging in <math>2/3</math> for <math>x\sin{x}</math> in the original equation (when <math>x\sin{x}=\frac23</math>; since <math>x\sin x</math> is continuous and increasing on the interval <math>0 \le x \le \frac{\pi}{2}</math> and its range on that interval is from <math>0 \le x\sin x \le \frac{\pi}{2}</math>, by the [[Intermediate Value Theorem]] this value is attainable).<br />
<br />
=== Solution 2 ===<br />
We can rewrite the numerator to be a perfect square by adding <math>-\dfrac{12x \sin x}{x \sin x}</math>. Thus, we must also add back <math>12</math>.<br />
<br />
This results in <math>\dfrac{(3x \sin x-2)^2}{x \sin x}+12</math>.<br />
<br />
Thus, if <math>3x \sin x-2=0</math>, then the minimum is obviously <math>12</math>. We show this possible with the same methods in Solution 1; thus the answer is <math>\boxed{012}</math>.<br />
<br />
=== Solution 3 (uses calculus) ===<br />
<br />
Let <math>y = x\sin{x}</math> and rewrite the expression as <math>f(y) = 9y + \frac{4}{y}</math>, similar to the previous solution. To minimize <math>f(y)</math>, take the [[derivative]] of <math>f(y)</math> and set it equal to zero. <br />
<br />
The derivative of <math>f(y)</math>, using the Power Rule, is<br />
<br />
<math>f'(y)</math> = <math>9 - 4y^{-2}</math><br />
<br />
<math>f'(y)</math> is zero only when <math>y = \frac{2}{3}</math> or <math>y = -\frac{2}{3}</math>. It can further be verified that <math>\frac{2}{3}</math> and <math>-\frac{2}{3}</math> are relative minima by finding the derivatives of other points near the critical points. However, since <math>x \sin{x}</math> is always positive in the given domain, <math>y = \frac{2}{3}</math>. Therefore, <math>x\sin{x}</math> = <math>\frac{2}{3}</math>, and the answer is <math>\frac{(9)(\frac{2}{3})^2 + 4}{\frac{2}{3}} = \boxed{012}</math>.<br />
<br />
=== Solution 4 (also uses calculus) ===<br />
<br />
As above, let <math>y = x\sin{x}</math>. Add <math>\frac{12y}{y}</math> to the expression and subtract <math>12</math>, giving <math>f(x) = \frac{(3y+2)^2}{y} - 12</math>. Taking the [[derivative]] of <math>f(x)</math> using the [[Chain Rule]] and [[Quotient Rule]], we have <math>\frac{df}{dx} = \frac{6y(3y+2)-(3y+2)^2}{y^2}</math>. We find the minimum value by setting this to 0. Simplifying, we have <math>6y(3y+2) = (3y+2)^2</math> and <math>y = \pm{\frac{2}{3}} = x\sin{x}</math>. Since both <math>x</math> and <math>\sin{x}</math> are positive on the given interval, we can ignore the negative result. Plugging <math>y = \frac{2}{3}</math> into our expression for <math>f(x)</math>, we have <math>\frac{(3(\frac{2}{3})+2)^2}{y}-12 = \frac{16}{\frac{2}{3}}-12 = \boxed{012}</math>.<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
[[Category:Intermediate Trigonometry Problems]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_7&diff=999331983 AIME Problems/Problem 72019-01-01T23:40:13Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let <math>P</math> be the [[probability]] that at least two of the three had been sitting next to each other. If <math>P</math> is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Solution 1 ===<br />
We can use [[Complementary counting]] by finding the probability that none are sitting next to each other and subtracting it from <math>1</math>.<br />
<br />
Imagine the <math>22</math> other (indistinguishable) people are already seated, and fixed into place. <br />
<br />
We will place <math>A</math>, <math>B</math>, and <math>C</math> with and without the restriction.<br />
<br />
There are <math>22</math> places to place <math>A</math>, followed by <math>21</math> places to place <math>B</math>, and <math>20</math> places to place <math>C</math> after <math>A</math> and <math>B</math>. Hence, there are <math>22\cdot21\cdot20</math> ways to place <math>A, B, C</math> in between these people with restrictions.<br />
<br />
Without restrictions, there are <math>22</math> places to place <math>A</math>, followed by <math>23</math> places to place <math>B</math>, and <math>24</math> places to place <math>C</math> after <math>A</math> and <math>B</math>. Hence, there are <math>22\cdot23\cdot24</math> ways to place <math>A,B,C</math> in between these people without restrictions.<br />
<br />
Thus, the desired amount is <math>1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}</math>, and the answer is <math>11+46=\boxed{057}</math>.<br />
<br />
=== Solution 2 ===<br />
There are <math>(25-1)! = 24!</math> configurations for the knights about the table. <br />
<br />
There are <math>{3\choose 2} = 3</math> ways to pick a pair of knights from the trio, and there are <math>2! = 2</math> ways to determine which order they are seated. Since these two knights must be attached, we let them be a single entity, so there are <math>(24-1)! = 23!</math> configurations for the entities. <br />
<br />
However, this [[Principle of Inclusion-Exclusion|overcounts]] the instances in which the trio sits together; when all three knights sit together, then two of the pairs from the previous case are counted. However, we only want to count this as one case, so we need to subtract the number of instances in which the trio sits together (as a single entity). There are <math>3! = 6</math> ways to determine their order, and there are <math>(23-1)! = 22!</math> configurations.<br />
<br />
Thus, the answer is <math>\frac{2 \times 3 \times 23! - 6 \times 22!}{24!} = \frac{11}{46}</math>, and the answer is <math>\boxed{057}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Number the knights around the table 1-25. There are two possibilities: All three sit next to each other, or two sit next to each other and one is not sitting next to the other two.<br />
<br />
Case 1: All three sit next to each other. In this case, you are picking <math>(1,2,3)</math>, <math>(2,3,4)</math>, <math>(4,5,6)</math>...<math>(25,1,2)</math>. This makes <math>25</math> combinations.<br />
<br />
Case 2: Like above, there are <math>25</math> ways to pick the pair of knights sitting next to each other. Once a pair is picked, you cannot pick either of the two adjacent knights. (i.e. if you pick <math>(5,6)</math>, you may not pick 4 or 7). Thus, there are <math>25-4=21</math> ways to pick the third knight, for a total of <math>25\cdot21</math> combinations.<br />
<br />
Thus, you have a total of <math>25 + (25\cdot21) = 25\cdot22</math> allowable ways to pick the knights. The total number of ways to pick the knights is <math>{25\choose 3} = \frac{25\cdot24\cdot23}{3\cdot2\cdot1} = 25\cdot23\cdot4</math>. <br />
<br />
The probability is <math>\frac{25\cdot22}{25\cdot23\cdot4} = \frac{11}{46}</math>, and the answer is <math>\boxed{057}</math>.<br />
<br />
=== Solution 4 ===<br />
<br />
Pick an arbitrary spot for the first knight. Then pick spots for the next two knights in order.<br />
<br />
Case 1: The second knight sits next to the first knight. There are 2 possible places for this out of 24, so the probability of this is <math>\frac{1}{12}</math>. We do not need to consider the third knight.<br />
<br />
Case 2: The second knight sits two spaces from the first knight. There are 2 possible places for this out of 24, so the probability is <math>\frac{1}{12}</math>. Then there are 3 places out of a remaining 23 for the third knight to sit, so the total probability for this case is <math>\frac{1}{12} \times \frac{3}{23}</math><br />
<br />
Case 3: The second knight sits 3 or more spaces from the first knight. There are 20 possible places for this out of 24, so the probability is <math>\frac{5}{6}</math>. Then there are four places to put the last knight out of 23, so the total probability for this case is <math>\frac{5}{6}\times\frac{4}{23}</math><br />
<br />
So add the probabilities to get the total: <br />
<br />
<br />
<cmath>\frac{1}{12}+\frac{1}{12}\times\frac{3}{23}+\frac{5}{6}\times\frac{4}{23}=\frac{1}{12}\times\frac{1}{23}\left(23+3+40\right)=\frac{66}{12\times 23}</cmath><br />
<cmath>=\frac{6\times 11}{6\times 2 \times 23}=\frac{11}{46}\to\boxed{057}</cmath><br />
<br />
=== Solution 5 ===<br />
<br />
We simplify this problem by using complementary counting and fixing one knight in place. Then, either a knight can sit two away from the fixed knight or a knight can sit more than two away from the fixed knight. The probability is then <math>\frac{24\left(23\right)-\left[2\left(20\right)+20\left(19\right)\right]}{24\left(23\right)}=\frac{11}{46}</math>, so the answer is <math>11+46=\boxed{057}</math>.<br />
<br />
== See Also ==<br />
{{AIME box|year=1983|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2005_Canadian_MO_Problems&diff=997692005 Canadian MO Problems2018-12-26T15:45:37Z<p>Flyhawkeye: /* Problem 2 */</p>
<hr />
<div>==Problem 1==<br />
Consider an equilateral triangle of side length <math>n</math>, which is divided into unit triangles, as shown. Let <math>f(n)</math> be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example of one such path is illustrated below for <math>n=5</math>. Determine the value of <math>f(2005)</math>.<br />
<br />
[[Image:CanMO_2005_1.png]]<br />
<br />
[[2005 Canadian MO Problems/Problem 1 | Solution]]<br />
==Problem 2==<br />
Let <math>(a,b,c)</math> be a Pythagorean triple, ''i.e.'', a triplet of positive integers with <math>{a}^2+{b}^2={c}^2</math>.<br />
<br />
* Prove that <math>\left(\frac{c}{a}+\frac{c}{b}\right)^2>8</math>.<br />
* Prove that there are no integer <math>n</math> and Pythagorean triple <math>(a,b,c)</math> satisfying <math>\left(\frac{c}{a}+\frac{c}{b}\right)^2=n</math>.<br />
<br />
[[2005 Canadian MO Problems/Problem 2 | Solution]]<br />
<br />
==Problem 3==<br />
Let <math>S</math> be a set of <math>n\ge 3</math> points in the interior of a circle. <br />
<br />
* Show that there are three distinct points <math>a,b,c\in S</math> and three distinct points <math>A,B,C</math> on the circle such that <math>a</math> is (strictly) closer to <math>A</math> than any other point in <math>S</math>, <math>b</math> is closer to <math>B</math> than any other point in <math>S</math> and <math>c</math> is closer to <math>C</math> than any other point in <math>S</math>.<br />
* Show that for no value of <math>n</math> can four such points in <math>S</math> (and corresponding points on the circle) be guaranteed.<br />
<br />
[[2005 Canadian MO Problems/Problem 3 | Solution]]<br />
==Problem 4==<br />
Let <math>ABC</math> be a triangle with circumradius <math>R</math>, perimeter <math>P</math> and area <math>K</math>. Determine the maximum value of <math>KP/R^3</math>.<br />
<br />
[[2005 Canadian MO Problems/Problem 4 | Solution]]<br />
==Problem 5==<br />
Let's say that an ordered triple of positive integers <math>(a,b,c)</math> is <math>n</math>-''powerful'' if <math>a \le b \le c</math>, <math>\gcd(a,b,c) = 1</math>, and <math>a^n + b^n + c^n</math> is divisible by <math>a+b+c</math>. For example, <math>(1,2,2)</math> is 5-powerful.<br />
<br />
* Determine all ordered triples (if any) which are <math>n</math>-powerful for all <math>n \ge 1</math>.<br />
* Determine all ordered triples (if any) which are 2004-powerful and 2005-powerful, but not 2007-powerful.<br />
<br />
[[2005 Canadian MO Problems/Problem 5 | Solution]]<br />
<br />
== Resources ==<br />
[[2005 Canadian MO]]</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_24&diff=991062005 AMC 12A Problems/Problem 242018-11-27T02:57:02Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>P(x)=(x-1)(x-2)(x-3)</math>. For how many [[polynomial]]s <math>Q(x)</math> does there exist a polynomial <math>R(x)</math> of degree 3 such that <math>P(Q(x))=P(x)* R(x)</math>?<br />
<br />
<br />
<math>\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32</math><br />
<br />
== Solution ==<br />
We can write the problem as<br />
<div style="text-align:center;"><br />
<math>P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x)</math>.<br />
</div><br />
<br />
<br />
Since <math>\deg P(x) = 3</math> and <math>\deg R(x) = 3</math>, <math>\deg P(x)\cdot R(x) = 6</math>. Thus, <math>\deg P(Q(x)) = 6</math>, so <math>\deg Q(x) = 2</math>.<br />
<div style="text-align:center;"><br />
<math><br />
P(Q(1))=(Q(1)-1)(Q(1)-2)(Q(1)-3)=P(1)\cdot R(1)=0,<br />
</math><br /><math><br />
P(Q(2))=(Q(2)-1)(Q(2)-2)(Q(2)-3)=P(2)\cdot R(2)=0,<br />
</math><br /><math><br />
P(Q(3))=(Q(3)-1)(Q(3)-2)(Q(3)-3)=P(3)\cdot R(3)=0.<br />
</math><br />
</div><br />
Hence, we conclude <math>Q(1)</math>, <math>Q(2)</math>, and <math>Q(3)</math> must each be <math>1</math>, <math>2</math>, or <math>3</math>. Since a [[quadratic equation|quadratic]] is uniquely determined by three points, there can be <math>3*3*3 = 27</math> different quadratics <math>Q(x)</math> after each of the values of <math>Q(1)</math>, <math>Q(2)</math>, and <math>Q(3)</math> are chosen.<br />
<br />
<br />
However, we have included <math>Q(x)</math> which are not quadratics: lines. Namely,<br />
<div style="text-align:center;"><br />
<math><br />
Q(1)=Q(2)=Q(3)=1 \Rightarrow Q(x)=1,<br />
</math><br /><math><br />
Q(1)=Q(2)=Q(3)=2 \Rightarrow Q(x)=2,<br />
</math><br /><math><br />
Q(1)=Q(2)=Q(3)=3 \Rightarrow Q(x)=3,<br />
</math><br /><math><br />
Q(1)=1, Q(2)=2, Q(3)=3 \Rightarrow Q(x)=x,<br />
</math><br /><math><br />
Q(1)=3, Q(2)=2, Q(3)=1 \Rightarrow Q(x)=4-x.<br />
</math><br />
</div><br />
Clearly, we could not have included any other constant functions. For any linear function, we have <math>2\cdot Q(2) = Q(1) + Q(3)</math> because <math>Q(2)</math> is y-value of the midpoint of <math>(1, Q(1))</math> and <math>(3, Q(3))</math>. So we have not included any other linear functions. Therefore, the desired answer is <math>27 - 5 = \boxed{\textbf{(B) }22}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2005|ab=A|num-b=23|num-a=25}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10A_Problems/Problem_19&diff=977672002 AMC 10A Problems/Problem 192018-09-09T20:34:45Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>==Problem==<br />
Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?<br />
<br />
<math>\text{(A)}\ 2\pi/3 \qquad \text{(B)}\ 2\pi \qquad \text{(C)}\ 5\pi/2 \qquad \text{(D)}\ 8\pi/3 \qquad \text{(E)}\ 3\pi</math><br />
<br />
==Solution==<br />
Part of what Spot can reach is <math>\frac{240}{360}=\frac{2}{3}</math> of a circle with radius 2, which <br />
gives him <math>\frac{8\pi}{3}</math>. He can also reach two <math>\frac{60}{360}</math> parts of a unit circle, which combines to give <math>\frac{\pi}{3}</math>. The total area is then <math>3\pi</math>, which gives <math>\boxed{\text{(E)}}</math>.<br />
<br />
===Note===<br />
We can clearly see that the area must be more than <math>\frac{8\pi}{3}</math>, and the only such answer is <math>\boxed{\text{(E)}}</math>.<br />
<br />
==See Also==<br />
{{AMC10 box|ab=A|year=2002|num-b=18|num-a=20}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_5&diff=975632011 AIME I Problems/Problem 52018-08-31T02:10:11Z<p>Flyhawkeye: /* Quick Solution */</p>
<hr />
<div>== Problem ==<br />
The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements.<br />
<br />
== Solution ==<br />
=== Quick Solution ===<br />
Notice that there are three pairs of congruent integers mod 3 (<math>(1,4,7),(2,5,8),(3,6,9)</math>). There are <math>3!</math> ways to order each pair individually and <math>3!</math> ways to order the pairs as a group. Rotations are indistinguishable, so in total there are <math>6^4/9=\boxed{144}</math> ways.<br />
<br />
=== Solution 1 ===<br />
First, we determine which possible combinations of digits <math>1</math> through <math>9</math> will yield sums that are multiples of <math>3</math>. It is simplest to do this by looking at each of the digits <math>\bmod{3}</math>.<br />
<br />
We see that the numbers <math>1, 4,</math> and <math>7</math> are congruent to <math>1 \pmod{3}</math>, that the numbers <math>2, 5,</math> and <math>8</math> are congruent to <math>2 \pmod{3}</math>, and that the numbers <math>3, 6,</math> and <math>9</math> are congruent to <math>0 \pmod{3}</math>. In order for a sum of three of these numbers to be a multiple of three, the mod <math>3</math> sum must be congruent to <math>0</math>. Quick inspection reveals that the only possible combinations are <math>0+0+0, 1+1+1, 2+2+2,</math> and <math>0+1+2</math>. However, every set of three consecutive vertices must sum to a multiple of three, so using any of <math>0+0+0, 1+1+1</math>, or <math>2+2+2</math> would cause an adjacent sum to include exactly 2 digits with the same mod 3 value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different <math> \bmod{3} </math> values. <br />
<br />
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to <math> 1 \pmod{3} </math> can be located counterclockwise of a digit congruent to <math> 0 </math> and clockwise of a digit congruent to <math> 2 \pmod{3} </math>, or the reverse can be true. <br />
<br />
The nonagon can be rotated, so if we find all possible strings beginning with one particular digit, we have found all indistinguishable arrangements. For each of the two trio arrangements, we find <math> 72 </math> possible strings: <br />
<br />
