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2006 AIME A Problems/Problem 6
#REDIRECT [[2006 AIME I Problems/Problem 6]]

44 bytes (5 words) - 12:05, 28 June 2009

Page text matches

Pythagorean Theorem
<math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>. {{Halmos}} <cmath>3-4-5</cmath>

5 KB (886 words) - 13:51, 15 May 2024
2005 AMC 12A Problems/Problem 16
<math> (\mathrm {A}) \ 5 \qquad (\mathrm {B}) \ 6 \qquad (\mathrm {C})\ 8 \qquad (\mathrm {D}) \ 9 \qquad (\mathrm {E})\ 10 < [[Image:2005_12A_AMC-16b.png]]

2 KB (307 words) - 15:30, 30 March 2024
2016 AMC 10A Problems/Problem 5
https://youtu.be/bc-somFWrbg https://youtu.be/Msaux-erFJ0

1 KB (184 words) - 13:58, 22 August 2023
2016 AMC 10A Problems/Problem 7
...using the mean in your reasoning, you can just take the mean of the other 6 numbers, and it'll solve it marginally faster. -[[User:Integralarefun|Integ {{AMC10 box|year=2016|ab=A|num-b=6|num-a=8}}

2 KB (268 words) - 18:19, 27 September 2023
2016 AMC 10A Problems/Problem 9
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf ...trying to find the value of <math>N</math> such that <cmath>1+2+3\cdots+(N-1)+N=\frac{N(N+1)}{2}=2016.</cmath> Noticing that <math>\frac{63\cdot 64}{2}

2 KB (315 words) - 15:34, 18 June 2022
2016 AMC 10A Problems/Problem 10
filldraw(rectangle((1,1),(6,4)),gray(0.75)); ...(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8</math>

2 KB (337 words) - 14:56, 25 June 2023
2016 AMC 10A Problems/Problem 11
...uad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8</math> ...</math>, which add to the area of the shaded region, which is <math>\boxed{6\frac{1}{2}}</math>.

8 KB (1,016 words) - 00:17, 31 December 2023
List of United States elementary school mathematics competitions
* [http://www.mindresearch.org/gameathon/ Game-a-thon] challenges students to design and build math games in this national co ...roblem solving contest for elementary and middle school students. [Grades 2-8]

4 KB (473 words) - 16:14, 1 May 2024
List of United States middle school mathematics competitions
...large middle school math contest taken in-school. ([http://www.maa.org/math-competitions website]) *[http://www.imc-impea.org IMC-IMPEA] is an offline/online math contest for all grades level. The contest o

7 KB (792 words) - 10:14, 23 April 2024
Alabama mathematics competitions
* [[Ala-La-Miss Mathematics League]] * [[Alabama School of Fine Arts]] -- 6th grade, Pre-Algebra, Algebra I. Geometry (coming 2015)

3 KB (411 words) - 21:32, 8 December 2014
MathILy (serious mathematics infused with levity)
'''{MathILy, MathILy-Er}''' are a pair of intensive 5-week-long residential summer math programs for high school students. ...ate math students. The math content covered is undergraduate- and graduate-level.

5 KB (706 words) - 23:49, 29 January 2024
MATHCOUNTS
...ter), 2-2.5 (State/National) Target: 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}} ...dents have 40 minutes to complete the Sprint Round. This round is very fast-paced and requires speed and accuracy as well. The first 20 problems are usu

10 KB (1,506 words) - 21:31, 14 May 2024
California mathematics competitions
* [[San Jose JACL-Tokutomi Mathematics Contest]]- for grades 7-12. [http://www.sanjosejacl.org/math.html website] ...Trimathlon] is a fun filled Math Competition for kids in grades 3 through 6.

4 KB (514 words) - 04:02, 21 September 2023
ICTM Math Contest
...thematics offers two areas of math contests: Grade School (Grades 3, 4, 5, 6, 7, 8 + Algebra 1) and High School (Regional and State Finals). ...Committee offers in-school contests at six different grade levels (grades 3-8). The season consists of three contests to be given at your school. Each s

8 KB (1,182 words) - 14:26, 3 April 2024
Kansas mathematics competitions
...], (grades 4-12). It is divided into three divisions, grades 4-6, 7-8 and 9-12. The contest features an individual competition, where participants take ..., (grades 5-12). It is divided into three divisions, grades 5-6, 7-9 and 10-12, and is a 40 minute, 5 question test. First, second and third in each div

3 KB (473 words) - 16:11, 16 June 2020
Math textbooks
* Getting Started is recommended for students grades 6 to 9. ...ing Started with Competition Math], a textbook meant for true beginners (on-target middle school students, or advanced elementary school students). It i

7 KB (901 words) - 14:11, 6 January 2022
William Lowell Putnam Mathematical Competition
...am Lowell Putnam Mathematical Competition''' is a highly challenging, proof-oriented [[mathematics competition]] for undergraduate students in North Ame ...Problem A/B, 1/2: 7 Problem A/B, 3/4: 8 Problem A/B, 5/6: 9}}

4 KB (623 words) - 13:11, 20 February 2024
Computer science books
* [http://www.amazon.com/exec/obidos/ASIN/0262581116/artofproblems-20 Adaptation in Natural and Artificial Systems] by John H. Holland. * [http://www.amazon.com/exec/obidos/ASIN/0780334264/artofproblems-20 Ones and Zeros: Understanding Boolean Algebra, Digital Circuits and the L

2 KB (251 words) - 00:45, 17 November 2023
2005 Alabama ARML TST Problems/Problem 7
=\frac{2}{2}+\frac{4}{4}+\frac{6}{8}+\frac{8}{16}+\cdots\\ 2y=2+\frac{4}{2}+\frac{6}{4}+\frac{8}{8}+\cdots\\

1 KB (193 words) - 21:13, 18 May 2021
Mathematics
==Overview=={{asy image|<math>1\,2\,3\,4\,5\,6\,7\,8\,9\,0</math>|right|The ten [[digit]]s making up the base ten n

6 KB (902 words) - 12:53, 3 September 2019
AMC historical results
...s qualify for AIME each year, rather than a fixed percentage. Usually, 6000-7000 competitors from the AMC 10 and 12 qualify for the AIME. *AIME Floor: 105 (top ~6%)

17 KB (1,921 words) - 20:53, 10 May 2024
Inequality
...b</math> if <math>a</math> is greater than <math>b</math>, that is, <math>a-b</math> is positive. ...b</math> if <math>a</math> is smaller than <math>b</math>, that is, <math>a-b</math> is negative.