The first digit is predetermined as <math> 3 </math> because we want to avoid strings that rotate to become indistinguishable, so we have one option as a choice for the first digit. The other two <math> 0 \pmod{3} </math> numbers can be arranged in <math> 2!=2 </math> ways. The three <math> 1 \pmod{3}</math> and three <math> 2 \pmod{3} </math> can both be arranged in <math>3!=6</math> ways. Therefore, the desired result is <math> 2(2 \times 6 \times 6)=\boxed{144} </math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_5&diff=975622011 AIME I Problems/Problem 52018-08-31T02:08:51Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3. Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane. Find the number of distinguishable acceptable arrangements.<br />
<br />
== Solution ==<br />
=== Quick Solution ===<br />
Notice that there are three pairs of congruent integers mod 3 (<math>(1,4,7),(2,5,8),(3,6,9)</math>). There are <math>3!</math> ways to order each pair and rotations are indistinguishable, so in total there are <math>6^3/9=\boxed{144}</math> ways.<br />
<br />
=== Solution 1 ===<br />
First, we determine which possible combinations of digits <math>1</math> through <math>9</math> will yield sums that are multiples of <math>3</math>. It is simplest to do this by looking at each of the digits <math>\bmod{3}</math>.<br />
<br />
We see that the numbers <math>1, 4,</math> and <math>7</math> are congruent to <math>1 \pmod{3}</math>, that the numbers <math>2, 5,</math> and <math>8</math> are congruent to <math>2 \pmod{3}</math>, and that the numbers <math>3, 6,</math> and <math>9</math> are congruent to <math>0 \pmod{3}</math>. In order for a sum of three of these numbers to be a multiple of three, the mod <math>3</math> sum must be congruent to <math>0</math>. Quick inspection reveals that the only possible combinations are <math>0+0+0, 1+1+1, 2+2+2,</math> and <math>0+1+2</math>. However, every set of three consecutive vertices must sum to a multiple of three, so using any of <math>0+0+0, 1+1+1</math>, or <math>2+2+2</math> would cause an adjacent sum to include exactly 2 digits with the same mod 3 value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different <math> \bmod{3} </math> values. <br />
<br />
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to <math> 1 \pmod{3} </math> can be located counterclockwise of a digit congruent to <math> 0 </math> and clockwise of a digit congruent to <math> 2 \pmod{3} </math>, or the reverse can be true. <br />
<br />
The nonagon can be rotated, so if we find all possible strings beginning with one particular digit, we have found all indistinguishable arrangements. For each of the two trio arrangements, we find <math> 72 </math> possible strings: <br />
<br />
The first digit is predetermined as <math> 3 </math> because we want to avoid strings that rotate to become indistinguishable, so we have one option as a choice for the first digit. The other two <math> 0 \pmod{3} </math> numbers can be arranged in <math> 2!=2 </math> ways. The three <math> 1 \pmod{3}</math> and three <math> 2 \pmod{3} </math> can both be arranged in <math>3!=6</math> ways. Therefore, the desired result is <math> 2(2 \times 6 \times 6)=\boxed{144} </math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_25&diff=973292003 AMC 10A Problems/Problem 252018-08-20T20:24:48Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Let <math>n</math> be a <math>5</math>-digit number, and let <math>q</math> and <math>r</math> be the quotient and the remainder, respectively, when <math>n</math> is divided by <math>100</math>. For how many values of <math>n</math> is <math>q+r</math> divisible by <math>11</math>? <br />
<br />
<math> \mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090 </math><br />
<br />
== Solution ==<br />
=== Simple Solution ===<br />
<math>11|q+r</math> implies that <math>11|100q+r</math>, so <math>11|n</math>. Then, <math>n</math> can range from <math>910</math> to <math>9090</math> for a total of <math>\boxed{8181\Rightarrow \mathrm{(B)}}</math> numbers.<br />
<br />
=== Solution 1 ===<br />
When a <math>5</math>-digit number is divided by <math>100</math>, the first <math>3</math> digits become the quotient, <math>q</math>, and the last <math>2</math> digits become the remainder, <math>r</math>. <br />
<br />
Therefore, <math>q</math> can be any integer from <math>100</math> to <math>999</math> inclusive, and <math>r</math> can be any integer from <math>0</math> to <math>99</math> inclusive. <br />
<br />
For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\left\lfloor \frac{100}{11} \right\rfloor = 9</math> possible values of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Since there is <math>1</math> "extra" possible value of <math>r</math> that is congruent to <math>0\pmod{11}</math>, each of the <math>\left\lfloor \frac{900}{11} \right\rfloor = 81</math> values of <math>q</math> that are congruent to <math>0\pmod{11}</math> have <math>1</math> more possible value of <math>r</math> such that <math>q+r \equiv 0\pmod{11}</math>. <br />
<br />
Therefore, the number of possible values of <math>n</math> such that <math>q+r \equiv 0\pmod{11}</math> is <math>900\cdot9+81\cdot1=8181 \Rightarrow B</math>. <br />
<br />
=== Solution 2 ===<br />
Let <math>n</math> equal <math>\overline{abcde}</math>, where <math>a</math> through <math>e</math> are digits. Therefore,<br />
<br />
<math>q=\overline{abc}=100a+10b+c</math><br />
<br />
<math>r=\overline{de}=10d+e</math><br />
<br />
We now take <math>q+r\bmod{11}</math>:<br />
<br />
<math>q+r=100a+10b+c+10d+e\equiv a-b+c-d+e\equiv 0\bmod{11}</math><br />
<br />
The divisor trick for 11 is as follows:<br />
<br />
"Let <math>n=\overline{a_1a_2a_3\cdots a_x}</math> be an <math>x</math> digit integer. If <math>a_1-a_2+a_3-\cdots +(-1)^{x-1} a_x</math> is divisible by <math>11</math>, then <math>n</math> is also divisible by <math>11</math>."<br />
<br />
Therefore, the five digit number <math>n</math> is divisible by 11. The 5-digit multiples of 11 range from <math>910\cdot 11</math> to <math>9090\cdot 11</math>. There are <math>8181\Rightarrow \mathrm{(B)}</math> divisors of 11 between those inclusive.<br />
<br />
=== Solution 3 ===<br />
<br />
Since <math>q</math> is a quotient and <math>r</math> is a remainder when <math>n</math> is divided by <math>100</math>. So we have <math>n=100q+r</math>. Since we are counting choices where <math>q+r</math> is divisible by <math>11</math>, we have <math>n=99q+q+r=99q+11k</math> for some <math>k</math>. This means that <math>n</math> is the sum of two multiples of <math>11</math> and would thus itself be a multiple of <math>11</math>. Then we can count all the five digit multiples of <math>11</math> as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)<br />
<br />
==== Notes ====<br />
<br />
The part labeled "divisor trick" actually follows from the same observation we made in the previous step: <math>10\equiv (-1)\pmod{11}</math>, therefore <math>10^{2k}\equiv 1</math> and <math>10^{2k+1}\equiv (-1)</math> for all <math>k</math>. <br />
For a <math>5-</math>digit number <math>\overline{abcde}</math> we get <math>\overline{abcde}\equiv a\cdot 1 + b\cdot(-1) + c\cdot 1 + d\cdot(-1) + e\cdot 1 = a-b+c-d+e</math>, as claimed.<br />
<br />
Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a <math>+</math>, the result will have the same remainder modulo <math>11</math> as the original number.<br />
<br />
== See Also ==<br />
{{AMC10 box|year=2003|ab=A|num-b=24|after=None}}<br />
{{AMC12 box|year=2003|ab=A|num-b=17|num-a=19}}<br />
[[Category:Introductory Number Theory Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_7&diff=972911998 AIME Problems/Problem 72018-08-18T19:16:05Z<p>Flyhawkeye: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Let <math>n</math> be the number of ordered quadruples <math>(x_1,x_2,x_3,x_4)</math> of positive odd [[integer]]s that satisfy <math>\sum_{i = 1}^4 x_i = 98.</math> Find <math>\frac n{100}.</math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Define <math>x_i = 2y_i - 1</math>. Then <math>2\left(\sum_{i = 1}^4 y_i\right) - 4 = 98</math>, so <math>\sum_{i = 1}^4 y_i = 51</math>. <br />
<br />
So we want to find four natural numbers that sum up to 51; we can imagine this as trying to split up 51 on the number line into 4 ranges. This is equivalent to trying to place 3 markers on the numbers 1 through 50; thus the answer is <math>n = {50\choose3} = \frac{50 * 49 * 48}{3 * 2} = 19600</math>, and <math>\frac n{100} = \boxed{196}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
<br />
Another way we can approach this problem is by imagining a line of 98 stones. We want to place the stones into <math>4</math> boxes so that each box has an odd number of stones. We then proceed by placing one stone in each box to begin with, ensuring that we have a positive number in every box. Now we have <math>94</math> stones left. Because we want an odd number in each box, we pair the stones, creating <math>47</math> sets of <math>2</math>. Every time we add a pair to one of the boxes, the number of stones in the box remains odd, because (an odd #) + (an even #) = (an odd #).<br />
<br />
Our problem can now be restated: how many ways are there to partition a line of <math>47</math> stones? We can easily solve this by using <math>3</math> sticks to separate the stones into <math>4</math> groups, and this is the same as arranging a line of <math>3</math> sticks and <math>47</math> stones. <cmath>\frac{50!}{47! \cdot 3!} = 19600</cmath> <cmath>\frac{50 * 49 * 48}{3 * 2} = 19600</cmath> Our answer is therefore <math>\frac{19600}{100} = \boxed{196}</math><br />
<br />
== See also ==<br />
{{AIME box|year=1998|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_1&diff=971752007 AIME II Problems/Problem 12018-08-11T18:51:33Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains N license plates. Find N/<math>10</math>.<br />
<br />
== Solution ==<br />
There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.<br />
<br />
*If <math>0</math> appears 0 or 1 times amongst the sequence, there are <math>\frac{7!}{(7-5)!} = 2520</math> sequences possible.<br />
*If <math>0</math> appears twice in the sequence, there are <math>{5\choose2} = 10</math> places to place the <math>0</math>s. There are <math>\frac{6!}{(6-3)!} = 120</math> ways to place the remaining three characters. Totally, that gives us <math>10 \cdot 120 = 1200</math>.<br />
<br />
Thus, <math>N = 2520 + 1200 = 3720</math>, and <math>\frac{N}{10} = \boxed{372}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2007|n=II|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_4&diff=971731994 AIME Problems/Problem 42018-08-11T17:33:45Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Find the positive integer <math>n\,</math> for which<br />
<cmath><br />
\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994<br />
</cmath><br />
(For real <math>x\,</math>, <math>\lfloor x\rfloor\,</math> is the greatest integer <math>\le x.\,</math>)<br />
<br />
== Solution ==<br />
Note that if <math>2^x \le a<2^{x+1}</math> for some <math>x\in\mathbb{Z}</math>, then <math>\lfloor\log_2{a}\rfloor=\log_2{2^{x}}=x</math>. <br />
<br />
Thus, there are <math>2^{x+1}-2^{x}=2^{x}</math> integers <math>a</math> such that <math>\lfloor\log_2{a}\rfloor=x</math>. So the sum of <math>\lfloor\log_2{a}\rfloor</math> for all such <math>a</math> is <math>x\cdot2^x</math>. <br />
<br />
Let <math>k</math> be the integer such that <math>2^k \le n<2^{k+1}</math>. So for each integer <math>j<k</math>, there are <math>2^j</math> integers <math>a\le n</math> such that <math>\lfloor\log_2{a}\rfloor=j</math>, and there are <math>n-2^k+1</math> such integers such that <math>\lfloor\log_2{a}\rfloor=k</math>. <br />
<br />
Therefore, <math>\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor= \sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1994</math>. <br />
<br />
Through computation: <math>\sum_{j=0}^{7}(j\cdot2^j)=1538<1994</math> and <math>\sum_{j=0}^{8}(j\cdot2^j)=3586>1994</math>. Thus, <math>k=8</math>. <br />
<br />
So, <math>\sum_{j=0}^{k-1}(j\cdot2^j) + k(n-2^k+1) = 1538+8(n-2^8+1)=1994 \Rightarrow n = \boxed{312}</math>.<br />
<br />
Alternatively, one could notice this is an [[arithmetico-geometric series]] and avoid a lot of computation.<br />
<br />
== See also ==<br />
{{AIME box|year=1994|num-b=3|num-a=5}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_7&diff=967992004 AIME II Problems/Problem 72018-08-05T19:45:07Z<p>Flyhawkeye: /* Problem */ Added diagram</p>
<hr />
<div>== Problem ==<br />
<math> ABCD </math> is a rectangular sheet of paper that has been folded so that corner <math> B </math> is matched with point <math> B' </math> on edge <math> AD. </math> The crease is <math> EF, </math> where <math> E </math> is on <math> AB </math> and <math> F </math> is on <math> CD. </math> The dimensions <math> AE=8, BE=17, </math> and <math> CF=3 </math> are given. The perimeter of rectangle <math> ABCD </math> is <math> m/n, </math> where <math> m </math> and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math><br />
<br />
<asy><br />
size(200);<br />
defaultpen(linewidth(0.7)+fontsize(10));<br />
pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C;<br />
draw(F--D--A--E);<br />
draw(E--B--C--F, linetype("4 4"));<br />
filldraw(E--F--Cp--Bp--cycle, white, black);<br />
pair point=( 12.5, 35/3 );<br />
label("$A$", A, dir(point--A));<br />
label("$B$", B, dir(point--B));<br />
label("$C$", C, dir(point--C));<br />
label("$D$", D, dir(point--D));<br />
label("$E$", E, dir(point--E));<br />
label("$F$", F, dir(point--F));<br />
label("$B^\prime$", Bp, dir(point--Bp));<br />
label("$C^\prime$", Cp, dir(point--Cp));</asy><br />
<br />
__TOC__<br />
<br />
== Solution ==<br />
=== Solution 1 (synthetic) ===<br />
<center><asy><br />
pointpen = black; pathpen = black +linewidth(0.7);<br />
pair A=(0,0),B=(0,25),C=(70/3,25),D=(70/3,0),E=(0,8),F=(70/3,22),G=(15,0);<br />
D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle);<br />
D(MP("E",E,W)--MP("F",F,(1,0))); D(B--G);<br />
D(E--MP("B'",G)--F--B,dashed);<br />
MP("8",(A+E)/2,W);MP("17",(B+E)/2,W);MP("22",(D+F)/2,(1,0));<br />
</asy></center><br />
Since <math>EF</math> is the perpendicular bisector of <math>\overline{BB'}</math>, it follows that <math>BE = B'E</math> (by SAS). By the [[Pythagorean Theorem]], we have <math>AB' = 15</math>. Similarly, from <math>BF = B'F</math>, we have<br />
<cmath><br />
\begin{align*}<br />
BC^2 + CF^2 = B'D^2 + DF^2 &\Longrightarrow BC^2 + 9 = (BC - 15)^2 + 484 \\<br />
BC &= \frac{70}{3}<br />
\end{align*}<br />
</cmath><br />
Thus the perimeter of <math>ABCD</math> is <math>2\left(25 + \frac{70}{3}\right) = \frac{290}{3}</math>, and the answer is <math>m+n=\boxed{293}</math>.<br />
<br />
=== Solution 2 (analytic) ===<br />
Let <math>A = (0,0), B=(0,25)</math>, so <math>E = (0,8)</math> and <math>F = (l,22)</math>, and let <math>l = AD</math> be the length of the rectangle. The [[slope]] of <math>EF</math> is <math>\frac{14}{l}</math> and so the equation of <math>EF</math> is <math>y -8 = \frac{14}{l}x</math>. We know that <math>EF</math> is perpendicular to and bisects <math>BB'</math>. The slope of <math>BB'</math> is thus <math>\frac{-l}{14}</math>, and so the equation of <math>BB'</math> is <math>y -25 = \frac{-l}{14}x</math>. Let the point of intersection of <math>EF, BB'</math> be <math>G</math>. Then the y-coordinate of <math>G</math> is <math>\frac{25}{2}</math>, so<br />
<cmath><br />
\begin{align*}<br />
\frac{14}{l}x &= y-8 = \frac{9}{2}\\<br />
\frac{-l}{14}x &= y-25 = -\frac{25}{2}\\<br />
\end{align*}<br />
</cmath><br />
Dividing the two equations yields<br />
<center><math>l^2 = \frac{25 \cdot 14^2}{9} \Longrightarrow l = \frac{70}{3}</math></center><br />
The answer is <math>\boxed{293}</math> as above.<br />
<br />
===Solution 3 (Coordinate Bashing)===<br />
Firstly, observe that if we are given that <math>AE=8</math> and <math>BE=17</math>, the length of the triangle is given and the height depends solely on the length of <math>CF</math>. Let Point <math>A = (0,0)</math>. Since <math>AE=8</math>, point E is at (8,0). Next, point <math>B</math> is at <math>(25,0)</math> since <math>BE=17</math> and point <math>B'</math> is at <math>(0,-15)</math> since <math>BE=AE</math> by symmetry. Draw line segment <math>BB'</math>. Notice that this is perpendicular to <math>EF</math> by symmetry. Next, find the slope of EB, which is <math>\frac{15}{25}=\frac{3}{5}</math>. Then, the slope of <math>EF</math> is -<math>\frac{5}{3}</math>. <br />
<br />
Line EF can be written as y=<math>-\frac{5}{3}x+b</math>. Plug in the point <math>(8,0)</math>, and we get the equation of EF to be y=<math>_\frac{5}{3}x+\frac{40}{3}</math>. Since the length of <math>AB</math>=25, a point on line <math>EF</math> lies on <math>DC</math> when <math>x=25-3=22</math>. Plug in <math>x=22</math> into our equation to get <math>y=-\frac{70}{3}</math>. <math>|y|=BC=\frac{70}{3}</math>. Therefore, our answer is <math>2(AB+BC)=2\left(25+\frac{70}{3}\right)=2\left(\frac{145}{3}\right)=\frac{290}{3}= \boxed{293}</math>.<br />
<br />
===Solution 4 (Trig)===<br />
Firstly, note that <math>B'E=BE=17</math>, so <math>AB'=\sqrt{17^2-8^2}=15</math>. Then let <math>\angle BEF=\angle B'EF=\theta</math>, so <math>\angle B'EA = \pi-2\theta</math>. Then <math>\tan(\pi-2\theta)=\frac{15}{8}</math>, or <br />
<br />
<cmath>\frac{2\tan(\theta)}{\tan^2(\theta)-1}=\frac{15}{8}</cmath> using supplementary and double angle identities. Multiplying though and factoring yields <br />
<br />
<cmath>(3\tan(\theta)-5)(5\tan(\theta)+3)=0</cmath><br />
<br />
It is clear from the problem setup that <math>0<\theta<\frac\pi2</math>, so the correct value is <math>\tan(\theta)=\frac53</math>. Next, extend rays <math>\overrightarrow{BC}</math> and <math>\overrightarrow{EF}</math> to intersect at <math>C'</math>. Then <math>\tan(\theta)=\frac{BC'}{17}=\frac53</math>, so <math>BC'=\frac{85}{3}</math>. By similar triangles, <math>CC'=\frac{3}{17}BC'=\frac{15}{3}</math>, so <math>BC=\frac{70}{3}</math>. The perimeter is <math>\frac{140}{3}+50=\frac{290}{3}\Longrightarrow \boxed{293}</math><br />
<br />
<br />
An even faster way to finish is, to draw a line segment <math>FF'</math> where <math>F'</math> is a point on <math>EB</math> such that <math>FF'</math> is perpendicular to <math>EB</math>. This makes right triangle <math>FF'E</math>, Also, note that <math>F'B</math> has length of <math>3</math> (draw the diagram out, and note the <math>F'B =FC</math>). From here, through <math>\tan \theta = \frac{5}{3}</math>, we can note that <math>\frac{FF'}{EF'} = \frac{5}{3} \implies \frac{FF'}{14} = \frac{5}{3} \implies FF' = \frac{70}{3}</math>. <math>FF'</math> is parallel and congrurent to <math>CB</math> and <math>AD</math>, and hence we can use this to calculate the perimeter. The perimeter is simply <math>\frac{70}{3} + \frac{70}{3} + 25 + 25 = \frac{290}{3} \Longrightarrow \boxed{293}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2004|n=II|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_9&diff=967131986 AIME Problems/Problem 92018-08-03T15:10:50Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In <math>\triangle ABC</math>, <math>AB= 425</math>, <math>BC=450</math>, and <math>AC=510</math>. An interior [[point]] <math>P</math> is then drawn, and [[segment]]s are drawn through <math>P</math> [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length <math>d</math>, find <math>d</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Simple Solution ===<br />
<center><asy><br />
size(200);<br />
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br />
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br />
/* construct remaining points */<br />
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br />
pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br />
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);<br />
dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s));<br />
D(D--Ea);D(Da--F);D(Fa--E);<br />
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);<br />
/*P copied from above solution*/<br />
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); <br />
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --><br />
<br />
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.<br />
<br />
By similar triangles, <math>BE'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Since <math>FD'=BC-EE'</math>, we have <math>900-\frac{33}{17}d=d</math>, so <math>d=\boxed{306}</math>.<br />
<br />
===Solution 1 ===<br />
<asy><br />
size(200);<br />
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br />
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br />
/* construct remaining points */<br />
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br />
pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br />
/* Construct P */<br />
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,NE)); <br />
pair X = IP(L(A,P,4), B--C); dot(MP("X",X,NW));<br />
pair Y = IP(L(B,P,4), C--A); dot(MP("Y",Y,NE));<br />
pair Z = IP(L(C,P,4), A--B); dot(MP("Z",Z,N));<br />
<br />
D(A--X); D(B--Y); D(C--Z);<br />
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);<br />
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);<br />
</asy><br />
<br />
Construct cevians <math>AX</math>, <math>BY</math> and <math>CZ</math> through <math>P</math>. Place masses of <math>x,y,z</math> on <math>A</math>, <math>B</math> and <math>C</math> respectively; then <math>P</math> has mass <math>x+y+z</math>.<br />
<br />
Notice that <math>Z</math> has mass <math>x+y</math>. On the other hand, by similar triangles, <math>\frac{CP}{CZ} = \frac{d}{AB}</math>. Hence by mass points we find that <cmath> \frac{x+y}{x+y+z} = \frac{d}{AB} </cmath> Similarly, we obtain <cmath> \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} </cmath> Summing these three equations yields <cmath> \frac{d}{AB} + \frac{d}{BC} + \frac{d}{CA} = \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} = \frac{2x+2y+2z}{x+y+z} = 2 </cmath><br />
<br />
Hence, <center><math> d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}</math><math>= \frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}</math></center><br />
<br />
=== Solution 2 ===<br />
<center><asy><br />
size(200);<br />
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br />
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br />
/* construct remaining points */<br />
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br />
pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br />
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);<br />
dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s));<br />
D(D--Ea);D(Da--F);D(Fa--E);<br />
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);<br />
/*P copied from above solution*/<br />
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); <br />
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --><br />
<br />
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.<br />
<br />
Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. <br />
<br />
Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]:<br />
<br />
<center><math>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></center><br />
<br />
Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>.<br />
<br />
=== Solution 3 ===<br />
Define the points the same as above.<br />
<br />
Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>[BEPD'] = c</math>, <math>[D'PD] = d</math>, <math>[DAF'P] = e</math> and <math>[F'D'P] = f</math><br />
<br />
The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.<br />
<br />
Let the length of the segment be <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get:<br />
<br />
<math>\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2</math>, <math>\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2</math>, <math>\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2</math>.<br />
Adding all these together and using <math>a + b + c + d + e + f = A</math> we get<br />
<math>\frac {f + d + b}{A} + 1 = x^2 \cdot \left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math><br />
<br />
Using [[corresponding angles]] from parallel lines, it is easy to show that <math>\triangle ABC \sim \triangle F'PF</math>; since <math>ADPF'</math> and <math>CFPE'</math> are parallelograms, it is easy to show that <math>FF' = AC - x</math><br />
<br />
Now we have the side length [[ratio]], so we have the area ratio<br />
<math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>. By symmetry, we have<br />
<math>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math><br />
<br />
Substituting these into our initial equation, we have<br />
<math>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</math><br />
<math>\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0</math><br />
<math>\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math><br />
and the answer follows after some hideous computation.<br />
<br />
===Solution 4===<br />
Refer to the diagram in solution 2; let <math>a^2=[E'EP]</math>, <math>b^2=[D'DP]</math>, and <math>c^2=[F'FP]</math>. Now, note that <math>[E'BD]</math>, <math>[D'DP]</math>, and <math>[E'EP]</math> are similar, so through some similarities we find that <math>\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2</math>. Similarly, we find that <math>[D'AF]=(b+c)^2</math> and <math>[F'CE]=(c+a)^2</math>, so <math>[ABC]=(a+b+c)^2</math>. Now, again from similarity, it follows that <math>\frac{d}{510}=\frac{a+b}{a+b+c}</math>, <math>\frac{d}{450}=\frac{b+c}{a+b+c}</math>, and <math>\frac{d}{425}=\frac{c+a}{a+b+c}</math>, so adding these together, simplifying, and solving gives <math>d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}</math><br />
<math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>.<br />
<br />
=== Solution 5 ===<br />
Refer to the diagram from Solution 2. Notice that because <math>CE'PF</math>, <math>AF'PD</math>, and <math>BD'PE</math> are parallelograms, <math>\overline{DD'} = 425-d</math>, <math>\overline{EE'} = 450-d</math>, and <math>\overline{FF'} = 510-d</math>. <br />
<br />
Let <math>F'P = x</math>. Then, because <math>\triangle ABC \sim \triangle F'PF</math>, <math>\frac{AB}{AC}=\frac{F'P}{F'F}</math>, so <math>\frac{425}{510}=\frac{x}{510-d}</math>. Simplifying the LHS and cross-multiplying, we have <math>6x=2550-5d</math>. From the same triangles, we can find that <math>FP=\frac{18}{17}x</math>.<br />
<br />
<math>\triangle PEE'</math> is also similar to <math>\triangle F'PF</math>. Since <math>EF'=d</math>, <math>EP=d-x</math>. We now have <math>\frac{PE}{EE'}=\frac{F'P}{FP}</math>, and <math>\frac{d-x}{450-d}=\frac{17}{18}</math>. Cross multiplying, we have <math>18d-18x=450 \cdot 17-17d</math>. Using the previous equation to substitute for <math>x</math>, we have: <cmath>18d-3\cdot2550+15d=450\cdot17-17d</cmath> This is a linear equation in one variable, and we can solve to get <math>d=\boxed{306}</math><br />
<br />
*I did not show the multiplication in the last equation because most of it cancels out when solving.<br />
<br />
(Note: I chose <math>F'P</math> to be <math>x</math> only because that is what I had written when originally solving. The solution would work with other choices for <math>x</math>.)<br />
<br />
== See also ==<br />
{{AIME box|year=1986|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1986_AIME_Problems/Problem_9&diff=967121986 AIME Problems/Problem 92018-08-03T15:07:07Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
In <math>\triangle ABC</math>, <math>AB= 425</math>, <math>BC=450</math>, and <math>AC=510</math>. An interior [[point]] <math>P</math> is then drawn, and [[segment]]s are drawn through <math>P</math> [[parallel]] to the sides of the [[triangle]]. If these three segments are of an equal length <math>d</math>, find <math>d</math>.<br />
<br />
__TOC__<br />
== Solution ==<br />
=== Simple Solution ===<br />
<center><asy><br />
size(200);<br />
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br />
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br />
/* construct remaining points */<br />
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br />
pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br />
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);<br />
dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s));<br />
D(D--Ea);D(Da--F);D(Fa--E);<br />
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);<br />
/*P copied from above solution*/<br />
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); <br />
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --><br />
<br />
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.<br />
<br />
By similar triangles, <math>CF'=\frac{d}{510}\cdot450=\frac{15}{17}d</math> and <math>FA=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Since <math>E'D=CA-FF'</math>, we have <math>900-\frac{33}{17}d=d</math>, so <math>d=\boxed{306}</math>.<br />
<br />
===Solution 1 ===<br />
<asy><br />
size(200);<br />
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br />
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br />
/* construct remaining points */<br />
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br />
pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br />
/* Construct P */<br />
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,NE)); <br />
pair X = IP(L(A,P,4), B--C); dot(MP("X",X,NW));<br />
pair Y = IP(L(B,P,4), C--A); dot(MP("Y",Y,NE));<br />
pair Z = IP(L(C,P,4), A--B); dot(MP("Z",Z,N));<br />
<br />
D(A--X); D(B--Y); D(C--Z);<br />
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);<br />
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);<br />
</asy><br />
<br />
Construct cevians <math>AX</math>, <math>BY</math> and <math>CZ</math> through <math>P</math>. Place masses of <math>x,y,z</math> on <math>A</math>, <math>B</math> and <math>C</math> respectively; then <math>P</math> has mass <math>x+y+z</math>.<br />
<br />
Notice that <math>Z</math> has mass <math>x+y</math>. On the other hand, by similar triangles, <math>\frac{CP}{CZ} = \frac{d}{AB}</math>. Hence by mass points we find that <cmath> \frac{x+y}{x+y+z} = \frac{d}{AB} </cmath> Similarly, we obtain <cmath> \frac{y+z}{x+y+z} = \frac{d}{BC} \qquad \text{and} \qquad \frac{z+x}{x+y+z} = \frac{d}{CA} </cmath> Summing these three equations yields <cmath> \frac{d}{AB} + \frac{d}{BC} + \frac{d}{CA} = \frac{x+y}{x+y+z} + \frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} = \frac{2x+2y+2z}{x+y+z} = 2 </cmath><br />
<br />
Hence, <center><math> d = \frac{2}{\frac{1}{AB} + \frac{1}{BC} + \frac{1}{CA}} = \frac{2}{\frac{1}{510} + \frac{1}{450} + \frac{1}{425}} = \frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}</math><math>= \frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}=\frac{10}{\frac{10}{306}} = \boxed{306}</math></center><br />
<br />
=== Solution 2 ===<br />
<center><asy><br />
size(200);<br />
pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);<br />
pair C=(0,0),A=(510,0),B=IP(circle(C,450),circle(A,425));<br />
/* construct remaining points */<br />
pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);<br />
pair D=IP(Ea--(Ea+A-C),A--B),F=IP(Da--(Da+C-B),A--C),Fa=IP(E--(E+A-B),A--C); <br />
D(MP("A",A,s)--MP("B",B,N,s)--MP("C",C,s)--cycle);<br />
dot(MP("D",D,NE,s));dot(MP("E",E,NW,s));dot(MP("F",F,s));dot(MP("D'",Da,NE,s));dot(MP("E'",Ea,NW,s));dot(MP("F'",Fa,s));<br />
D(D--Ea);D(Da--F);D(Fa--E);<br />
MP("450",(B+C)/2,NW);MP("425",(A+B)/2,NE);MP("510",(A+C)/2);<br />
/*P copied from above solution*/<br />
pair P = IP(D--Ea,E--Fa); dot(MP("P",P,N)); <br />
</asy></center> <!-- Asymptote replacement for Image:1986_AIME-9.png by azjps --><br />
<br />
Let the points at which the segments hit the triangle be called <math>D, D', E, E', F, F'</math> as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are [[similar triangles|similar]] (<math>\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF</math>). The remaining three sections are [[parallelogram]]s.<br />
<br />
Since <math>PDAF'</math> is a parallelogram, we find <math>PF' = AD</math>, and similarly <math>PE = BD'</math>. So <math>d = PF' + PE = AD + BD' = 425 - DD'</math>. Thus <math>DD' = 425 - d</math>. By the same logic, <math>EE' = 450 - d</math>. <br />
<br />
Since <math>\triangle DPD' \sim \triangle ABC</math>, we have the [[proportion]]:<br />
<br />
<center><math>\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d</math></center><br />
<br />
Doing the same with <math>\triangle PEE'</math>, we find that <math>PE' =510 - \frac{17}{15}d</math>. Now, <math>d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = \boxed{306}</math>.<br />
<br />
=== Solution 3 ===<br />
Define the points the same as above.<br />
<br />
Let <math>[CE'PF] = a</math>, <math>[E'EP] = b</math>, <math>[BEPD'] = c</math>, <math>[D'PD] = d</math>, <math>[DAF'P] = e</math> and <math>[F'D'P] = f</math><br />
<br />
The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.<br />
<br />
Let the length of the segment be <math>x</math> and the area of the triangle be <math>A</math>, using the theorem, we get:<br />
<br />
<math>\frac {d + e + f}{A} = \left(\frac {x}{BC}\right)^2</math>, <math>\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2</math>, <math>\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2</math>.<br />
Adding all these together and using <math>a + b + c + d + e + f = A</math> we get<br />
<math>\frac {f + d + b}{A} + 1 = x^2 \cdot \left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)</math><br />
<br />
Using [[corresponding angles]] from parallel lines, it is easy to show that <math>\triangle ABC \sim \triangle F'PF</math>; since <math>ADPF'</math> and <math>CFPE'</math> are parallelograms, it is easy to show that <math>FF' = AC - x</math><br />
<br />
Now we have the side length [[ratio]], so we have the area ratio<br />
<math>\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2</math>. By symmetry, we have<br />
<math>\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2</math> and <math>\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2</math><br />