12 KB (1,798 words) - 16:20, 14 March 2023
United States of America Mathematical Talent Search
...|difficulty=3-6|breakdown=Problem 1-2: 3-4 Problem 3-5: 5-6}} ...ed to high scorers at the end of the year. These typically include a free t-shirt, along with other prizes like books or software of the participant's c

4 KB (613 words) - 13:08, 18 July 2023
AMC 10
...|breakdown=Problem 1-5: 1 Problem 6-20: 2 Problem 21-25: 3}} ...contest are invited to take the [[AIME]].[http://www.unl.edu/amc/e-exams/e7-aime/adminaime.html]

4 KB (574 words) - 15:28, 22 February 2024
AMC 12
...reakdown=Problem 1-10: 2 Problem 11-20: 3 Problem 21-25: 4}} ...C 12 is scored in a way that penalizes guessing. Correct answers are worth 6 points, incorrect answers are worth 0 points, and unanswered questions are

4 KB (520 words) - 12:11, 13 March 2024
American Invitational Mathematics Examination
...Problem 6-10: 4 Problem 10-12: 5 Problem 12-15: 6}} ...www.unl.edu/amc/ AMC homepage] and their [http://www.unl.edu/amc/e-exams/e7-aime/aime.shtml AIME page]

8 KB (1,057 words) - 12:02, 25 February 2024
1954 AHSME Problems/Problem 46
draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2));

3 KB (415 words) - 18:01, 24 May 2020
2015 IMO Problems
<math>\mathcal{S}</math> is centre-free if for any three points <math>A</math>, <math>B</math>, <math>C</ma <ol style="list-style-type: lower-latin;">

4 KB (692 words) - 22:33, 15 February 2021
United States of America Mathematical Olympiad
...9|breakdown=Problem 1/4: 7 Problem 2/5: 8 Problem 3/6: 9}} The USAMO is an invitation-only proof-type examination administered to approximately 500 of the best and brightest

6 KB (869 words) - 12:52, 20 February 2024
Informatics competitions
...r Computer Science]] Learning, Practice and Challenges tailored for Grade 6-8 ...r Science]] (5-9) Weekly Challenge Online [https://www.beestar.org/computer-science/?&mid=1016&div=1 website]

7 KB (932 words) - 12:13, 15 January 2024
American Regions Mathematics League
{{WotWAnnounce|week=June 6-12}} ...roblem 6/8), 6 (Problem 10) Team: 3.5 (Problem 1-5), 5 (Problem 6-10)}}

2 KB (267 words) - 17:06, 7 March 2020
Joining an ARML team
...as well as practices of previous years' team rounds. Please email Xinke Guo-Xue at xinkeguoxue@gmail.com, or message Xinke's AoPS account "hurdler", if ...room 2112, on Thursdays at 7pm. For more information, e-mail Eric Brattain-Morrin at [mailto:eric.brattain@gmail.com eric.brattain@gmail.com] and visit

21 KB (3,500 words) - 18:41, 23 April 2024
Engineering competitions
* [[First Lego League]] -- Ages 9-14 [http://www.firstlegoleague.org/ Website] * [[First Lego League]] -- Ages 9-14 [http://www.firstlegoleague.org/ Website]

6 KB (706 words) - 21:50, 20 March 2022
Heron's Formula
<math>A=\sqrt{s(s-a)(s-b)(s-c)}</math> where the [[semi-perimeter]] <math>s=\frac{a+b+c}{2}</math>.

4 KB (675 words) - 00:05, 22 January 2024
Simon's Favorite Factoring Trick
...bles in a linear term on the other side. An example would be: <cmath>xy+66x-88y=23333</cmath>where <math>23333</math> is the constant term, <math>xy</ma ...e previous example, <math>xy+66x-88y=23333</math> is the same as: <cmath>(x-88)(y+66)=(23333)+(-88)(66)</cmath>

7 KB (1,107 words) - 07:35, 26 March 2024
USAMO historical results
Since 1996, a perfect score on the [[USAMO]] has been 42 points for 6 problems. Prior to 1996 a perfect score was 100 points across 5 problems. **Po-Ru Loh

10 KB (1,317 words) - 08:16, 23 April 2024
Mathematical problem solving
<math>\frac{\pi^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\cdots</math> 

2 KB (314 words) - 06:45, 1 May 2014
Manhattan Mathematical Olympiad
...U) during April. It is divided into three divisions, grades 5-6, 7-8, and 9-12, and is a three hour, 4 question test, modeled after other math olympiads * [http://www.math.ksu.edu/~soibel/olympiad/ Past tests from 2000-2006]

578 bytes (81 words) - 19:22, 23 November 2015
Geometric mean
The geometric mean of the numbers 6, 4, 1 and 2 is <math>\sqrt[4]{6\cdot 4\cdot 1 \cdot 2} = \sqrt[4]{48} = 2\sqrt[4]{3}</math>. The geometric mean features prominently in the [[Arithmetic Mean-Geometric Mean Inequality]].

2 KB (282 words) - 22:04, 11 July 2008
Principle of Inclusion-Exclusion
The '''Principle of Inclusion-Exclusion''' (abbreviated PIE) provides an organized method/formula to find ...then <math>3!</math> ways to order the other three guys for <math>3!\binom{6}{3}=120</math>. Same goes for <math>|B|</math>, <math>|C|</math>, and <math

9 KB (1,703 words) - 07:25, 24 March 2024
2020 CIME II Problems/Problem 10
First suppose that <math>c=1</math>. Then <math>a+b+1=ab\Rightarrow (a-1)(b-1)=2</math> from whence we have <math>(a,b,c)\in\{(2,3,1),(3,2,1)\}</math>. ...,3c,4c,5c\}</math>, for if <math>a+b\geq 6c</math> we have <math>6c\leq 2c+6\Rightarrow c\leq3/2\Rightarrow c=1</math>; a contradiction since <math>c\ge

2 KB (332 words) - 09:37, 30 December 2021
Factorial
...[[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>3! = 6</math>

10 KB (809 words) - 16:40, 17 March 2024
Menelaus' theorem
pair A=(7,6), B=(0,0), C=(10,0), P=(4,0), Q=(6,8), R; draw((0,0)--(10,0)--(7,6)--(0,0),blue+0.75);

5 KB (804 words) - 03:01, 12 June 2023
San Diego ARML
...or 2007-2009, and [[user:MCrawford | Mathew Crawford]] was a coach for 2006-2007. ...e for most of the members of that team, who were, for the most part, middle-schoolers.

2 KB (378 words) - 16:34, 5 January 2010
Fermat's Little Theorem
...ath> is not [[divisibility|divisible]] by <math>{p}</math>, then <math>a^{p-1}\equiv 1 \pmod {p}</math>. ...> denotes [[Euler's totient function]]. In particular, <math>\varphi(p) = p-1</math> for prime numbers <math>p</math>. In turn, this is a special case o

16 KB (2,658 words) - 16:02, 8 May 2024
Parabola
...second is [[completing the square | completed square]] form, or <math>y=a(x-h)^2+k</math> where a, h, and k are constants and the [[vertex]] is (h,k). T <center> <math>3x^2-14x+8</math> (<math>a</math> is positive)

3 KB (551 words) - 16:22, 13 September 2023
Euler's Totient Theorem
...e. For this reason it is also known as Euler's generalization or the Fermat-Euler theorem. ...ainder when <math>a_1^3+a_2^3+\cdots+a_{2018}^3</math> is divided by <math>6</math>?