<br />
Substituting these into our initial equation, we have<br />
<math>1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0</math><br />
<math>\implies 1 + \sum_{cyc}1 - 2 \cdot \frac {x}{AB} = 0</math><br />
<math>\implies \frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x</math><br />
and the answer follows after some hideous computation.<br />
<br />
===Solution 4===<br />
Refer to the diagram in solution 2; let <math>a^2=[E'EP]</math>, <math>b^2=[D'DP]</math>, and <math>c^2=[F'FP]</math>. Now, note that <math>[E'BD]</math>, <math>[D'DP]</math>, and <math>[E'EP]</math> are similar, so through some similarities we find that <math>\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2</math>. Similarly, we find that <math>[D'AF]=(b+c)^2</math> and <math>[F'CE]=(c+a)^2</math>, so <math>[ABC]=(a+b+c)^2</math>. Now, again from similarity, it follows that <math>\frac{d}{510}=\frac{a+b}{a+b+c}</math>, <math>\frac{d}{450}=\frac{b+c}{a+b+c}</math>, and <math>\frac{d}{425}=\frac{c+a}{a+b+c}</math>, so adding these together, simplifying, and solving gives <math>d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5}\left(\frac{1}{17}+\frac{1}{18}\right)+\frac{1}{102}}=\frac{10}{\frac{1}{5}\cdot\frac{35}{306}+\frac{3}{306}}</math><br />
<math>=\frac{10}{\frac{10}{306}}=\boxed{306}</math>.<br />
<br />
=== Solution 5 ===<br />
Refer to the diagram from Solution 2. Notice that because <math>CE'PF</math>, <math>AF'PD</math>, and <math>BD'PE</math> are parallelograms, <math>\overline{DD'} = 425-d</math>, <math>\overline{EE'} = 450-d</math>, and <math>\overline{FF'} = 510-d</math>. <br />
<br />
Let <math>F'P = x</math>. Then, because <math>\triangle ABC \sim \triangle F'PF</math>, <math>\frac{AB}{AC}=\frac{F'P}{F'F}</math>, so <math>\frac{425}{510}=\frac{x}{510-d}</math>. Simplifying the LHS and cross-multiplying, we have <math>6x=2550-5d</math>. From the same triangles, we can find that <math>FP=\frac{18}{17}x</math>.<br />
<br />
<math>\triangle PEE'</math> is also similar to <math>\triangle F'PF</math>. Since <math>EF'=d</math>, <math>EP=d-x</math>. We now have <math>\frac{PE}{EE'}=\frac{F'P}{FP}</math>, and <math>\frac{d-x}{450-d}=\frac{17}{18}</math>. Cross multiplying, we have <math>18d-18x=450 \cdot 17-17d</math>. Using the previous equation to substitute for <math>x</math>, we have: <cmath>18d-3\cdot2550+15d=450\cdot17-17d</cmath> This is a linear equation in one variable, and we can solve to get <math>d=\boxed{306}</math><br />
<br />
*I did not show the multiplication in the last equation because most of it cancels out when solving.<br />
<br />
(Note: I chose <math>F'P</math> to be <math>x</math> only because that is what I had written when originally solving. The solution would work with other choices for <math>x</math>.)<br />
<br />
== See also ==<br />
{{AIME box|year=1986|num-b=8|num-a=10}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=File:2009aimei4.PNG&diff=96695File:2009aimei4.PNG2018-08-02T22:25:34Z<p>Flyhawkeye: </p>
<hr />
<div></div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=File:Abcd38571092.PNG&diff=96694File:Abcd38571092.PNG2018-08-02T22:24:38Z<p>Flyhawkeye: </p>
<hr />
<div></div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1999_AIME_Problems/Problem_4&diff=966261999 AIME Problems/Problem 42018-07-31T16:57:50Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
The two [[square]]s shown share the same [[center]] <math>O_{}</math> and have sides of length 1. The length of <math>\overline{AB}</math> is <math>43/99</math> and the [[area]] of octagon <math>ABCDEFGH</math> is <math>m/n,</math> where <math>m_{}</math> and <math>n_{}</math> are [[relatively prime]] [[positive]] [[integer]]s. Find <math>m+n.</math><br />
<br />
<asy><br />
//code taken from thread for problem<br />
real alpha = 25;<br />
pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin;<br />
pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z;<br />
draw(W--X--Y--Z--cycle^^w--x--y--z--cycle);<br />
pair A=intersectionpoint(Y--Z, y--z), <br />
C=intersectionpoint(Y--X, y--x), <br />
E=intersectionpoint(W--X, w--x), <br />
G=intersectionpoint(W--Z, w--z), <br />
B=intersectionpoint(Y--Z, y--x), <br />
D=intersectionpoint(Y--X, w--x), <br />
F=intersectionpoint(W--X, w--z), <br />
H=intersectionpoint(W--Z, y--z);<br />
dot(O);<br />
label("$O$", O, SE);<br />
label("$A$", A, dir(O--A));<br />
label("$B$", B, dir(O--B));<br />
label("$C$", C, dir(O--C));<br />
label("$D$", D, dir(O--D));<br />
label("$E$", E, dir(O--E));<br />
label("$F$", F, dir(O--F));<br />
label("$G$", G, dir(O--G));<br />
label("$H$", H, dir(O--H));</asy><br />
__TOC__<br />
== Solution ==<br />
=== Simple Solution ===<br />
Triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, and each area is <math>\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}</math>. Since the area of a triangle is <math>bh/2</math>, the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.<br />
<br />
<br />
=== Other Solution ===<br />
Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>. <br />
<br />
By the [[Pythagorean theorem]],<br />
<cmath>x^2 + y^2 = \left(\frac{43}{99}\right)^2</cmath><br />
<br />
Also,<br />
<cmath>\begin{align*}x + y + \frac{43}{99} &= 1\\<br />
x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}</cmath><br />
<br />
Substituting,<br />
<cmath>\begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\<br />
2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}</cmath><br />
<br />
Thus, the area of the octagon is <math>1 - \frac{13}{99} = \frac{86}{99}</math>, so <math>m + n = \boxed{185}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1999|num-b=3|num-a=5}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_6&diff=966251989 AIME Problems/Problem 62018-07-31T14:59:27Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Two skaters, Allie and Billie, are at [[point]]s <math>A</math> and <math>B</math>, respectively, on a flat, frozen lake. The [[distance]] between <math>A</math> and <math>B</math> is <math>100</math> meters. Allie leaves <math>A</math> and skates at a [[speed]] of <math>8</math> meters per second on a straight line that makes a <math>60^\circ</math> angle with <math>AB</math>. At the same time Allie leaves <math>A</math>, Billie leaves <math>B</math> at a speed of <math>7</math> meters per second and follows the [[straight]] path that produces the earliest possible meeting of the two skaters, given their speeds. How many meters does Allie skate before meeting Billie?<br />
<center><asy><br />
pointpen=black; pathpen=black+linewidth(0.7); <br />
pair A=(0,0),B=(10,0),C=6*expi(pi/3);<br />
D(B--A); D(A--C,EndArrow); MP("A",A,SW);MP("B",B,SE);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);<br />
</asy></center><!-- Minsoen's image: [[Image:AIME_1989_Problem_6.png]] --><br />
<br />
== Solution ==<br />
Label the point of [[intersection]] as <math>C</math>. Since <math>d = rt</math>, <math>AC = 8t</math> and <math>BC = 7t</math>. According to the [[law of cosines]],<br />
<br />
<center><asy><br />
pointpen=black; pathpen=black+linewidth(0.7); <br />
pair A=(0,0),B=(10,0),C=16*expi(pi/3);<br />
D(B--A); D(A--C); D(B--C,dashed); MP("A",A,SW);MP("B",B,SE);MP("C",C,N);MP("60^{\circ}",A+(0.3,0),NE);MP("100",(A+B)/2);MP("8t",(A+C)/2,NW);MP("7t",(B+C)/2,NE);<br />
</asy></center><br />
<br />
<cmath>\begin{align*}(7t)^2 &= (8t)^2 + 100^2 - 2 \cdot 8t \cdot 100 \cdot \cos 60^\circ\\<br />
0 &= 15t^2 - 800t + 10000 = 3t^2 - 160t + 2000\\<br />
t &= \frac{160 \pm \sqrt{160^2 - 4\cdot 3 \cdot 2000}}{6} = 20, \frac{100}{3}.\end{align*}</cmath><br />
<br />
Since we are looking for the earliest possible intersection, <math>20</math> seconds are needed. Thus, <math>8 \cdot 20 = \boxed{160}</math> meters is the solution.<br />
<br />
Alternatively, we can drop an altitude from <math>C</math> and arrive at the same answer.<br />
<br />
== See also ==<br />
{{AIME box|year=1989|num-b=5|num-a=7}}<br />
<br />
[[Category:Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_I_Problems/Problem_6&diff=965132009 AIME I Problems/Problem 62018-07-27T01:17:32Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
How many positive integers <math>N</math> less than <math>1000</math> are there such that the equation <math>x^{\lfloor x\rfloor} = N</math> has a solution for <math>x</math>? (The notation <math>\lfloor x\rfloor</math> denotes the greatest integer that is less than or equal to <math>x</math>.)<br />
<br />
== Solution ==<br />
First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>. <br />
<br />
Because <math>{\lfloor x\rfloor}</math> must be an integer, we can do some simple case work:<br />
<br />
For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> as long as <math>x \neq 0</math>. This gives us <math>1</math> value of <math>N</math>.<br />
<br />
For <math>{\lfloor x\rfloor}=1</math>, <math>N</math> can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math><br />
<br />
Therefore, <math>N=1</math>. However, we got N=1 in case 1 so it got counted twice.<br />
<br />
For <math>{\lfloor x\rfloor}=2</math>, <math>N</math> can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math><br />
<br />
This gives us <math>3^2-2^2=5</math> <math>N</math>'s<br />
<br />
For <math>{\lfloor x\rfloor}=3</math>, <math>N</math> can be anything between <math>3^3</math> to <math>4^3</math> excluding <math>4^3</math><br />
<br />
This gives us <math>4^3-3^3=37</math> <math>N</math>'s<br />
<br />
For <math>{\lfloor x\rfloor}=4</math>, <math>N</math> can be anything between <math>4^4</math> to <math>5^4</math> excluding <math>5^4</math><br />
<br />
This gives us <math>5^4-4^4=369</math> <math>N</math>'s<br />
<br />
Since <math>x</math> must be less than <math>5</math>, we can stop here and the answer is <math>1+5+37+369= \boxed {412}</math> possible values for <math>N</math>.<br />
<br />
Alternatively, one could find that the values which work are <math>1^1,\ 2^2,\ 3^3,\ 4^4,\ \sqrt{5}^{\lfloor\sqrt{5}\rfloor},\ \sqrt{6}^{\lfloor\sqrt{6}\rfloor},\ \sqrt{7}^{\lfloor\sqrt{7}\rfloor},\ \sqrt{8}^{\lfloor\sqrt{8}\rfloor},\ \sqrt[3]{28}^{\lfloor\sqrt[3]{28}\rfloor},\ \sqrt[3]{29}^{\lfloor\sqrt[3]{29}\rfloor},\ \sqrt[3]{30}^{\lfloor\sqrt[3]{30}\rfloor},\ ...,\ \sqrt[3]{63}^{\lfloor\sqrt[3]{63}\rfloor},\ \sqrt[4]{257}^{\lfloor\sqrt[4]{257}\rfloor},\ \sqrt[4]{258}^{\lfloor\sqrt[4]{258}\rfloor},\ ...,\ \sqrt[4]{624}^{\lfloor\sqrt[4]{624}\rfloor}</math> to get the same answer.<br />
<br />
== See also ==<br />
{{AIME box|year=2009|n=I|num-b=5|num-a=7}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2002_AIME_II_Problems/Problem_3&diff=964802002 AIME II Problems/Problem 32018-07-24T15:03:43Z<p>Flyhawkeye: /* Solution */ Added LaTeX and clarified solution</p>
<hr />
<div>== Problem ==<br />
It is given that <math>\log_{6}a + \log_{6}b + \log_{6}c = 6,</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[positive]] [[integer]]s that form an increasing [[geometric sequence]] and <math>b - a</math> is the [[Perfect square|square]] of an integer. Find <math>a + b + c.</math><br />
<br />
== Solution ==<br />
<math>abc=6^6</math>. Since they form an increasing geometric sequence, <math>b</math> is the [[geometric mean]] of the [[product]] <math>abc</math>. <math>b=\sqrt[3]{abc}=6^2=36</math>.<br />
<br />
Since <math>b-a</math> is the square of an integer, we can find a few values of <math>a</math> that work: <math>11, 20, 27, 32,</math> and <math>35</math>. Out of these, the only value of <math>a</math> that works is <math>a=27</math>, from which we can deduce that <math>c=\dfrac{36}{27}\cdot 36=\dfrac{4}{3}\cdot 36=48</math>.<br />
<br />
Thus, <math>a+b+c=27+36+48=\boxed{111}</math><br />
<br />
== See also ==<br />
{{AIME box|year=2002|n=II|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_7&diff=963922000 AIME I Problems/Problem 72018-07-21T15:24:19Z<p>Flyhawkeye: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.<br />
<br />
<br />
===Solution 1===<br />
We can rewrite <math>xyz=1</math> as <math>\frac{1}{z}=xy</math>.<br />
<br />
Substituting into one of the given equations, we have <br />
<cmath>x+xy=5</cmath><br />
<cmath>x(1+y)=5</cmath><br />
<cmath>\frac{1}{x}=\frac{1+y}{5}.</cmath><br />
<br />
We can substitute back into <math>y+\frac{1}{x}=29</math> to obtain<br />
<cmath>y+\frac{1+y}{5}=29</cmath><br />
<cmath>5y+1+y=145</cmath><br />
<cmath>y=24.</cmath><br />
<br />
We can then substitute once again to get<br />
<cmath>x=\frac15</cmath><br />
<cmath>y=\frac{5}{24}.</cmath><br />
Thus, <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>.<br />
<br />
<br />
===Solution 2===<br />
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>.<br />
<br />
<cmath><br />
\begin{align*}<br />
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\<br />
&=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\<br />
&=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\<br />
&=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\<br />
&=2 + 5 + 29 + r\\<br />
&=36 + r<br />
\end{align*}<br />
</cmath><br />
<br />
Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>.<br />
<br />
=== Solution 3 ===<br />
<br />
Since <math>x+(1/z)=5, 1=z(5-x)=xyz</math>, so <math>5-x=xy</math>. Also, <math>y=29-(1/x)</math> by the second equation. Substitution gives <math>x=1/5</math>, <math>y=24</math>, and <math>z=5/24</math>, so the answer is 4+1 which is equal to <math>5</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_7&diff=963912000 AIME I Problems/Problem 72018-07-21T15:23:14Z<p>Flyhawkeye: Added Solution</p>
<hr />
<div>== Problem ==<br />
Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>.<br />
<br />
<br />
===Solution 1===<br />
We can rewrite <math>xyz=1</math> as <math>\frac{1}{z}=xy</math>.<br />
<br />
Substituting into one of the given equations, we have <br />
<cmath>x+xy=5</cmath><br />
<cmath>x(1+y)=5</cmath><br />
<cmath>\frac{1}{x}=\frac{1+y}{5}.</cmath><br />
<br />
We can substitute back into <math>y+\frac{1}{x}=29</math> to obtain<br />
<cmath>y+\frac{1+y}{5}=29</cmath><br />
<cmath>5y+1+y=145</cmath><br />
<cmath>y=24.</cmath><br />
<br />
We can then substitute once again to get<br />
<cmath>x=\frac15</cmath><br />
<cmath>y=\frac{5}{24}.</cmath><br />
Thus, <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>.<br />
<br />
<br />
===Solution 2===<br />
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>.<br />
<br />
<cmath><br />
\begin{align*}<br />
(5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\<br />
&=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\<br />
&=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\<br />
&=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\<br />
&=2 + 5 + 29 + r\\<br />
&=36 + r<br />
\end{align*}<br />
</cmath><br />
<br />
Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
Since <math>x+(1/z)=5, 1=z(5-x)=xyz</math>, so <math>5-x=xy</math>. Also, <math>y=29-(1/x)</math> by the second equation. Substitution gives <math>x=1/5</math>, <math>y=24</math>, and <math>z=5/24</math>, so the answer is 4+1 which is equal to <math>5</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2000|n=I|num-b=6|num-a=8}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2010_AIME_I_Problems&diff=963562010 AIME I Problems2018-07-19T15:47:38Z<p>Flyhawkeye: /* Problem 3 */</p>
<hr />
<div>{{AIME Problems|year=2010|n=I}}<br />
<br />
== Problem 1 ==<br />
Maya lists all the positive divisors of <math>2010^2</math>. She then randomly selects two distinct divisors from this list. Let <math>p</math> be the probability that exactly one of the selected divisors is a perfect square. The probability <math>p</math> can be expressed in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
Find the remainder when <math>9 \times 99 \times 999 \times \cdots \times \underbrace{99\cdots9}_{\text{999 9's}}</math> is divided by <math>1000</math>.<br />
<br />
[[2010 AIME I Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
Suppose that <math>y = \frac34x</math> and <math>x^y = y^x</math>. The quantity <math>x + y</math> can be expressed as a rational number <math>\frac {r}{s}</math>, where <math>r</math> and <math>s</math> are relatively prime positive integers. Find <math>r + s</math>.<br />
<br />
[[2010 AIME I Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
Jackie and Phil have two fair coins and a third coin that comes up heads with probability <math>\frac47</math>. Jackie flips the three coins, and then Phil flips the three coins. Let <math>\frac {m}{n}</math> be the probability that Jackie gets the same number of heads as Phil, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>.<br />
<br />
[[2010 AIME I Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
Let <math>P(x)</math> be a quadratic polynomial with real coefficients satisfying <math>x^2 - 2x + 2 \le P(x) \le 2x^2 - 4x + 3</math> for all real numbers <math>x</math>, and suppose <math>P(11) = 181</math>. Find <math>P(16)</math>.<br />
<br />
[[2010 AIME I Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