3 KB (542 words) - 17:45, 21 March 2023
Number theory/Advanced
...mous problems in algebraic number theory include the [[Birch and Swinnerton-Dyer Conjecture]] and [[Fermat's Last Theorem]]. ...[analysis]] and [[calculus]] are particularly well-suited to studying large-scale properties of prime numbers. The most famous problem in analytic numbe

5 KB (849 words) - 16:14, 18 May 2021
Harmonic mean
...instance, if we tried to take the harmonic mean of the set <math>\{-2, 3, 6\}</math> we would be trying to calculate <math>\frac 3{\frac 13 + \frac 16 ...uently used inequality, the [[RMS-AM-GM-HM | Arithmetic mean-Geometric mean-Harmonic mean inequality]]. The Inequality states that for a set of positive

1 KB (196 words) - 00:49, 6 January 2021
Alabama ARML
...[[Alabama ARML TST]]'s that are written to help decide the team. From 2005-2008, the TST has been written by former AoPS class writer, instructor, and ...l alternates who wish to compete in 2022. In both 2021 and 2022, there are 6 Spring practices to help decide the team.

2 KB (258 words) - 00:50, 28 December 2021
LaTeX
...metimes "Lay-tek", "Lah-tekh", or "Lah-tex") is widely used to produce well-formatted [[math|mathematical]] and scientific writing. <math>\text{\LaTeX}< <cmath>1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.</cmath>

1 KB (164 words) - 19:09, 14 February 2024
Divisibility
...math>48</math> as <math>6 \times 8 = 48</math>, so we may write that <math>6 \mid 48</math>. ...h>6</math> as <math>6/(-2) = -3</math>, so we may write that <math>-2 \mid 6</math>.

2 KB (277 words) - 16:21, 29 April 2023
Divisibility rules
...cular other [[integer]]s. All of these rules apply for [[Base number| base-10]] ''only'' -- other bases have their own, different versions of these rul ...if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8.

8 KB (1,315 words) - 18:18, 2 March 2024
Induction
...d [[International Mathematics Olympiad | IMO]]. A good example of an upper-level problem that can be solved with induction is [http://www.artofproblems ...=1</math>, <math>F_{2}=1</math>, and the recursion <math>F_{n}=F_{n-1}+F_{n-2}</math> when <math>n \geq 3</math>. Prove that <math>F_{1}+F_{2}+ \ldots +

5 KB (768 words) - 20:45, 1 September 2022
Overcounting
...nt counting tools such as [[combinations]] and the [[Principle of Inclusion-Exclusion]]. ...divisible by 6, and there are 16 such numbers. Thus, there are <math>50+33-16=\boxed{67}</math> numbers that are divisible by either 2 or 3.

4 KB (635 words) - 12:19, 2 January 2022
Constructive counting
''How many four-digit numbers are there?'' '''Solution''': We can construct a four-digit by picking the first digit, then the second, and so on until the fourt

12 KB (1,896 words) - 23:55, 27 December 2023
Euclidean algorithm
If we have two non-negative integers <math>a,b</math> with <math>a|b</math> and <math>b\ne0</ma <math>r_{n-1} \pmod {r_n} \equiv 0</math> 

6 KB (924 words) - 21:50, 8 May 2022
Generating function
...tric sequence#Infinite|sum formula for geometric series]] <cmath>\frac{1}{1-x} = \sum_{k=0}^{\infty} x^k = 1 + x + x^2 + x^3 + \dots</cmath> for <math> For example, using a change of variables, <cmath>\frac{1}{1-ax} = \sum_{k=0}^{\infty} (ax)^k = \sum_{k=0}^{\infty} a^kx^k = 1 + ax + a^2

4 KB (659 words) - 12:54, 7 March 2022
Binomial Theorem
...that for [[real]] or [[complex]] <math>a</math>, <math>b</math>, and [[non-negative]] [[integer]] <math>n</math>, <center><math>(a+b)^n = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k</math></center>

5 KB (935 words) - 13:11, 20 February 2024
Prime number
...a number not equal to <math>2,3</math> is congruent to <math>0,2,3,4 \pmod{6}</math>, then the number is not prime i.e a prime ≥ 5 is of the form 6N± ...1)\cdot (2b+1)=2(2ab+a+b)+1</cmath> and if <math>a,b>0</math> this is a non-trivial factorization.

6 KB (985 words) - 12:38, 25 February 2024
Function
...defined when its argument is nonnegative. The range is the set of all non-negative real numbers, because the square root can never return a negative v ...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. Therefore, <math>g

10 KB (1,761 words) - 03:16, 12 May 2023
Complementary counting
...see that there are <math>19</math> of them. Thus, our answer is is <math>99-19 = 80</math>. <math>\square</math> ...2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is a 2 or a 3?''

8 KB (1,192 words) - 17:20, 16 June 2023
Bijection
...] (or ''onto''). A function has a [[Function#The_Inverse_of_a_Function|two-sided inverse]] exactly when it is a bijection between its [[domain]] and [[ [[2001 AIME I Problems/Problem 6]]

1,016 bytes (141 words) - 03:39, 29 November 2021
Combinatorial identity
<math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math> ...gs added to the number of ways to choose <math>k</math> things from <math>n-1</math> things.

12 KB (1,996 words) - 12:01, 18 May 2024
Composite number
...tself. Some composite numbers are <math>4=2^2</math> and <math>12=2\times 6=3\times 4</math>. Composite numbers '''atleast have 2 distinct [[prime]] [[ 4 6 8 9 10 12 14 15 16 18 20 21 22 24 25 26 27 28 30 32 33 34 35 36 38 39 40 42

6 KB (350 words) - 12:58, 26 September 2023
Natural number
...mathbb{Z}_{\geq0}</math> and <math>\mathbb{Z}_0^+</math> for the set of non-negative integers, and <math>\mathbb{Z}_{>0}</math> and <math>\mathbb{Z}^+</ * [[Well Ordering Principle|Well-ordering principle]]

1 KB (162 words) - 21:44, 13 March 2022
Circle
...- 3)^2 + (y + 6)^2 = 25</math> represents the circle with center <math>(3,-6)</math> and radius 5 units. ...polygon that is closest to the circle's area. Therefore we have <math>A-P<A-T</math> so <math>P>T</math>. Let the apothem be <math>a</math> and the peri

9 KB (1,581 words) - 18:59, 9 May 2024
Base numbers
...umber can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...of the decimal place (recall that the decimal place is to the right of the 6, i.e. 2746.0) tells us that there are six <math>10^0</math>'s, the second d

4 KB (547 words) - 17:23, 30 December 2020
Base numbers/Common bases
[[Hexadecimal]] is base 16. The digits in hexadecimal are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. One of its common uses is for color charts

2 KB (351 words) - 10:39, 1 October 2015
2001 AMC 12 Problems/Problem 1
...d \textbf{(B)}\ 3S + 2\qquad \textbf{(C)}\ 3S + 6 \qquad\textbf{(D)}\ 2S + 6 \qquad \textbf{(E)}\ 2S + 12</math> {{AMC12 box|year=2001|before=First question|num-a=2}}

788 bytes (120 words) - 10:32, 8 November 2021
2001 AMC 12 Problems/Problem 2
...1 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} of the integer <math>n</math>. For example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a

1,007 bytes (165 words) - 00:28, 30 December 2023
American Scholastic Mathematics Association
== 2019-2020 Contest Dates == Contest #6 - March 12, 2020

1 KB (153 words) - 13:11, 14 May 2019
Least common multiple
...of two numbers, 4 and 6. We would begin by listing the multiples of 4 and 6 until we find the smallest number in both lists, as shown below. 6 '''12'''

2 KB (383 words) - 10:49, 4 September 2022
Math Bee
* Level III: For grades 6, 7, and 8. [[MATHCOUNTS]]/[[AMC 8]]-type problems can be found on Level 3. ...se questions to be solved in 35 minutes, while the second phase has 20 free-response questions to be solved in 15 minutes.

1 KB (197 words) - 10:59, 14 April 2024
Math Day at the Beach
...oth individual and team competition. Teams represent high schools and have 6 members each. The competition takes place on a Saturday in March. This round lasts 45 minutes and consists of 15 multiple-choice questions. Scoring consists of:

4 KB (644 words) - 12:56, 29 March 2017
Pi
...ces, such as in the sum <math>\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations for pi are <math>\frac{ ...e leave to the reader), so, it will suffice to show that <math>\frac{\pi^2}6</math>

8 KB (1,469 words) - 21:11, 16 September 2022
Fibonacci sequence
...sion|recursively]] as <math>F_1 = F_2 = 1</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n \geq 3</math>. This is the simplest nontrivial exampl .... <math>\varphi</math> is a solution of the quadratic equation <math>x^2-x-1=0</math>. The other root is <math>-\varphi^{-1} = \frac{1-\sqrt{5}}{2}</m

6 KB (957 words) - 23:49, 7 March 2024
1988 AJHSME Problems/Problem 5
label("160",(1.6,.5),NE); {{AJHSME box|year=1988|num-b=4|num-a=6}}

1 KB (160 words) - 16:53, 17 December 2020
Fundamental Theorem of Calculus
...th>e^{t^2}</math> because I wanted a function that doesn't have a nice anti-derivative.) Interpret the distance that the object travels between times < ...een broken up into <math>n</math> subintervals, so <math>\Delta x = \frac{b-a}{n}</math>. Let <math>x_0 = a, x_1 = a + \Delta x, x_2 = a + 2\Delta x, \

11 KB (2,082 words) - 15:23, 2 January 2022
Polygon
...e number of sides. Thus in regular polygons, any angle is <math>\frac{180(n-2)}{n}^\circ</math>. |6

2 KB (372 words) - 19:04, 30 May 2015
Vector
...ected line segment. In many situations, a vector is best considered as an n-tuple of numbers (often real or complex). Most generally, but also most abst ...h (or magnitude) and the angle it makes with some fixed line (usually the x-axis) or by describing it as an arrow beginning at the origin and ending at

7 KB (1,265 words) - 13:22, 14 July 2021
Logarithm
...down to never having to deal with massive numbers. ex. :<cmath>((((((3^5)^6)^7)^8)^9)^{10})^{11}=\underbrace{1177\ldots 1}_{\text{793549 digits}}</cmat <cmath>\log_4(6)\approx 1.292481250360578090726869472\ldots</cmath>

4 KB (680 words) - 12:54, 16 October 2023
Law of Cosines
...les with two sides and an include angle are [[congruent]] by the Side-Angle-Side congruence postulate. <cmath>a^2+b^2-2f^2=d^2+e^2</cmath>

6 KB (1,003 words) - 09:11, 7 June 2023
Vocabulary and spelling competitions
...cripps Howard National Spelling Bee | National Spelling Bee]] The most well-known middle school competition, at least in the US. [http://spellingbee.com ...hallenge | Word Power Challenge]] The premier vocabulary competition, for 6-8th grades. Over 3 million students participate each year. 2 AoPSers have wo

786 bytes (99 words) - 17:53, 22 June 2006
Vocabulary competitions
...hallenge | Word Power Challenge]] The premier vocabulary competition, for 6-8th grades. Over 3 million students participate each year. 2 AoPSers have wo

392 bytes (50 words) - 02:36, 29 November 2018
Angle bisector
{{WotWAnnounce|week=June 6-12}} X=A+expi((angle(A-B)+angle(C-A))/2);

3 KB (575 words) - 15:27, 19 March 2023
Mathew Crawford
* Two-time perfect scorer on the [[AHSME]]. * Three-time invitee to the [[Mathematical Olympiad Summer Program]].

2 KB (360 words) - 02:20, 2 December 2010
Law of Sines
\qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2

4 KB (658 words) - 16:19, 28 April 2024
2005 USAMO Problems/Problem 4
...math>, where <math>n</math> is a positive integer. For how many ordered 4-tuples <math>(k_1, k_2, k_3, k_4)</math> of nonnegative integers can we cut ...we can biject the stable tables of <math>x</math> to the stable of <math>2n-x</math> by replacing all of the legs with the cut off portions.

7 KB (1,276 words) - 20:51, 6 January 2024
Functional equation
...function: <math>f(x) = x^2 + 6</math>. The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>. In this case, <mat

2 KB (361 words) - 14:40, 24 August 2021
Floor function
...floor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> ...math>n</math> satisfy the equation <math>\left[\frac{n}{5}\right]=\frac{n}{6}</math>.

3 KB (508 words) - 21:05, 26 February 2024
Pascal's triangle
...math>{4 \choose 2} = 6</math>. Pascal's Triangle thus can serve as a "look-up table" for binomial expansion values. Also, many of the characteristics ...above it. For example, <math>{5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}</math>. This property allows the easy creation of the first few

5 KB (838 words) - 17:20, 3 January 2023
1985 IMO Problems/Problem 5
{{IMO box|year=1985|num-b=4|num-a=6}}

3 KB (496 words) - 13:35, 18 January 2023
Trigonometry
A common mnemonic to remember this is '''SOH-CAH-TOA''', where '''S'''ine = '''O'''pposite / '''H'''ypotenuse, '''C'''osine = ...on for acute angles, but extends their domain from acute angles to all real-valued angles. As such, this definition is usually preferred in intermediate

8 KB (1,217 words) - 20:15, 7 September 2023
Harmonic series
There are several sub-types of '''harmonic series'''. The [[zeta-function]] is a harmonic series when the input is one.

2 KB (334 words) - 20:52, 13 March 2022
Goldbach Conjecture
<math>6 = 3 + 3</math> ...s: for instance, it assumes that the events that <math>m</math> and <math>n-m</math> are prime are statistically independent of each other. Nevertheless

7 KB (1,201 words) - 16:59, 19 February 2024
Riemann Hypothesis
...of the seven [http://www.claymath.org/millennium-problems/millennium-prize-problems millenium questions]. * All nontrivial zeros of all [[L-series]] have real part one half where an L-series is of the form <math>\sum_{n=1}^\infty \frac{a_n}{n^s}</math>. This

2 KB (425 words) - 12:01, 20 October 2016
International Mathematical Olympiad
...c competitions. Each year, countries from around the world send a team of 6 students to compete in a grueling competition. ...lem SL1-2: 5.5-7 Problem SL3-4: 7-8 Problem SL5+: 8-10}}