Define an ordered triple <math>(A, B, C)</math> of sets to be <math>\textit{minimally intersecting}</math> if <math>|A \cap B| = |B \cap C| = |C \cap A| = 1</math> and <math>A \cap B \cap C = \emptyset</math>. For example, <math>(\{1,2\},\{2,3\},\{1,3,4\})</math> is a minimally intersecting triple. Let <math>N</math> be the number of minimally intersecting ordered triples of sets for which each set is a subset of <math>\{1,2,3,4,5,6,7\}</math>. Find the remainder when <math>N</math> is divided by <math>1000</math>.<br />
<br />
'''Note''': <math>|S|</math> represents the number of elements in the set <math>S</math>.<br />
<br />
[[2010 AIME I Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
For a real number <math>a</math>, let <math>\lfloor a \rfloor</math> denote the greatest integer less than or equal to <math>a</math>. Let <math>\mathcal{R}</math> denote the region in the coordinate plane consisting of points <math>(x,y)</math> such that <math>\lfloor x \rfloor ^2 + \lfloor y \rfloor ^2 = 25</math>. The region <math>\mathcal{R}</math> is completely contained in a disk of radius <math>r</math> (a disk is the union of a circle and its interior). The minimum value of <math>r</math> can be written as <math>\frac {\sqrt {m}}{n}</math>, where <math>m</math> and <math>n</math> are integers and <math>m</math> is not divisible by the square of any prime. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
Let <math>(a,b,c)</math> be a real solution of the system of equations <math>x^3 - xyz = 2</math>, <math>y^3 - xyz = 6</math>, <math>z^3 - xyz = 20</math>. The greatest possible value of <math>a^3 + b^3 + c^3</math> can be written in the form <math>\frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
Let <math>N</math> be the number of ways to write <math>2010</math> in the form <math>2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0</math>, where the <math>a_i</math>'s are integers, and <math>0 \le a_i \le 99</math>. An example of such a representation is <math>1\cdot 10^3 + 3\cdot 10^2 + 67\cdot 10^1 + 40\cdot 10^0</math>. Find <math>N</math>.<br />
<br />
[[2010 AIME I Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
Let <math>\mathcal{R}</math> be the region consisting of the set of points in the coordinate plane that satisfy both <math>|8 - x| + y \le 10</math> and <math>3y - x \ge 15</math>. When <math>\mathcal{R}</math> is revolved around the line whose equation is <math>3y - x = 15</math>, the volume of the resulting solid is <math>\frac {m\pi}{n\sqrt {p}}</math>, where <math>m</math>, <math>n</math>, and <math>p</math> are positive integers, <math>m</math> and <math>n</math> are relatively prime, and <math>p</math> is not divisible by the square of any prime. Find <math>m + n + p</math>.<br />
<br />
[[2010 AIME I Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
Let <math>m \ge 3</math> be an integer and let <math>S = \{3,4,5,\ldots,m\}</math>. Find the smallest value of <math>m</math> such that for every partition of <math>S</math> into two subsets, at least one of the subsets contains integers <math>a</math>, <math>b</math>, and <math>c</math> (not necessarily distinct) such that <math>ab = c</math>.<br />
<br />
'''Note''': a partition of <math>S</math> is a pair of sets <math>A</math>, <math>B</math> such that <math>A \cap B = \emptyset</math>, <math>A \cup B = S</math>.<br />
<br />
[[2010 AIME I Problems/Problem 12|Solution]]<br />
<br />
== Problem 13 ==<br />
Rectangle <math>ABCD</math> and a semicircle with diameter <math>AB</math> are coplanar and have nonoverlapping interiors. Let <math>\mathcal{R}</math> denote the region enclosed by the semicircle and the rectangle. Line <math>\ell</math> meets the semicircle, segment <math>AB</math>, and segment <math>CD</math> at distinct points <math>N</math>, <math>U</math>, and <math>T</math>, respectively. Line <math>\ell</math> divides region <math>\mathcal{R}</math> into two regions with areas in the ratio <math>1: 2</math>. Suppose that <math>AU = 84</math>, <math>AN = 126</math>, and <math>UB = 168</math>. Then <math>DA</math> can be represented as <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m + n</math>.<br />
<br />
[[2010 AIME I Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
For each positive integer n, let <math>f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor</math>. Find the largest value of n for which <math>f(n) \le 300</math>.<br />
<br />
'''Note:''' <math>\lfloor x \rfloor</math> is the greatest integer less than or equal to <math>x</math>.<br />
<br />
[[2010 AIME I Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the incircles of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal radii. Let <math>p</math> and <math>q</math> be positive relatively prime integers such that <math>\frac {AM}{CM} = \frac {p}{q}</math>. Find <math>p + q</math>.<br />
<br />
[[2010 AIME I Problems/Problem 15|Solution]]<br />
<br />
== See also ==<br />
* [[American Invitational Mathematics Examination]]<br />
* [[AIME Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_5&diff=963552004 AIME II Problems/Problem 52018-07-19T15:17:06Z<p>Flyhawkeye: Added Solution 2</p>
<hr />
<div>==Problem==<br />
In order to complete a large job, <math>1000</math> workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then <math>100</math> workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional <math>100</math> workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the <math>800</math> workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?<br />
<br />
==Solution==<br />
A train is traveling at <math>1000</math> miles per hour and has one hour to reach its destination <math>1000</math> miles away. After <math>15</math> minutes and <math>250</math> miles it slows to <math>900</math> mph, and thus takes <math>\frac{250}{900}(60)=\frac{50}{3}</math> minutes to travel the next <math>250</math> miles. Then it slows to <math>800</math> mph, so the next quarter takes <math>\frac{250}{800}(60)=\frac{150}{8}</math>. The train then has <math>60-15-\frac{50}{3}-\frac{150}{8}=230/24</math> minutes left to travel 250 miles, and doing the arithmetic shows that during this last stretch it must travel more than <math>1565</math> mph to make it on time. Therefore the company needs to add <math>1566-800 = \boxed{766}</math> more workers.<br />
<br />
Solution by rocketscience<br />
<br />
==Solution 2==<br />
Let each worker's speed be <math>w</math>, the entire time be <math>t</math>, and the total work be <math>1</math>.<br />
<br />
<br />
From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.<br />
<br />
<br />
We need to find the time the next quarter takes to complete the same amount of work, which is <math>\frac{1000}{900}\cdot\frac{1}{4}\cdot t=\frac{5}{18}t</math>.<br />
<br />
<br />
Similarly, we find that the time the third quarter takes is <math>\frac{1000}{800}\cdot\frac{1}{4}\cdot t=\frac{5}{16}t</math>.<br />
<br />
<br />
Finally, the time the last quarter takes is <math>t\left[1-\left(\frac{1}{4}+\frac{5}{18}+\frac{5}{16}\right)\right]=\frac{23}{144}t</math>.<br />
<br />
<br />
We then let the number of workers needed be <math>x</math>, so we have the equation <math>\left(800+x\right)w\cdot\frac{23}{144}t=\frac{1}{4}</math>. Dividing by the first equation, we have<br />
<br />
<cmath>\frac{800+x}{1000}\cdot\frac{\frac{23}{144}}{\frac{1}{4}}=1</cmath><br />
<cmath>\frac{800+x}{1000}\cdot\frac{23}{36}=1</cmath><br />
<cmath>800+x=\frac{36000}{23}</cmath><br />
<cmath>x=\frac{17600}{23}.</cmath><br />
<br />
We can't have a part of a worker, so we take the ceiling of <math>x</math>, which we find to be <math>\boxed{766}</math>.<br />
<br />
-flyhawkeye<br />
<br />
== See also ==<br />
{{AIME box|year=2004|num-b=4|num-a=6|n=II}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=963432011 AIME I Problems/Problem 12018-07-18T14:35:39Z<p>Flyhawkeye: /* Solution 3 */</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}</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{085}</math><br />
<br />
== Solution 3 ==<br />
One may first combine all three jars in to a single container. That container will have <math>10</math> liters of liquid, and it should be <math>50\%</math> acidic. Thus there must be <math>5</math> liters of acid. <br />
<br />
Jug A contained <math>45\% \cdot 4L</math>, or <math>1.8L</math> of acid, and jug B <math>48\% \cdot 5L</math> or <math>2.4L</math>. Solving for the amount of acid in jug C, <math>k = (5 - 2.4 - 1.8) = .8</math>, or <math>80\%</math>.<br />
<br />
Once one knows that the jug C is <math>80\%</math> acid, use solution 1 to figure out m and n for <math>k+m+n=80+2+3=\boxed{085}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_1&diff=963422011 AIME I Problems/Problem 12018-07-18T14:35:32Z<p>Flyhawkeye: /* Solution 3 */</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}</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{085}</math><br />
<br />
== Solution 3 ==<br />
One may first combine all three jars in to a single container. That container will have <math>10</math> liters of liquid, and it should be <math>50\%</math> acidic. Thus there must be <math>5</math> liters of acid. <br />
<br />
Jug A contained <math>45\% \cdot 4L</math>, or <math>1.8L</math> of acid, and jug B <math>48\% \cdot 5L</math> or <math>2.4L</math>. Solving for the amount of acid in jug C, <math>k = (5 - 2.4 - 1.8) = .8</math>, or <math>80\%</math>.<br />
<br />
Once one knows that the jug C is <math>80\%</math> acid, use solution 1 to figure out m and n for <math>k+m+n=80+2+3=\boxed{085}</math>..<br />
<br />
== See also ==<br />
{{AIME box|year=2011|n=I|before=First Problem|num-a=2}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2008_AIME_II_Problems/Problem_2&diff=963412008 AIME II Problems/Problem 22018-07-18T14:22:54Z<p>Flyhawkeye: Added Simple Solution</p>
<hr />
<div>== Problem ==<br />
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Rudolph bikes at a [[constant]] rate and stops for a five-minute break at the end of every mile. Jennifer bikes at a constant rate which is three-quarters the rate that Rudolph bikes, but Jennifer takes a five-minute break at the end of every two miles. Jennifer and Rudolph begin biking at the same time and arrive at the <math>50</math>-mile mark at exactly the same time. How many minutes has it taken them?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude><br />
<br />
== Simple Solution ==<br />
Let <math>r</math> be the time Rudolph takes disregarding breaks and <math>\frac{4}{3}r</math> be the time Jennifer takes disregarding breaks. We have the equation<br />
<cmath>r+5\left(49\right)=\frac{4}{3}r+5\left(24\right)</cmath><br />
<cmath>123=\frac13r</cmath><br />
<cmath>r=375.</cmath><br />
Thus, the total time they take is <math>375 + 5(49) = \boxed{620}</math> minutes.<br />
<br />
== Solution ==<br />
Let Rudolf bike at a rate <math>r</math>, so Jennifer bikes at the rate <math>\dfrac 34r</math>. Let the time both take be <math>t</math>. <br />
<br />
Then Rudolf stops <math>49</math> times (because the rest after he reaches the finish does not count), losing a total of <math>49 \cdot 5 = 245</math> minutes, while Jennifer stops <math>24</math> times, losing a total of <math>24 \cdot 5 = 120</math> minutes. The time Rudolf and Jennifer actually take biking is then <math>t - 245,\, t-120</math> respectively. <br />
<br />
Using the formula <math>r = \frac dt</math>, since both Jennifer and Rudolf bike <math>50</math> miles,<br />
<center><cmath>\begin{align}r &= \frac{50}{t-245}\\<br />
\frac{3}{4}r &= \frac{50}{t-120}<br />
\end{align}</cmath></center><br />
Substituting equation <math>(1)</math> into equation <math>(2)</math> and simplifying, we find <br />
<center><cmath>\begin{align*}50 \cdot \frac{3}{4(t-245)} &= 50 \cdot \frac{1}{t-120}\\<br />
\frac{1}{3}t &= \frac{245 \cdot 4}{3} - 120\\<br />
t &= \boxed{620}\ \text{minutes}<br />
\end{align*}</cmath></center><br />
<br />
== See also ==<br />
{{AIME box|year=2008|n=II|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2009_AIME_II_Problems/Problem_1&diff=962962009 AIME II Problems/Problem 12018-07-16T20:27:17Z<p>Flyhawkeye: /* Solution */ Added solution</p>
<hr />
<div>== Problem ==<br />
Before starting to paint, Bill had <math>130</math> ounces of blue paint, <math>164</math> ounces of red paint, and <math>188</math> ounces of white paint. Bill painted four equally sized stripes on a wall, making a blue stripe, a red stripe, a white stripe, and a pink stripe. Pink is a mixture of red and white, not necessarily in equal amounts. When Bill finished, he had equal amounts of blue, red, and white paint left. Find the total number of ounces of paint Bill had left.<br />
<br />
== Solution ==<br />
<br />
=== Simple Solution ===<br />
<br />
Let the stripes be <math>b, r, w,</math> and <math>p</math>, respectively. Let the red part of the pink be <math>\frac{r_p}{p}</math> and the white part be <math>\frac{w_p}{p}</math> for <math>\frac{r_p+w_p}{p}=p</math>.<br />
<br />
Since the stripes are of equal size, we have <math>b=r=w=p</math>. Since the amounts of paint end equal, we have <math>130-b=164-r-\frac{r_p}{p}=188-w-\frac{w_p}{p}</math>. Thus, we know that<br />
<cmath>130-p=164-p-\frac{r_p}{p}=188-p-\frac{w_p}{p}</cmath><br />
<cmath>130=164-\frac{r_p}{p}=188-\frac{w_p}{p}</cmath><br />
<cmath>r_p=34p, w_p=58p</cmath><br />
<cmath>\frac{r_p+w_p}{p}=92=p=b.</cmath><br />
Each paint must end with <math>130-92=38</math> oz left, for a total of <math>3 \cdot 38 = \boxed{114}</math> oz.<br />
<br />
=== Solution 1 ===<br />
<br />
After the pink stripe is drawn, all three colors will be used equally so the pink stripe must bring the amount of red and white paint down to <math>130</math> ounces each. Say <math>a</math> is the fraction of the pink paint that is red paint and <math>x</math> is the size of each stripe. Then equations can be written: <math>ax = 164 - 130 = 34</math> and <math>(1-a)x = 188 - 130 = 58</math>. The second equation becomes <math>x - ax = 58</math> and substituting the first equation into this one we get <math>x - 34 = 58</math> so <math>x = 92</math>. The amount of each color left over at the end is thus <math>130 - 92 = 38</math> and <math>38 * 3 = \boxed{114}</math>.<br />
<br />
=== Solution 2 ===<br />
<br />
We know that all the stripes are of equal size. We can then say that <math>r</math> is the amount of paint per stripe. Then <math>130 - r</math> will be the amount of blue paint left. Now for the other two stripes. The amount of white paint left after the white stripe and the amount of red paint left after the blue stripe are <math>188 - r</math> and <math>164 - r</math> respectively. The pink stripe is also r ounces of paint, but let there be <math>k</math> ounces of red paint in the mixture and <math>r - k</math> ounces of white paint. We now have two equations: <math>164 - r - k = 188 - r - (r-k)</math> and <math>164 - r - k = 130 - r</math>. Solving yields k = 34 and r = 92. We now see that there will be <math>130 - 92 = 38</math> ounces of paint left in each can. <math>38 * 3 = \boxed{114}</math><br />
<br />
=== Solution 3 ===<br />
<br />
Let the amount of paint each stripe painted used be <math>x</math>. Also, let the amount of paint of each color left be <math>y</math>. 1 stripe is drawn with the blue paint, and 3 stripes are drawn with the red and white paints. Add together the amount of red and white paint, <math>164 + 188 = 352</math> and obtain the following equations : <math>352 - 3x = 2y</math> and <math>130 - x = y</math>. Solve to obtain <math>x = 92</math>. Therefore <math>y</math> is <math>130 - 92 = 38</math>, with three cans of equal amount of paint, the answer is <math>38 * 3 = \boxed{114}</math>.<br />
<br />
== See Also ==<br />
<br />
{{AIME box|year=2009|n=II|before=First Question|num-a=2}}<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_2&diff=962922004 AIME I Problems/Problem 22018-07-16T15:32:52Z<p>Flyhawkeye: /* Simple Solution */ Fixed solution and added LaTeX</p>
<hr />
<div>== Problem ==<br />
[[Set]] <math>A</math> consists of <math>m</math> consecutive integers whose sum is <math>2m</math>, and set <math>B</math> consists of <math>2m</math> consecutive integers whose sum is <math>m.</math> The absolute value of the difference between the greatest element of <math>A</math> and the greatest element of <math>B</math> is <math>99</math>. Find <math>m.</math><br />
<br />
==Simple Solution==<br />
<br />
Look at the problem... consecutive integers. Now, since set <math>A</math> has the properties of <math>m</math> integers that sum to <math>2m</math>, it's obvious that the middle integer is just <math>2</math> (the average that "balances out" everything), and the largest is <math>2 + \frac{m-1}{2}</math>. That's because there are <math>m-1</math> to go after taking <math>2</math>, and they are "evenly balanced" on either side. <br />
<br />
From there, we see that set <math>B</math>'s average is <math>0.5</math>. How can that happen? Only if the middle TWO values are <math>0</math> and <math>1</math>. Since set <math>B</math> has <math>2m</math> integers, its maximum must be <math>99</math> larger than the maximum of set <math>A</math> unless <math>m=1</math>, which is impossible.<br />