3 KB (490 words) - 03:32, 23 July 2023
Euler's Polyhedral Formula
...f [[vertex|vertices]], [[edge]]s, and [[face]]s, respectively. Then <math>V-E+F=2</math>. ...on &4 &6 & 4 \\ \hline Cube/Hexahedron & 8 & 12 & 6\\ \hline Octahedron & 6 & 12 & 8\\ \hline Dodecahedron & 20 & 30 & 12\\ \hline Icosahedron & 12 & 3

1,006 bytes (134 words) - 14:15, 6 March 2022
Mock AMC
Listed below are the [higher-quality] Mock AMCs that have been hosted over AoPS in the past. Feel free to {| class="wikitable" style="text-align:center;width:100%"

51 KB (6,175 words) - 20:58, 6 December 2023
Mock AIME
...listed under the range of their respective competition season, whereas non-recurring competitions will be listed in the year of their release.) ** [[Mock AIME 1 2006-2007]]

8 KB (906 words) - 17:30, 26 April 2024
Riemann zeta function
<math>\frac{\pi^2}{6}</math>. Euler also found that since every number is the product (1-p^{-s})^{-1}.</cmath>

9 KB (1,547 words) - 03:04, 13 January 2021
Modular arithmetic/Introduction
<math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math> <math>0, 1, 2, \ldots, n-1. </math>

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Perfect square
...square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>

954 bytes (155 words) - 01:14, 29 November 2023
Modular arithmetic/Intermediate
.... This structure gives us a useful tool for solving a wide range of number-theoretic problems, including finding solutions to [[Diophantine equation|Di * Subtraction: <math>a-c\equiv b-d\pmod {n}</math>.

14 KB (2,317 words) - 19:01, 29 October 2021
Right triangle
label("$45^{\circ}$", A, 6*dir(290));

3 KB (499 words) - 23:41, 11 June 2022
2006 AMC 10A
* [[2006 AMC 10A Problems/Problem 6]]

2 KB (180 words) - 18:06, 6 October 2014
2006 AIME I Problems
Let set <math> \mathcal{A} </math> be a 90-element subset of <math> \{1,2,3,\ldots,100\}, </math> and let <math> S </ma The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+

7 KB (1,173 words) - 03:31, 4 January 2023
2006 AIME I Problems/Problem 15
Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possi ...sequence. The only thing that will be left will be a sequence <math>\{0,3,6,9,\cdots,3k\}</math> for some even <math>k</math>. Since we started with 20

6 KB (910 words) - 19:31, 24 October 2023
2006 AIME I Problems/Problem 13
...For each positive integer <math> n, </math> let <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 ...math> but not by <math>2^n, \ldots,</math> and <math>2^{n-1}-2^{n-2} = 2^{n-2}</math> elements of <math>S</math> that are divisible by <math>2^1</math>

10 KB (1,702 words) - 00:45, 16 November 2023
2006 AIME I Problems/Problem 11
...mediately on top of a cube with edge-length <math> k </math> must have edge-length at most <math> k+2. </math> ...e 2 towers which use blocks <math>1, 2</math>, so there are <math>2\cdot 3^6 = 1458</math> towers using blocks <math>1, 2, \ldots, 8</math>, so the answ

3 KB (436 words) - 05:40, 4 November 2022
2006 AIME I Problems/Problem 10
draw((6.5,0)--origin--(0,6.5), Arrows(5)); draw((6.5,0)--origin--(0,6.5), Arrows(5));

4 KB (731 words) - 17:59, 4 January 2022
2006 AIME I Problems/Problem 9
...th> [[positive integer]]s. <math>a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}</math> so <math>a^{2}r^{11}=2^{1003}</math>. y&=&\frac{1003-2x}{11} \end{eqnarray*}</cmath>

4 KB (651 words) - 18:27, 22 May 2021
2006 AIME I Problems/Problem 7
fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white);

4 KB (709 words) - 01:50, 10 January 2022
2006 AIME I Problems/Problem 5
The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+ <cmath> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath>

3 KB (439 words) - 18:24, 10 March 2015
2021 JMC 10
<math>\textbf{(A) }\pi-e \qquad\textbf{(B) }2\pi-2e\qquad\textbf{(C) }2e\qquad\textbf{(D) }2\pi \qquad\textbf{(E) }2\pi +e</m ==Problem 6==

12 KB (1,784 words) - 16:49, 1 April 2021
2006 AMC 12B Problems
...1.99</math>, and <math>\textdollar0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <ma == Problem 6 ==

13 KB (2,058 words) - 12:36, 4 July 2023
2006 AMC 12A
* [[2006 AMC 12A Problems/Problem 6]]

1 KB (168 words) - 21:51, 6 October 2014
2004 AMC 12A
* [[2004 AMC 12A Problems/Problem 6]]

2 KB (186 words) - 17:35, 16 December 2019
2005 AMC 12A
** [[2005 AMC 12A Problems/Problem 6|Problem 6]]

2 KB (202 words) - 21:30, 6 October 2014
2002 AMC 12A
* [[2002 AMC 12A Problems/Problem 6]]

1 KB (158 words) - 21:33, 6 October 2014
2003 AMC 12A
* [[2003 AMC 12A Problems/Problem 6]]

1 KB (162 words) - 21:52, 6 October 2014
2006 AMC 12A Problems
Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>? Doug and Dave shared a pizza with <math>8</math> equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half t

15 KB (2,223 words) - 13:43, 28 December 2020
2005 AMC 12A Problems
== Problem 6 == ...they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?

13 KB (1,971 words) - 13:03, 19 February 2020
2004 AMC 12A Problems
On the AMC 12, each correct answer is worth <math>6</math> points, each incorrect answer is worth <math>0</math> points, and ea ...ters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?

13 KB (1,953 words) - 00:31, 26 January 2023
2003 AMC 12A Problems
...Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 more than a pair of socks. Each member needs one pair of The sum of the two 5-digit numbers <math>AMC10</math> and <math>AMC12</math> is <math>123422</mat

13 KB (1,955 words) - 21:06, 19 August 2023
2002 AMC 12A Problems
<math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> ..., and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

12 KB (1,792 words) - 13:06, 19 February 2020
2000 AMC 12 Problems
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9</math> ...qquad \textbf{(C)}\ 2-2p \qquad \textbf{(D)}\ 2p-2 \qquad \textbf{(E)}\ |2p-2|</math>

13 KB (1,948 words) - 12:26, 1 April 2022
2001 AMC 12 Problems
...3\qquad \text{(B)}\ 3S + 2\qquad \text{(C)}\ 3S + 6 \qquad\text{(D)} 2S + 6 \qquad \text{(E)}\ 2S + 12</math> of the integer <math>n</math>. For example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a

13 KB (1,957 words) - 12:53, 24 January 2024
2002 AMC 12B Problems
\qquad\mathrm{(D)}\ 6 \qquad\mathrm{(C)}\ 6\ \text{divides\ }n

10 KB (1,547 words) - 04:20, 9 October 2022
2003 AMC 12B Problems
<cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> Moe uses a mower to cut his rectangular 90-foot by 150-foot lawn. The swath he cuts is 28 inches wide, but he overlaps each cut by