<br />
From there, the largest element of set <math>B</math> is <math>1 + (m-1) = m</math>.<br />
<br />
Solving, we get <br />
<cmath>m - 2 - \frac{m-1}{2} = 99</cmath><br />
<cmath>m-\frac{m}{2}+\frac{1}{2}=101</cmath><br />
<cmath>\frac{m}{2}=100\frac{1}{2}.</cmath><br />
There we go- <math>m</math> is equal to none other than <math>\boxed{201}</math>.<br />
<br />
== Solution ==<br />
Let us give the [[element]]s of our sets names:<br />
<math>A = \{x, x + 1, x + 2, \ldots, x + m - 1\}</math> and <math>B = \{y, y + 1, \ldots, y + 2m - 1\}</math>. So we are given that<br />
<cmath>2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,</cmath> <br />
so <math>2 = x + \frac{m - 1}2</math> and <math>x + (m - 1) = \frac{m + 3}2</math>. Also,<br />
<cmath>m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,</cmath><br />
so <math>1 = 2y + (2m - 1)</math> so <math>2m = 2(y + 2m - 1)</math> and <math>m = y + 2m - 1</math>.<br />
<br />
Then by the given, <math>99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|</math>. <math>m</math> is a [[positive integer]] so we must have <math>99 = \frac{m - 3}2</math> and so <math>m = \boxed{201}</math>.<br />
<br />
== Solution 2 == <br />
<br />
The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.<br />
<br />
First, we note that for set <math>A</math><br />
<br />
<cmath>\frac{m(f + l)}{2} = 2m</cmath><br />
<br />
Where <math>f</math> and <math>l</math> represent the first and last terms of <math>A</math>. This comes from the sum of an arithmetic sequence.<br />
<br />
Solving for <math>f+l</math>, we find the sum of the two terms is <math>4</math>.<br />
<br />
Doing the same for set B, and setting up the equation with <math>b</math> and <math>e</math> being the first and last terms of set <math>B</math>, <br />
<br />
<cmath>m(b+e) = m</cmath><br />
<br />
and so <math>b+e = 1</math>.<br />
<br />
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set <math>A</math> has half the number of elements as set <math>B</math>, and the difference between the greatest terms of the two two sequences is <math>99</math> (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where <math>x</math> is the last term of set A:<br />
<br />
<cmath>2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)</cmath><br />
<br />
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to <math>4</math> and <math>1</math> respectively (add <math>x</math> and <math>(-x+4)</math> to see what i mean).<br />
<br />
Solving this equation we find <math>x = 102</math>. We know the first and last terms have to sum to <math>4</math> so we find the first term of the sequence is <math>-98</math>. Now, the solution is in clear sight, we just find the number of integers between <math>-98</math> and <math>102</math>, inclusive, and it is <math>201</math>.<br />
<br />
Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.<br />
<br />
== See also ==<br />
{{AIME box|year=2004|n=I|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_2&diff=962912004 AIME I Problems/Problem 22018-07-16T15:16:09Z<p>Flyhawkeye: /* Problem */ Fixed spacing</p>
<hr />
<div>== Problem ==<br />
[[Set]] <math>A</math> consists of <math>m</math> consecutive integers whose sum is <math>2m</math>, and set <math>B</math> consists of <math>2m</math> consecutive integers whose sum is <math>m.</math> The absolute value of the difference between the greatest element of <math>A</math> and the greatest element of <math>B</math> is <math>99</math>. Find <math>m.</math><br />
<br />
==Simple Solution==<br />
<br />
Look at the problem... consecutive integers. Now, since set A has the properties of m integers that sum to 2m, it's obvious that the middle integer is just 2 (the average that "balances out" everything), and the largest is 2 + (m-1)/2. That's because there are m-1 to go after taking 2, and they are "evenly balanced" on either side. <br />
<br />
Now, since 2m consecutive integers only sum to m (set B) and just m integers already got to 2m (set A), it's clear that max({A}) is 99 larger than max({B}).<br />
<br />
From there, we see that set B's average is 0.5. How can that happen? Only if the middle TWO values are 0 and 1. <br />
<br />
From there, the largest element of set B is 1 + (m-1) = m.<br />
<br />
Solving, we get:<br />
<br />
1.5 + 0.5m + 99 = m. There we go- m is equal to none other than 201.<br />
<br />
== Solution ==<br />
Let us give the [[element]]s of our sets names:<br />
<math>A = \{x, x + 1, x + 2, \ldots, x + m - 1\}</math> and <math>B = \{y, y + 1, \ldots, y + 2m - 1\}</math>. So we are given that<br />
<cmath>2m = x + (x + 1) + \ldots + (x + m - 1) = mx + (1 + 2 + \ldots + (m - 1)) = mx + \frac{m(m -1)}2,</cmath> <br />
so <math>2 = x + \frac{m - 1}2</math> and <math>x + (m - 1) = \frac{m + 3}2</math>. Also,<br />
<cmath>m = y + (y + 1) + \ldots + (y + 2m - 1) = 2my + \frac{2m(2m - 1)}2,</cmath><br />
so <math>1 = 2y + (2m - 1)</math> so <math>2m = 2(y + 2m - 1)</math> and <math>m = y + 2m - 1</math>.<br />
<br />
Then by the given, <math>99 = |(x + m - 1) - (y + 2m - 1)| = \left|\frac{m + 3}2 - m\right| = \left|\frac{m - 3}2\right|</math>. <math>m</math> is a [[positive integer]] so we must have <math>99 = \frac{m - 3}2</math> and so <math>m = \boxed{201}</math>.<br />
<br />
== Solution 2 == <br />
<br />
The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.<br />
<br />
First, we note that for set <math>A</math><br />
<br />
<cmath>\frac{m(f + l)}{2} = 2m</cmath><br />
<br />
Where <math>f</math> and <math>l</math> represent the first and last terms of <math>A</math>. This comes from the sum of an arithmetic sequence.<br />
<br />
Solving for <math>f+l</math>, we find the sum of the two terms is <math>4</math>.<br />
<br />
Doing the same for set B, and setting up the equation with <math>b</math> and <math>e</math> being the first and last terms of set <math>B</math>, <br />
<br />
<cmath>m(b+e) = m</cmath><br />
<br />
and so <math>b+e = 1</math>.<br />
<br />
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set <math>A</math> has half the number of elements as set <math>B</math>, and the difference between the greatest terms of the two two sequences is <math>99</math> (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where <math>x</math> is the last term of set A:<br />
<br />
<cmath>2(x-(-x+4)+1) = 1+(x+99)-(-x-99+1)</cmath><br />
<br />
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to <math>4</math> and <math>1</math> respectively (add <math>x</math> and <math>(-x+4)</math> to see what i mean).<br />
<br />
Solving this equation we find <math>x = 102</math>. We know the first and last terms have to sum to <math>4</math> so we find the first term of the sequence is <math>-98</math>. Now, the solution is in clear sight, we just find the number of integers between <math>-98</math> and <math>102</math>, inclusive, and it is <math>201</math>.<br />
<br />
Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.<br />
<br />
== See also ==<br />
{{AIME box|year=2004|n=I|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1995_AIME_Problems/Problem_2&diff=962481995 AIME Problems/Problem 22018-07-15T17:11:10Z<p>Flyhawkeye: Added Solution 3</p>
<hr />
<div>== Problem ==<br />
Find the last three digits of the product of the [[positive root]]s of<br />
<math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>.<br />
<br />
===Solution 1===<br />
Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields the [[quadratic equation]] <math>2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0</math>. Applying the [[quadratic formula]] yields that <math>\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}</math>. Thus, the product of the two roots (both of which are positive) is <math>1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>.<br />
<br />
===Solution 2===<br />
Instead of taking <math>\log_{1995}</math>, we take <math>\log_x</math> of both sides and simplify:<br />
<br />
<math> \log_x(\sqrt{1995}x^{\log_{1995}x})=\log_x(x^{2}) </math><br />
<br />
<math> \log_x\sqrt{1995}+\log_x x^{\log_{1995}x}=2 </math><br />
<br />
<math>\dfrac{1}{2} \log_x 1995 + \log_{1995} x = 2</math><br />
<br />
We know that <math>\log_x 1995</math> and <math>\log_{1995} x</math> are reciprocals, so let <math>a=\log_{1995} x</math>. Then we have <math>\dfrac{1}{2}\left(\dfrac{1}{a}\right) + a = 2</math>. Multiplying by <math>2a</math> and simplifying gives us <math>2a^2-4a+1=0</math>, as shown above.<br />
<br />
Because <math>a=\log_{1995} x</math>, <math>x=1995^a</math>. By the quadratic formula, the two roots of our equation are <math>a=\frac{2\pm\sqrt2}{2}</math>. This means our two roots in terms of <math>x</math> are <math>1995^\frac{2+\sqrt2}{2}</math> and <math>1995^\frac{2-\sqrt2}{2}.</math> Multiplying these gives <math>1995^2</math><br />
<br />
<math>1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>.<br />
<br />
===Solution 3===<br />
Let <math>y=\log_{1995}x</math>. Rewriting the equation in terms of <math>y</math>, we have<br />
<cmath> \sqrt{1995}\left(1995^y\right)^y=1995^{2y}</cmath><br />
<cmath> 1995^{y^2-\frac{1}{2}}=1995^{2y}</cmath><br />
<cmath> y^2-\frac{1}{2}=2y</cmath><br />
<cmath> 2y^2-4y-1=0</cmath><br />
<cmath> y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(-1\right)}}{4}=\frac{4\pm\sqrt{24}}{4}=\frac{2\pm\sqrt{6}}{2}</cmath><br />
Thus, the product of the positive roots is <math>\left(1995^{\frac{2+\sqrt{6}}{2}}\right)\left(1995^{\frac{2-\sqrt{6}}{2}}\right)=1995^2=\left(2000-5\right)^2</math>, so the last three digits are <math>\boxed{025}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1995|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_2&diff=962221990 AIME Problems/Problem 22018-07-14T18:25:12Z<p>Flyhawkeye: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.<br />
<br />
== Solution 1 ==<br />
Suppose that <math>52+6\sqrt{43}</math> is in the form of <math>(a + b\sqrt{43})^2</math>. [[FOIL]]ing yields that <math>52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}</math>. This implies that <math>a</math> and <math>b</math> equal one of <math>\pm1, \pm3</math>. The possible [[set]]s are <math>(3,1)</math> and <math>(-3,-1)</math>; the latter can be discarded since the [[square root]] must be positive. This means that <math>52 + 6\sqrt{43} = (\sqrt{43} + 3)^2</math>. Repeating this for <math>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>.<br />
<br />
Rewriting, we get <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3</math>. Using the difference of [[cube]]s, we get that <math>[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]</math> <math> = (6)(3 \cdot 43 + 9) = \boxed{828}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
The <math>3/2</math> power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that. <br />
Let <math>S</math> be the sum of the given expression. <br />
<cmath>S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2</cmath><br />
<cmath>S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}</cmath><br />
After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at <math>S^2 = 685584</math> which gives <math>S=\boxed{828}</math>.<br />
<br />
== Solution 3 ==<br />
<br />
Factor as a difference of cubes.<br />
<cmath>\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right)^2\right)\right] = </cmath><br />
<cmath>\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+\left(52^2-\left(36\right)\left(43\right)\right)^{\frac{1}{2}}\right] = </cmath><br />
<cmath>\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right].</cmath><br />
We can simplify the left factor as follows.<br />
<cmath>\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x</cmath><br />
<cmath>104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} = x^2</cmath><br />
<cmath>104-68 = x^2</cmath><br />
<cmath>36 = x^2.</cmath><br />
Since <math>\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}</math>, we know that <math>x=6</math>, so our final answer is <math>(6)(138) = \boxed{828}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1990|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_2&diff=962211990 AIME Problems/Problem 22018-07-14T18:22:34Z<p>Flyhawkeye: Added Solution 3</p>
<hr />
<div>== Problem ==<br />
Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.<br />
<br />
== Solution 1 ==<br />
Suppose that <math>52+6\sqrt{43}</math> is in the form of <math>(a + b\sqrt{43})^2</math>. [[FOIL]]ing yields that <math>52 + 6\sqrt{43} = a^2 + 43b^2 + 2ab\sqrt{43}</math>. This implies that <math>a</math> and <math>b</math> equal one of <math>\pm1, \pm3</math>. The possible [[set]]s are <math>(3,1)</math> and <math>(-3,-1)</math>; the latter can be discarded since the [[square root]] must be positive. This means that <math>52 + 6\sqrt{43} = (\sqrt{43} + 3)^2</math>. Repeating this for <math>52-6\sqrt{43}</math>, the only feasible possibility is <math>(\sqrt{43} - 3)^2</math>.<br />
<br />
Rewriting, we get <math>(\sqrt{43} + 3)^3 - (\sqrt{43} - 3)^3</math>. Using the difference of [[cube]]s, we get that <math>[\sqrt{43} + 3\ - \sqrt{43} + 3]\ [(43 + 6\sqrt{43} + 9) + (43 - 9) + (43 - 6\sqrt{43} + 9)]</math> <math> = (6)(3 \cdot 43 + 9) = \boxed{828}</math>.<br />
<br />
== Solution 2 ==<br />
<br />
The <math>3/2</math> power is quite irritating to work with so we look for a way to eliminate that. Notice that squaring the expression will accomplish that. <br />
Let <math>S</math> be the sum of the given expression. <br />
<cmath>S^2= ((52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2})^2</cmath><br />
<cmath>S^2 = (52+6\sqrt{43})^{3} + (52-6\sqrt{43})^{3} - 2((52+6\sqrt{43})(52-6\sqrt{43}))^{3/2}</cmath><br />
After doing the arithmetic (note that the first two terms will have some cancellation and that the last term will simplify quickly using difference of squares), we arrive at <math>S^2 = 685584</math> which gives <math>S=\boxed{828}</math>.<br />
<br />
== Solution 3 ==<br />
<br />
Factor as a difference of cubes.<br />
\begin{align*}<br />
\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[\left(\left(\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\right)^2+\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}}+\left(\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right)^2\right)\right] &= \\<br />
\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+\left(52^2-\left(36\right)\left(43\right)\right)^{\frac{1}{2}}\right] &= \\<br />
\left[\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}}\right]\left[104+34\right]<br />
\end{align*}<br />
We can simplify the left factor as follows.<br />
\begin{align*}<br />
\left(52+6\sqrt{43}\right)^{\frac{1}{2}}-\left(52-6\sqrt{43}\right)^{\frac{1}{2}} &= x \\<br />
104-2\left(52+6\sqrt{43}\right)^{\frac{1}{2}}\left(52-6\sqrt{43}\right)^{\frac{1}{2}} &= x^2 \\<br />
104-68 &= x^2 \\<br />
36 &= x^2<br />
\end{align*}<br />
Since <math>\left(52+6\sqrt{43}\right)^{\frac{1}{2}} > \left(52-6\sqrt{43}\right)^{\frac{1}{2}}</math>, we know that <math>x=6</math>, so our final answer is <math>(6)(138) = \boxed{828}</math>.<br />
<br />
== See also ==<br />
{{AIME box|year=1990|num-b=1|num-a=3}}<br />
<br />
[[Category:Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1987_AHSME_Problems/Problem_19&diff=961321987 AHSME Problems/Problem 192018-07-12T12:02:07Z<p>Flyhawkeye: /* Solution */ Corrected solution and added actual approximate value</p>
<hr />
<div>==Problem==<br />
<br />
Which of the following is closest to <math>\sqrt{65}-\sqrt{63}</math>?<br />
<br />
<math>\textbf{(A)}\ .12 \qquad<br />
\textbf{(B)}\ .13 \qquad<br />
\textbf{(C)}\ .14 \qquad<br />
\textbf{(D)}\ .15 \qquad<br />
\textbf{(E)}\ .16 </math> <br />
<br />
== Solution ==<br />
We have <math>\sqrt{65} > 8 > 7.5</math>. Also <math>7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63</math>, so <math>\sqrt{63} > 7.5</math>. Thus <math>\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15</math>. Now notice that <math>\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}}</math>, so <math>\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...</math>, so the answer must be <math>A</math> or <math>B</math>. To determine which, we write <math>\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 128 \cdot 0.0078125 + 0.0078125^2 = 4096 - 1 + 0.0078125^2</math> which is true. Hence as the expression is greater than <math>0.125</math>, and less than or equal to <math>0.13</math> (since we showed it is certainly less than <math>0.1333333...</math>), it is closest to <math>0.13</math>, which is answer <math>\boxed{\text{B}}</math>.<br />
<br />
(<math>\sqrt{65} - \sqrt{63}</math> is approximately equal to <math>0.125003815</math>)<br />
<br />
== See also ==<br />
{{AHSME box|year=1987|num-b=18|num-a=20}} <br />
<br />
[[Category: Intermediate Algebra Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=94365Gmaas2018-05-09T02:13:08Z<p>Flyhawkeye: Undo revision 94364 by Feedbackloop (talk)</p>
<hr />
<div>=== Gmaas Facts ===<br />
- None of these facts a true, except all of them<br />
<br />
- gm lion isn't even his final form.<br />
<br />
- piphi posts GMAAS pictures in his blog<br />
https://artofproblemsolving.com/community/c580181<br />
<br />
-Gmaas once farted. The result was the Big Bang.<br />
<br />
- Gmaas knows these digits of pi from memory: 3.1415926535897932384626433. Edit: He invented pi and pie.<br />
<br />
-since Gmaass invented pi, he knows that the digits are really 9.587979087879877897087r09780970987098790870987609870987908067897578786. digit-ese for "Gmaass is the ruler of the earth."<br />
<br />
- Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br />
<br />
- Gmaas has a summer house on Mars.<br />
<br />
-Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time.<br />
<br />
-Gmaas also attended Hogwarts and was a prefect.<br />
<br />
-Mrs.Norris is Gmaas's archenemy.<br />
<br />
-Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin.<br />
<br />
-Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br />
<br />
-Gmaas actually attended all the Ivy Leagues.<br />
<br />
- I am Gmaas<br />
<br />
-I too am Gmaas<br />
<br />
-But it is I who is Gmaas<br />
<br />
-Gmass is us all<br />
<br />
- Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br />
<br />
-Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br />
<br />
-Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br />
<br />
-Gmaas crossed the Delaware River with Washington.<br />
<br />
-Gmaas also crossed the Atlantic with the pilgrims.<br />
<br />
-if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br />
<br />
-Chuck norris makes Gmaas jokes.<br />
<br />
-Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br />
<br />
-Gmaas also owns Animal Farm. Napoleon was his servant.<br />
<br />
-Gmaas is the only one who knows where Amelia Earhart is.<br />
<br />
-Gmaas is the only cat that has been proven transcendental.<br />
<br />
- Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br />
<br />
-Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmass in't very happy about that, either.<br />
<br />
- The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br />
<br />
-Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br />
<br />
-Gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br />
<br />
- Gmaas does merely not use USD. He owns it.<br />
<br />
-Gmaas really knows that roblox is awful and does not play it seriously, thank god our lord is sane<br />
<br />
-In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br />
<br />
-"Actually, my name is spelled GMAAS"<br />
<br />
- Gmaas is the smartest living being in the universe.<br />
<br />
-It was Gmaas who helped Monkey King on the Journey to the West.<br />
<br />
-Gmaas is the real creator of Wikipedia.<br />
<br />
-It is said Gmaas could hack any website he desires.<br />
<br />
-Gmaas is the basis of Greek mythology<br />
<br />
-Gmaas once sold Google to a man for around <math>12</math> dollars!<br />
<br />
-Gmaas uses a HP printer.<br />
<br />
-Gmaas owns all AoPS staff including Richard Rusczyk<br />
<br />
-Gmaas was there when Yoda was born.<br />
<br />
- Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br />
<br />
- sseraj once spelled gmaas as gmaas on accident in Introduction to Geometry (1532).<br />
<br />
-Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br />
<br />
-Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br />
<br />
-Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br />
<br />
-"i am sand" destroyed Gmaas in FTW<br />
<br />
- sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br />
<br />
-Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also <math>\boxed{\text{loves}}</math> [b]Cat[/b]ch that fish.<br />
<br />
-Gmaas is Roy Moore's horse in the shape of a cat<br />
<br />
-Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over <math>289547987693</math> robux and <math>190348</math> in CPR.<br />
<br />
-This is all hypothetical EDIT: This is all factual <br />
<br />
-Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br />
(Warrior cats reference)<br />
<br />
- He is capable of salmon powers, according to PunSpark (ask him)<br />
<br />
The Gmaas told Richard Rusczyk to make AoPS<br />
<br />
-The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br />
<br />
-The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999th dimension<br />
<br />
-Certain theories provide evidence that he IS darth plagueis the wise<br />
<br />
-Gmaas is "TIRED OF PEOPLE ADDING TO HIS PAGE!!" (Maas 45).<br />
<br />
-Gmaas has multiple accounts; some of them are pifinity, cyumi, squareman, Electro3.0, and lakecomo224<br />
<br />
-Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br />
<br />
-Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br />
<br />
-Gmaas knows how to hack into top secret aops community pages<br />
<br />
-Gmaas was a river clant cat who crossed the event horizon of a black hole and came out the other end!<br />
<br />
- Gmaas is king of the first men, the anduls<br />
<br />
-Gmaas is a well known professor at MEOWston Academy<br />
<br />
-Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout<br />
<br />
- Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat (He also out grumpied grumpy cat!!!)<br />
<br />
-Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM<br />
<br />
- owner of sseraj, not pet<br />
<br />
- embodiment of life and universe and beyond <br />
<br />
- Watches memes<br />
<br />
-After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br />
<br />
-Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br />
<br />
-Gmaas is a certified Slytherin.<br />
<br />
-Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br />
<br />
-Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br />
<br />
- Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
<br />
-Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br />
<br />
-Gmaas is a supreme overlord who must be given <math>10^{1000000000000000000000^{1000000000000000000000}}</math> minecraft DIAMONDS<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br />
<br />
- Gmaas is everyone's favorite animal. <br />
<br />
- He lives with sseraj. <br />
<br />
-Gmaas is my favorite pokemon<br />
<br />
-Gmaas dislikes number theory but enjoys geometry.<br />
<br />
- Gmaas is cool<br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS</cmath> supercars, excluding the Purrari and the 138838383 Teslas. <br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
<br />
- He is distant relative of Mathcat1234.<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
<br />
~Alligator Swamp A 1072 <br />
<br />
~Alligator Swamp B 1073<br />
<br />
~Prealgebra A (1488)<br />
<br />
~Introduction to Algebra A (1170)<br />
<br />
~Introduction to Algebra B (1529)<br />
<br />
~Welcome to Panda Town Gate 1076<br />
<br />
~Welcome to Gmaas Town Gate 1221<br />
<br />
~Welcome to Gmaas Town Gate 1125<br />
<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
<br />
~The other side of the ice in Antarctica<br />
<br />
~Feisty Alligator Swamp 1115<br />
<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
<br />
~Introduction to Counting and Probability 1142 <br />
<br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
<br />
~Intermediate Counting and Probability 1137<br />
<br />
~Intermediate Counting and Probability 1207<br />
<br />
~Posting student surveys<br />
<br />
~USF Castle Walls - Elven Tribe 1203<br />
<br />
~Dark Lord's Hut 1210<br />
<br />
~AMC 10 Problem Series 1200<br />
<br />
~Intermediate Number Theory 1138<br />
<br />
~Intermediate Number Theory 1476<br />
<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
<br />
~Algebra B 1112<br />
<br />
~Intermediate Algebra 1561 7:17 PM 12/11/16<br />
<br />
~Nowhere Else, Tasmania<br />
<br />
~Earth Dimension C-137<br />
~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
EDIT: Nobody has yet seen him atop a tribal base yet.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: Gmaas rarely disguises himself as a penguin.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br />
<br />
-Aha!! An impostor!! <br />
http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br />
(look at the acronym).<br />
<br />
-EDIT. The above fact is slightly irrelevant.<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99999.\overline{9}}{100000}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- EDIT: Gmaas is a he.<br />
<br />
-Gmaas is love, Gmaas is life<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mew.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br />
https://artofproblemsolving.com/community/c287916h1291232<br />
<br />
- EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
-Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br />
<br />
- The original owner of Gmaas is Gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br />
<br />
- It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br />
<br />
- It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmewal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br />
<br />
- Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br />
<br />
-The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br />
<br />
- EDIT: The above fact is somewhat irrelevant.<br />
<br />
-EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has <math>57843504</math> regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
-Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
-Gmaas has a secret hidden garden full of turnips under sseraj's house.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br />
<br />
-Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br />
<br />
-Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
-Donald Trump once sang "All Hail the Chief" to Gmaas, 3 days after being sworn in as US President.<br />
<br />
- Gmaas likes to talk with rrusczyk from time to time.<br />
<br />
- Gmaas can shoot fire from his smelly butt.<br />
<br />
- Gmaas is the reason why the USF has the longest thread on AoPS.<br />
<br />
- Gmass is an avid watcher of the popular T.V. show "Bernie Sanders and the Gauntlet of DOOM"<br />
<br />
- sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying "Who wants to see 5space?" at around 5:16 PM Mountain Time, noting Gmaas was "also 5space"<br />
<br />
-EDIT: he also did it in Introduction to Algebra A once.<br />
<br />
- Gmaas is now my HD background on my Mac.<br />
<br />
- In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br />
<br />
- Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br />
<br />
- In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of <math>-GMAAS</math> seconds.<br />
<br />
-Gmaas beat Superman in a fight with ease<br />
<br />
-Gmaas was an admin of Roblox<br />
<br />
-Gmaas traveled around the world, paying so much <math>MONEY</math> just to eat :D<br />
<br />
-Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br />
Summary:<br />
<br />
-When Gmaas subtracts <math>0.\overline{99}</math> from <math>1</math>, the difference is greater than <math>0</math>.<br />
<br />
-Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br />
<br />
-Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br />
<br />
-The results of the revival are top secret, and nobody knows what happened.<br />
<br />
-sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br />
<br />
-sseraj likes to post a picture of gmaas in every class he passes by.<br />
<br />
-sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br />
EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br />
EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br />
<br />
-Gmaas is the lord of the pokemans<br />
<br />
-Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br />
<br />
-Picture of Gmaas http://i.imgur.com/PP9xi.png<br />
<br />
-Known by Mike Miller<br />
<br />
-Gmaas got mad at sseraj once, so he locked him in his own freezer<br />
<br />
-Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br />
<br />
-Gmaas is an obviously omnipotent cat.<br />
<br />
-ehawk11 met him<br />
<br />
-sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br />
<br />
-sseraj has posted pictures of gmaas in '"intro to algebra", before class started, with the title, "caption contest". anyone who posted a caption mysteriously vanished in the middle of the night. <br />
EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br />
<br />
- gmaas has once slept in your bed and made it wet<br />
<br />
-It is rumored that rrusczyk is actually Gmaas in disguise<br />
<br />
-Gmaas is suspected to be a Mewtwo in disguise<br />
<br />
-Gmaas is a cat but has characteristics of every other animal on Earth.<br />
<br />
-Gmaas is the ruler of the universe and has been known to be the creator of the species "Gmaasians".<br />
<br />
-There is a rumor that Gmaas is starting a poll<br />
<br />
-Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br />
<br />
-There is a rumored sport called "Gmaas Hunting" where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br />
<br />
- Gmaas burped and caused an earthquake.<br />
<br />
- Gmaas once drank from your pretty teacup.<br />
<br />
=== Gmaas photos ===<br />
http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".[/s]Sorry, this was a rick roll troll.<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath and themoocow are also writing a book about Gmaas<br />
<br />
-Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br />
<br />
-Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br />
<br />
- gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
<br />
-oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br />
<br />
-Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
<br />
-No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br />
<br />
-In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br />
https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
<br />
<br />
- Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br />
<br />
-<math>Another</math> sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br />
<br />
-Gmaas has been sighted several times on the Global Announcements forum<br />
<br />
-Gmaas uses the following transportation: <img> http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg </img><br />
<br />
- When Gmaas was mad, he started world wars 1 & 2. It is only because of Gmaas that we have not had World War 3.<br />
<br />
- Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br />
<br />
- Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br />
<br />
-Gmaas likes to whiz on the wilzo<br />
<br />
-Gmaas has been spotted in AMC 8 Basics<br />
<br />
-Gmaas is cool<br />
<br />
-Gmaas hemoon card that does over 9000000 dmg<br />
<br />
-Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br />
<br />
-Kirby once swallowed Gmaas. Gmaas had to spit him out.</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_34&diff=933481950 AHSME Problems/Problem 342018-03-21T00:59:36Z<p>Flyhawkeye: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:<br />
<br />
<math>\textbf{(A)}\ 5\text{ in} \qquad<br />
\textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad<br />
\textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad<br />
\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad<br />
\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math><br />
==Solution==<br />
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be <math>\pi</math> anymore)<br />
We see that the circumference was increased by <math>25\%</math>. This means the radius was also increased by <math>25\%</math>. The radius of the original balloon is <math>\frac{20}{2\pi}=\frac{10}{\pi}</math>. With the <math>25\%</math> increase, it becomes <math>\frac{12.5}{\pi}</math>. The increase is <math>\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math><br />
<br />
==Solution 2==<br />
The radii of the circles are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>.<br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1950|num-b=33|num-a=35}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=1950_AHSME_Problems/Problem_34&diff=933471950 AHSME Problems/Problem 342018-03-21T00:58:40Z<p>Flyhawkeye: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
When the circumference of a toy balloon is increased from <math>20</math> inches to <math>25</math> inches, the radius is increased by:<br />
<br />
<math>\textbf{(A)}\ 5\text{ in} \qquad<br />
\textbf{(B)}\ 2\dfrac{1}{2}\text{ in} \qquad<br />
\textbf{(C)}\ \dfrac{5}{\pi}\text{ in} \qquad<br />
\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in} \qquad<br />
\textbf{(E)}\ \dfrac{\pi}{5}\text{ in}</math><br />
==Solution==<br />
When the circumference of a circle is increased by a percentage, the radius is also increased by the same percentage (or else the ratio of the circumference to the diameter wouldn't be <math>\pi</math> anymore)<br />
We see that the circumference was increased by <math>25\%</math>. This means the radius was also increased by <math>25\%</math>. The radius of the original balloon is <math>\frac{20}{2\pi}=\frac{10}{\pi}</math>. With the <math>25\%</math> increase, it becomes <math>\frac{12.5}{\pi}</math>. The increase is <math>\frac{12.5-10}{\pi}=\frac{2.5}{\pi}=\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math><br />