13 KB (1,987 words) - 18:53, 10 December 2022
2004 AMC 12B Problems
<math>(\mathrm {A}) 3\qquad (\mathrm {B}) 6 \qquad (\mathrm {C}) 9 \qquad (\mathrm {D}) 12 \qquad (\mathrm {E}) 15</mat In the expression <math>c\cdot a^b-d</math>, the values of <math>a</math>, <math>b</math>, <math>c</math>, and

13 KB (2,049 words) - 13:03, 19 February 2020
2005 AMC 12B Problems
...sing part of the money she earned on her weekend job to buy several equally-priced CDs. She used one ﬁfth of her money to buy one third of the CDs. Wh \mathrm{(A)}\ 80-20\pi \qquad

12 KB (1,781 words) - 12:38, 14 July 2022
2006 AMC 12B Problems/Problem 5
The speed that Bob is catching up to John is <math>5-3=2</math> miles per hour. Since Bob is one mile behind John, it will take < {{AMC12 box|year=2006|ab=B|num-b=4|num-a=6}}

654 bytes (115 words) - 21:47, 1 August 2020
2006 AMC 12B Problems/Problem 7
{{AMC12 box|year=2006|ab=B|num-b=6|num-a=8}}

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2006 AMC 12B Problems/Problem 9
How many even three-digit integers have the property that their digits, all read from left to ri ...and for each choice there is one acceptable order. Similarly, for <math>c=6</math> and <math>c=8</math> there are, respectively, <math>\binom{5}{2}=10<

3 KB (409 words) - 17:10, 30 April 2024
2006 AMC 12B Problems/Problem 11
Joe and JoAnn each bought 12 ounces of coffee in a 16-ounce cup. Joe drank 2 ounces of his coffee and then added 2 ounces of crea ...she drank <math>\frac{1}{7}</math> of her cream, leaving her <math>2*\frac{6}{7}</math>.

927 bytes (137 words) - 10:45, 4 July 2013
2006 AMC 12B Problems/Problem 13
...qrt {3} \qquad \textrm{(C) } 8 \qquad \textrm{(D) } 9 \qquad \textrm{(E) } 6\sqrt {3}</math> ...eas is <math>(\frac{1}{\sqrt{3}})^2</math> or <math>\frac{1}{3}</math>. One-third the area of <math>ABCD</math> is equal to <math>8</math>. So the answe

3 KB (447 words) - 03:49, 16 January 2021
2006 AMC 12B Problems/Problem 15
O[1] = (6,0); A = (32/6,8*sqrt(2)/6);

3 KB (458 words) - 16:40, 6 October 2019
2006 AMC 12B Problems/Problem 16
...n is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>. ...ac{5\sqrt{2}}{2}</math>, yielding an area of <math>\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}</math>.

1 KB (203 words) - 16:36, 18 September 2023
2006 AMC 12B Problems/Problem 17
...h>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</math> on the ...first and <math>6</math> on the second die is <math>\frac 1{21}\cdot\frac 6{21}</math>. Similarly we can express the probabilities for the other five w

1 KB (188 words) - 22:10, 9 June 2016
2006 AMC 12B Problems/Problem 19
...On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times. "Look, daddy!" s \mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 5\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 8

4 KB (696 words) - 09:47, 10 August 2015
2006 AMC 12B Problems/Problem 21
\mathrm{(D)}\ 6\sqrt {2006} ...rrow a=2\sqrt{1003}</math> and because <math>a=\frac{l+w}{2}</math>, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of

2 KB (339 words) - 13:15, 12 July 2015
2006 AMC 12B Problems/Problem 23
...ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have pair P = IP(arc(B,7,180,360),arc(C,6,0,90));

7 KB (1,169 words) - 14:04, 10 June 2022
2006 AMC 12B Problems/Problem 24
\mathrm{(C)}\ \dfrac{\pi^2}{6} MP("\frac{\pi}{6}", (1,0), plain.S);

3 KB (563 words) - 22:45, 24 October 2021
2006 AMC 12A Problems/Problem 10
<math> \textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11</math> {{AMC12 box|year=2006|ab=A|num-b=9|num-a=11}}

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2006 AMC 12A Problems/Problem 13
The [[vertex|vertices]] of a <math>3-4-5</math> [[right triangle]] are the centers of three mutually externally tan <cmath>r_A + r_B + r_C = 6</cmath>

1 KB (184 words) - 13:57, 19 January 2021
2006 AMC 12A Problems/Problem 15
...{2}\qquad \mathrm{(D) \ } \frac{5\pi}{6} \quad \mathrm{(E) \ } \frac{7\pi}{6}</math> ...</math> will be that of <math>\frac{5\pi}{3} - \frac{3\pi}{2} = \frac{\pi}{6} \Rightarrow \mathrm{(A)}</math>.

919 bytes (138 words) - 12:45, 4 August 2017
2006 AMC 12A Problems/Problem 19
...-axis. The equation of <math>L_1</math> is <math>y=\frac{5}{12}x+\frac{19}{6}</math>, so the coordinate of this point is <math>\left(-\frac{38}{5},0\rig {{AMC12 box|year=2006|ab=A|num-b=18|num-a=20}}

2 KB (253 words) - 22:52, 29 December 2021
2006 AMC 12A Problems/Problem 20
Therefore, there is a <math>\frac{1}{243} \cdot \frac{6}{9} = \frac{6}{2187}</math> chance of finding a good path in this case. ...ply the probabilities for this case to get <math>\frac{2}{243} \cdot \frac{6}{9} = \frac{12}{2187}</math>.

5 KB (908 words) - 19:23, 22 September 2022
2006 AMC 12A Problems/Problem 22
...rt{6}+\sqrt{3}\qquad \rm{(D) \ } 3\sqrt{2}+\sqrt{6}\qquad \mathrm{(E) \ } 6\sqrt{2}-\sqrt{3}</math> ...{\circ} = 180^{\circ}</math>. Each arc must equal <math>\frac{180^{\circ}}{6} = 30^\mathrm{\circ}</math>.

2 KB (343 words) - 15:39, 14 June 2023
2005 AMC 12A Problems/Problem 25
=== Solution 1 (non-rigorous) === ...ill actually give an [[octahedron]], not a cube, because it only has <math>6</math> vertices.

4 KB (498 words) - 00:46, 4 August 2023
2005 AMC 12B Problems/Problem 3
...sing part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. Wha {{AMC10 box|year=2005|ab=B|num-b=4|num-a=6}}

978 bytes (156 words) - 14:14, 14 December 2021
2005 AMC 12B Problems/Problem 4
...C 12B Problems|2005 AMC 12B #4]] and [[2005 AMC 10B Problems|2005 AMC 10B #6]]}} ...zzes left, so she can afford to get less than an <math>A</math> on <math>20-18=\boxed{\textbf{(B) }2}</math> of them.

1 KB (197 words) - 14:16, 14 December 2021
2005 AMC 12B Problems/Problem 5
\textbf{(A) }\ 80-20\pi \qquad \textbf{(B) }\ 60-10\pi \qquad

2 KB (223 words) - 14:30, 15 December 2021
2005 AMC 12B Problems/Problem 6
{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #6]] and [[2005 AMC 10B Problems|2005 AMC 10B #10]]}} ...ath>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=\boxed{\textbf{(A) }3}</math>.

2 KB (299 words) - 15:29, 5 July 2022
2005 AMC 12B Problems/Problem 7
\mathrm{(A)}\ 6 \qquad <cmath>\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}</cmath>

2 KB (357 words) - 20:15, 27 December 2020
2005 AMC 12B Problems/Problem 9
...h>85</math>. The mean is <math>\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86</math>. The difference between the mean and media {{AMC10 box|year=2005|ab=B|num-b=18|num-a=20}}

2 KB (280 words) - 15:35, 16 December 2021
2005 AMC 12B Problems/Problem 11
...math>. There are <math>12</math> ways to choose a twenty and some other non-twenty bill. There is <math>1</math> way to choose both twenties, and also < ...nties, so you have <math>6</math> bills left. <math>\dbinom{6}{2} = \dfrac{6\times5}{2\times1} = 15</math> ways. However, you counted the case when you

4 KB (607 words) - 21:01, 20 May 2023
2005 AMC 12B Problems/Problem 13
Suppose that <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math>. What is <math>x_1x_2 ...that we can re-write <math>4^{x_1}=5</math>, <math>5^{x_2}=6</math>, <math>6^{x_3}=7</math>, ... , <math>127^{x_{124}}=128</math> as <math>\left(...\lef

1 KB (203 words) - 19:57, 24 December 2020
2005 AMC 12B Problems/Problem 14
...>, is tangent to the lines <math>y=x</math>, <math>y=-x</math> and <math>y=6</math>. What is the radius of this circle? \mathrm{(A)}\ 6\sqrt{2}-6 \qquad

2 KB (278 words) - 21:12, 24 December 2020
2005 AMC 12B Problems/Problem 15
The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math ...h> can have is <math>(1+2+3+4)=10</math>, and the greatest value is <math>(6+7+8+9)=30</math>. Therefore, <math>(e+f+g+h)</math> must equal <math>11</ma

2 KB (411 words) - 21:02, 21 December 2020
2005 AMC 12B Problems/Problem 18
f.p=fontsize(6); ..., which is <math>\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}</math>, which is approximately <math>51</math>. The answer

2 KB (262 words) - 21:20, 21 December 2020
2005 AMC 12B Problems/Problem 19
Let <math>x</math> and <math>y</math> be two-digit integers such that <math>y</math> is obtained by reversing the digits <cmath>x^2-y^2=(x+y)(x-y)</cmath>

2 KB (283 words) - 20:02, 24 December 2020
2005 AMC 12B Problems/Problem 20
...and <math>h</math> be distinct elements in the set <math>\{-7,-5,-3,-2,2,4,6,13\}.</math> ...5</math>, which can be achieved through <math>13-7-5+2</math> and <math>6-3-2+4</math>. So the minimum possible value is <math>3^2+5^2=34\Rightarrow\box

3 KB (463 words) - 19:28, 6 November 2022
2005 AMC 12B Problems/Problem 24
Our sum is simply <math>2 - 2\cdot\frac{8}{11} = \frac{6}{11}</math>, and thus we can divide by <math>2</math> to obtain <math>\frac ...ive x- axis, the answer is <math>\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}</math>.

4 KB (761 words) - 09:10, 1 August 2023
2005 AMC 12B Problems/Problem 25
...re are six ants, the probability of each case occuring is <math>\frac{1}{4^6} = \frac{1}{4096}</math>. Thus, the desired answer is <math>\frac{80}{4096} defaultpen(0.6);

10 KB (1,840 words) - 21:35, 7 September 2023
Power of a Point Theorem/Introductory Problem 2
Applying the Power of a Point Theorem gives <math> 6\cdot x = 4\cdot 1 </math>, so <math> x = \frac 23 </math>

289 bytes (45 words) - 13:14, 16 July 2017
2006 AMC 10A Problems
Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>? Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half

13 KB (2,028 words) - 16:32, 22 March 2022
2006 AMC 10A Problems/Problem 8
<math>b=-6</math> Then using <math>b=-6</math> in the first equation:

2 KB (348 words) - 23:10, 16 December 2021
2006 AMC 10A Problems/Problem 12
 Rolly wishes to secure his dog with an 8-foot rope to a [[square (geometry) | square]] shed that is 16 feet on each s

3 KB (424 words) - 10:14, 17 December 2021
2006 AMC 10A Problems/Problem 13
...}{6}</math>. The probability of winning is <math>\frac{1}{2}\cdot \frac{1}{6} =\frac{1}{12}</math>. If the game is to be fair, the amount paid, <math>5< {{AMC10 box|year=2006|ab=A|num-b=12|num-a=14}}

1 KB (207 words) - 09:39, 25 July 2023
2006 AMC 10A Problems/Problem 15
draw((6,0){up}..{left}(0,6),blue); draw((-6,0){up}..{right}(0,6),blue);

3 KB (532 words) - 17:49, 13 August 2023
2006 AMC 10A Problems/Problem 16
 D((2*t,2)--(2*t,4)); D((2*t,5)--(2*t,6));

5 KB (732 words) - 23:19, 19 September 2023
2006 AMC 10A Problems/Problem 18
So, there are <math>6 - 1 = 5</math> choices for the position of the letters. {{AMC10 box|year=2006|ab=A|num-b=17|num-a=19}}

2 KB (254 words) - 14:39, 5 April 2024
2006 AMC 10A Problems/Problem 20
...5</math> possible values in <math>\bmod{5}</math> and we are picking <math>6</math> numbers, by the [[Pigeonhole Principle]], two of the numbers must be Therefore the probability that some pair of the <math>6</math> integers has a difference that is a multiple of <math>5</math> is <m

1 KB (187 words) - 08:21, 17 March 2023
2006 AMC 10A Problems/Problem 21
How many four-digit positive integers have at least one digit that is a <math>2</math> or ...egers that have at least one <math>2</math> or <math>3</math> is <math>9000-3584=\boxed{\textbf{(E) }5416}.</math>

3 KB (525 words) - 20:25, 30 April 2024
2006 AMC 10A Problems/Problem 24
<math>\textbf{(A) } \frac{1}{8}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{1}{3}\qquad\textbf ...this volume, so <math>\frac{1}{12} \cdot 2 = \boxed{\textbf{(B) }\frac{1}{6}}</math>.

2 KB (292 words) - 10:19, 19 December 2021
Set
...s on the other hand. For the full collection of these axioms, see [[Zermelo-Fraenkel Axioms]]. In this article we shall present just a brief discussion ...tiple indistinct elements, such as the following: <math>\{1,4,5,3,24,4,4,5,6,2\}</math> Such an entity is actually called a multiset.