<br />
==Solution 2==<br />
The radii of the circle are <math>\frac{20}{2\pi}</math> and <math>\frac{25}{2\pi}</math>, respectively. The positive difference is therefore <math>\boxed{\textbf{(D)}\ \dfrac{5}{2\pi}\text{ in}}</math>.<br />
<br />
==See Also==<br />
{{AHSME 50p box|year=1950|num-b=33|num-a=35}}<br />
<br />
[[Category:Introductory Algebra Problems]]<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Flyhawkeyehttps://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=92439Gmaas2018-02-28T00:56:47Z<p>Flyhawkeye: /* Known Facts About Gmaas */</p>
<hr />
<div>=== Known Facts About Gmaas ===<br />
- The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br />
<br />
-Gmaas dueled Grumpy Cat and won<br />
<br />
gmaas sits on the statue of pallas and says Nevermore(the Raven refrence )<br />
<br />
- Gmaas does merely not use USD. He owns it.<br />
<br />
- Gmaas is the real reason of the "Trump" rally. One day he let Trump take over, and the Dow Jones plummeted.<br />
<br />
-Gmaas really knows that roblox is awful and does not play it seriously, thank god our lord is sane<br />
<br />
-In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br />
<br />
-"Actually, my name is spelled GMAAS"<br />
<br />
-Gmass owns all AoPS staff including Richard Rusczyk<br />
<br />
- Gmass's true number of lives left is unknown; however, Gmass recently confirmed that he had at least a million left. Why does Gmass have so many more lives than other cats? The power of Gmass.<br />
<br />
- sseraj once spelled gmaas as gmass on accident in Introduction to Geometry (1532).<br />
<br />
-Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br />
<br />
-Gmaas has beaten Chuck Norris in a fight...twice<br />
<br />
-Gmass is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br />
<br />
-"i am sand" destroyed Gmass in FTW<br />
<br />
- sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br />
<br />
-Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten<br />
<br />
-Gmaas is Roy Moore's horse in the shape of a cat<br />
<br />
-Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over <math>289547987693</math> robux and <math>190348</math> in CPR.<br />
<br />
-This is all hypothetical EDIT: This is all factual <br />
<br />
-Gmass's real name is Princess. He has a sibling named Rusty<br />
(Warrior cats reference)<br />
<br />
- He is capable of salmon powers, according to PunSpark (ask him)<br />
<br />
The Gmaas told Richard Rusczyk to make AoPS<br />
<br />
-The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br />
<br />
-The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999th dimension<br />
<br />
-Certain theories provide evidence that he IS darth plagueis the wise<br />
<br />
-Gmaas is "TIRED OF PEOPLE ADDING TO HIS PAGE!!" (Maas 45).<br />
<br />
-Gmaas has multiple accounts; some of them are pifinity, Lord_Baltimore, Spacehead1AU, squareman, and Electro3.0<br />
<br />
-Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br />
<br />
-Gmass owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br />
<br />
- Gmass made this page<br />
<br />
- Gmaas is king of the first men, the anduls<br />
<br />
-Gmass is a well known professor at MEOWston Academy<br />
<br />
-Gmass is a Tuna addict, along with other, more potent fish such as Salmon and Trout<br />
<br />
- Gmass won the reward of being cutest and fattest cat ever--he surpassed grumpy cat<br />
<br />
-Last sighting 1571-stretch-algebra-a 12/6/17<br />
<br />
- owner of sseraj, not pet<br />
<br />
- embodiment of life and universe and beyond <br />
<br />
- Watches memes<br />
<br />
-After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br />
<br />
-Gmass's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br />
<br />
-Gmass is a certified Slytherin and probably the cutest cat ever.<br />
<br />
-Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br />
<br />
-Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br />
<br />
- Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br />
<br />
-Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br />
<br />
-Gmaas is a supreme overlord who must be given <math>10^{1000000000000000000000^{1000000000000000000000}}</math> minecraft DIAMONDS<br />
<br />
- gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br />
<br />
- Gmaas is 5space's favorite animal. <br />
<br />
- He lives with sseraj. <br />
<br />
-Gmaas is my favorite pokemon<br />
<br />
-Gmaas dislikes number theory but enjoys geometry.<br />
<br />
- Gmaas is cool<br />
<br />
- He is often overfed (with probability <math>\frac{3972}{7891}</math>), or malnourished (with probability <math>\frac{3919}{7891}</math>) by sseraj.<br />
<br />
- He has <cmath>\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS</cmath> supercars, excluding the Purrari and the 138838383 Teslas. <br />
<br />
- He is an employee of AoPS.<br />
<br />
- He is a gmaas with yellow fur and white hypnotizing eyes.<br />
<br />
- He was born with a tail that is a completely different color from the rest of his fur.<br />
<br />
- His stare is very hypnotizing and effective at getting table scraps.<br />
<br />
- He sometimes appears several minutes before certain classes start as an admin. <br />
<br />
- He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br />
<br />
- It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br />
<br />
- Actually, he is a cat. He said so. And science also says so.<br />
<br />
- He is distant relative of Mathcat1234.<br />
<br />
- He is very famous now, and mods always talk about him before class starts.<br />
<br />
- His favorite food is AoPS textbooks because they help him digest problems.<br />
<br />
- Gmaas tends to reside in sseraj's fridge.<br />
<br />
- Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br />
<br />
- The fur of Gmaas can protect him from the harsh conditions of a freezer.<br />
<br />
- Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br />
<br />
- Gmaas is a sage omniscient cat.<br />
<br />
- He is looking for suitable places other than sseraj's fridge to live in.<br />
<br />
- Places where gmaas sightings have happened: <br />
~The Royal Scoop ice cream store in Bonita Beach Florida<br />
<br />
~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br />
<br />
~Alligator Swamp A 1072 <br />
<br />
~Alligator Swamp B 1073<br />
<br />
~Prealgebra A (1488)<br />
<br />
~Introduction to Algebra A (1170)<br />
<br />
~Introduction to Algebra B (1529)<br />
<br />
~Welcome to Panda Town Gate 1076<br />
<br />
~Welcome to Gmaas Town Gate 1221<br />
<br />
~Welcome to Gmaas Town Gate 1125<br />
<br />
~33°01'17.4"N 117°05'40.1"W (Rancho Bernardo Road, San Diego, CA)<br />
<br />
~The other side of the ice in Antarctica<br />
<br />
~Feisty Alligator Swamp 1115<br />
<br />
~Introduction to Geometry 1221 (Taught by sseraj)<br />
<br />
~Introduction to Counting and Probability 1142 <br />
<br />
~Feisty-ish Alligator Swamp 1115 (AGAIN)<br />
<br />
~Intermediate Counting and Probability 1137<br />
<br />
~Intermediate Counting and Probability 1207<br />
<br />
~Posting student surveys<br />
<br />
~USF Castle Walls - Elven Tribe 1203<br />
<br />
~Dark Lord's Hut 1210<br />
<br />
~AMC 10 Problem Series 1200<br />
<br />
~Intermediate Number Theory 1138<br />
<br />
~Intermediate Number Theory 1476<br />
<br />
~Introduction To Number Theory 1204. Date:7/27/16.<br />
<br />
~Algebra B 1112<br />
<br />
~Intermediate Algebra 1561 7:17 PM 12/11/16<br />
<br />
~Nowhere Else, Tasmania<br />
<br />
~Earth Dimension C-137<br />
<br />
<br />
<br />
- These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&B).<br />
<br />
- Gmaas has also been sighted in Olympiad Geometry 1148.<br />
<br />
- Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br />
<br />
- Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br />
<br />
- Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br />
<br />
EDIT: Nobody has yet seen him atop a tribal base yet.<br />
<br />
- Gmaas are often under the disguise of a penguin or cat. Look out for them.<br />
<br />
EDIT: Gmaas rarely disguises himself as a penguin.<br />
<br />
- He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br />
<br />
EDIT: He IS an AoPS site admin.<br />
<br />
- If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br />
<br />
- Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br />
<br />
-Aha!! An impostor!! <br />
http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br />
(look at the acronym).<br />
<br />
- Gmaas might have been viewing (with a <math>\frac{99999}{100000}</math> chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br />
<br />
- EDIT: Gmaas is a he.<br />
<br />
-Gmaas is love, Gmaas is life<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mew.<br />
<br />
- Gmaas is on the list of "Elusive Creatures." If you have questions or want the full list, contact moab33.<br />
<br />
- Gmaas can be summoned using the <math>\tan(90)</math> ritual. Draw a pentagram and write the numerical value of <math>\tan(90)</math> in the middle, and he will be summoned.<br />
<br />
- EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br />
https://artofproblemsolving.com/community/c287916h1291232<br />
<br />
- EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br />
<br />
- Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br />
<br />
- EDIT: That has never happened and thus it does not contain the singularity of a black hole. <br />
<br />
- Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br />
<br />
-Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br />
<br />
- The original owner of Gmaas is Gmaas.<br />
<br />
- Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.)<br />
<br />
- It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br />
<br />
- The name of Gmaas is so powerful, it radiates Deja Mu.<br />
<br />
- Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br />
<br />
- It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br />
<br />
- Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br />
<br />
- It has been confirmed that gmaas uses gmewal as his email service<br />
<br />
- Gmaas enjoys wearing gmean shorts<br />
<br />
- Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br />
<br />
- Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks.<br />
<br />
- Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br />
<br />
- Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br />
<br />
- Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br />
<br />
- In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seem the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br />
<br />
- The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br />
<br />
-The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br />
<br />
- EDIT: The above fact is somewhat irrelevant.<br />
<br />
- Gmaas is a Super Duper Uper Cat Time Lord. He has <math>57843504</math> regenerations and has used <math>3</math>. <cmath>9\cdot12\cdot2\cdot267794=57843504</cmath>. <br />
<br />
-Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br />
<br />
- Gmaas loves to eat turnips. At <math>\frac{13}{32}</math> of the sites he was spotted at, he was seen with a turnip.<br />
<br />
-Gmaas has a secret hidden garden full of turnips under sseraj's house.<br />
<br />
- Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br />
<br />
-Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br />
<br />
-Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows. (Except him.)<br />
<br />
-Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br />
<br />
-Gmaas is an excellent driver.<br />
<br />
-The native location of Gmaas is the twilight zone.<br />
<br />
-Donald Trump once sang "All Hail the Chief" to Gmaas, 3 days after being sworn in as US President.<br />
<br />
- Gmaas likes to talk with rrusczyk from time to time.<br />
<br />
- Gmaas can shoot fire from his paws.<br />
<br />
- Gmaas is the reason why the USF has the longest thread on AoPS.<br />
<br />
- He (or she) is an avid watcher of the popular T.V. show "Bernie Sanders and the Gauntlet of DOOM"<br />
<br />
- sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying "Who wants to see 5space?" at around 5:16 PM Mountain Time, noting Gmaas was "also 5space"<br />
<br />
-EDIT: he also did it in Introduction to Algebra A once<br />
<br />
- Gmaas is now my HD background on my Mac.<br />
<br />
- In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br />
<br />
- Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br />
<br />
- In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of <math>-GMAAS</math> seconds.<br />
<br />
-Gmass beat Superman in a fight with ease<br />
<br />
-Gmass was an admin of Roblox<br />
<br />
-Gmass traveled around the world, paying so much <math>MONEY</math> just to eat :D<br />
<br />
-Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br />
Summary:<br />
<br />
-When Gmaas subtracts <math>0.\overline{99}</math> from <math>1</math>, the difference is greater than <math>0</math>.<br />
<br />
-Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br />
<br />
-Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br />
<br />
-The results of the revival are top secret, and nobody knows what happened.<br />
<br />
-sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br />
<br />
-sseraj likes to post a picture of gmaas in every class he passes by.<br />
<br />
-sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Ender is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmass is now wandering space in search for a home.<br />
EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br />
<br />
-Gmaas is the lord of the pokemans<br />
<br />
-Gmass can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will<br />
<br />
-Picture of Gmass http://i.imgur.com/PP9xi.png<br />
<br />
-Known by Mike Miller<br />
<br />
-Gmaas got mad at sseraj once, so he locked him in his own freezer<br />
<br />
-Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br />
<br />
-Gmaas is an obviously omnipotent cat.<br />
<br />
-ehawk11 met him<br />
<br />
-sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br />
<br />
-sseraj has posted pictures of gmass in '"intro to algebra", before class started, with the title, "caption contest" anyone who posted a caption mysteriously vanished in the middle of the night. <br />
EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmass is typing this through his/her account...)<br />
<br />
- gmass has once slept in your bed and made it wet<br />
<br />
-It is rumored that rrusczyk is actually Gmaas in disguise<br />
<br />
-Gmaas is suspected to be a Mewtwo in disguise<br />
<br />
-Gmaas is a cat but has characteristics of every other animal on Earth.<br />
<br />
-Gmass is the ruler of the universe and has been known to be the creator of the species "Gmassians".<br />
<br />
-There is a rumor that Gmaas is starting a poll<br />
<br />
-Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br />
<br />
-There is a rumored sport called "Gmaas Hunting" where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br />
<br />
=== Gmaas photos ===<br />
http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br />
<br />
=== gmaas in Popular Culture ===<br />
<br />
- [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, "The Adventures of gmaas".[/s]Sorry, this was a rick roll troll.<br />
<br />
- BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,"What are gwali?" the customary answer "This is gwali" is returned. Scientist 5space is now looking into it.<br />
<br />
- Sullymath and themoocow are also writing a book about Gmaas<br />
<br />
-Ornx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br />
<br />
- Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br />
<br />
- Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br />
<br />
- Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br />
<br />
- gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br />
<br />
-oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br />
<br />
-Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br />
<br />
-No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br />
<br />
-In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br />
https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br />
https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br />
https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br />
https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br />
<br />
<br />
- Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br />
<br />
-<math>Another</math> sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br />
<br />
-Gmaas has been sighted several times on the Global Announcements forum<br />
<br />
-Gmaas uses the following transportation: [img]http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg[/img]</div>Flyhawkeye