11 KB (2,021 words) - 00:00, 17 July 2011
Newman's Tauberian Theorem
(which is well-defined by this formula for <math>\Re s>0</math>) admits an to some [[open set | open]] domain <math>E</math> containing the closed half-plane

6 KB (1,034 words) - 07:55, 12 August 2019
2006 USAMO Problems
if and only if <math>s</math> is not a divisor of <math>p-1</math>. === Problem 6 ===

3 KB (520 words) - 09:24, 14 May 2021
1991 AJHSME Problems
<math>1,000,000,000,000-777,777,777,777=</math> <math>\frac{16+8}{4-2}=</math>

17 KB (2,246 words) - 13:37, 19 February 2020
1990 AJHSME Problems/Problem 1
...t numbers that can be obtained by placing each of the six digits <math>4,5,6,7,8,9</math> in one of the six boxes in this addition problem? draw((1,1)--(3,1)--(3,3)--(1,3)--cycle); draw((1,4)--(3,4)--(3,6)--(1,6)--cycle);

1 KB (191 words) - 17:12, 29 October 2016
2022 USAJMO Problems/Problem 1
<math>\bullet</math> <math>a_n-g_n</math> is divisible by <math>m</math> for all integers <math>n>1</math>; <math>\bullet</math> <math>a_2-a_1</math> is not divisible by <math>m</math>.

4 KB (792 words) - 00:29, 13 April 2024
2005 AIME II Problems/Problem 1
...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math> ...{n \choose 6} = \frac{n\cdot(n-1)\cdot(n-2)\cdot(n-3)\cdot(n-4)\cdot(n-5)}{6\cdot5\cdot4\cdot3\cdot2\cdot1}</math>.

1 KB (239 words) - 11:54, 31 July 2023
2005 AIME II Problems/Problem 2
*Person 1: <math>\frac{9 \cdot 6 \cdot 3}{9 \cdot 8 \cdot 7} = \frac{9}{28}</math> *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math>

4 KB (628 words) - 11:28, 14 April 2024
2005 AIME II Problems/Problem 5
...d pair]]s <math> (a,b) </math> of [[integer]]s such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </mat ...<math> \frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a \log b}=5 </math> Multiplying through by <math>\log a \l

3 KB (547 words) - 19:15, 4 April 2024
2005 AIME II Problems/Problem 6
...xample, eight cards form a magical stack because cards number 3 and number 6 retain their original positions. Find the number of cards in the magical st 131=2n-2(131)+1\implies2n=\boxed{392}

2 KB (384 words) - 00:31, 26 July 2018
2005 AIME II Problems
...The number of possible sets of 6 cards that can be drawn from the deck is 6 times the number of possible sets of 3 cards that can be drawn. Find <math> ...ordered pairs <math> (a,b) </math> of integers such that <math> \log_a b + 6\log_b a=5, 2 \leq a \leq 2005, </math> and <math> 2 \leq b \leq 2005. </mat

7 KB (1,119 words) - 21:12, 28 February 2020
2005 AIME II Problems/Problem 7
...sqrt[2^n]{5} + 1)(\sqrt[2^n]{5} - 1) = (\sqrt[2^n]{5})^2 - 1^2 = \sqrt[2^{n-1}]{5} - 1 </math>. <cmath>(y^{12}+y^8+y^4+1)(y^2+1)=(y^{14}+y^{12}+y^{10}+y^8+y^6+y^4+y^2+1)</cmath>

2 KB (279 words) - 12:33, 27 October 2019
2005 AIME II Problems/Problem 8
pair C1 = (-10,0), C2 = (4,0), C3 = (0,0), H = (-10-28/3,0), T = 58/7*expi(pi-acos(3/7)); path cir1 = CR(C1,4.01), cir2 = CR(C2,10), cir3 = CR(C3,14), t = H--T+2*(T-H);

4 KB (693 words) - 13:03, 28 December 2021
2005 AIME II Problems/Problem 9
...})^n = i e^{-itn}</math>. Hence we must have <math>i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}</math>. Thus since <math>1000</math> ...and the RHS is <math>cis(90^{\circ}-nt)</math> So, <math>n(90-t) \equiv 90-nt \mod 360</math> Expanding and canceling the nt terms, we will get <math>9

6 KB (1,154 words) - 03:30, 11 January 2024
2005 AIME II Problems/Problem 12
...0)); pair A=(0,9), B=(9,9), C=(9,0), D=(0,0), E=(2.5-0.5*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), G=(4.5,9), O=(4.5,4.5); draw(A--B--C--D--A);draw(E--O--F);dr ...*sqrt(7),9), F=(6.5-0.5*sqrt(7),9), O=(4.5,4.5), G=O+(E-O)*dir(-90), J=O+(F-O)*dir(-90); draw(A--B--C--D--A);draw(E--O--F);draw(G--O--J);draw(F--G,linet

13 KB (2,080 words) - 21:20, 11 December 2022
Research Science Institute
The '''Research Science Institute''' (RSI) is a 6-week residential program held each summer at the [[Massachusetts Institute o ...Admission to RSI is extremely competitive. There is no tuition or room-and-board charge to attend RSI. Students in all branches of [[science]] and [[m

2 KB (284 words) - 16:37, 6 July 2023
2006 AMC 12A Problems/Problem 4
1+2&9&6&3\\ ...ays time in AM and PM, the value for the hours section varies from <math>00-12</math>. Therefore, the maximum value of the digits for the hours is when

2 KB (257 words) - 11:20, 2 January 2022
2005 AIME I Problems
== Problem 6 == ...math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math>

6 KB (983 words) - 05:06, 20 February 2019
2005 AIME I Problems/Problem 1
...h of <math>3r = 30</math>, and so <math>r = 10</math>. <math>K = 30^2\pi - 6(10^2\pi) = 300\pi</math>, so <math>\lfloor 300\pi \rfloor = \boxed{942}</ma {{AIME box|year=2005|n=I|before=First Question|num-a=2}}

1 KB (213 words) - 13:17, 22 July 2017
2005 AIME I Problems/Problem 2
...th>(6,334)</math>, <math>(12,167)</math>, <math>(167,12)</math>,<math>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> {{AIME box|year=2005|n=I|num-b=1|num-a=3}}

2 KB (303 words) - 01:31, 5 December 2022
2005 AIME I Problems/Problem 4
.../math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] <cmath>2n-2m+7 = 1</cmath>

8 KB (1,248 words) - 11:43, 16 August 2022
2005 AIME I Problems/Problem 5
...nly be one way to arrange the coins. We can the recurrence: <cmath>a_n=a_{n-1}+1.</cmath> Note <math>a_1=2</math>, so <math>a_8 = 9</math>. ...hoose 4</math> ways, and the rest of the silver coins can be put in <math>8-4</math>choose <math>4</math> ways. (We subtract the <math>4</math> from the

5 KB (830 words) - 01:51, 1 March 2023
2005 AIME I Problems/Problem 6
...math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math> The left-hand side of that [[equation]] is nearly equal to <math>(x - 1)^4</math>. T

4 KB (686 words) - 01:55, 5 December 2022
2005 AIME I Problems/Problem 7
draw((5,8.66)--(16.87,6.928)); ...angleup DAP</math> and <math>\bigtriangleup EBQ</math> are both <math>30-60-90</math>, so <math>AP=5</math> and <math>BQ=5</math>. <math>PQ=DE</math>, a

4 KB (567 words) - 20:20, 3 March 2020
2005 AIME I Problems/Problem 9
...ry) | cube]]s are painted orange on a set of four [[face]]s so that two non-painted faces share an [[edge]]. The 27 cubes are randomly arranged to form [[Image:2005_I_AIME-9.png]]

4 KB (600 words) - 21:44, 20 November 2023
2005 AIME I Problems/Problem 12
...05 </math> with <math> S(n) </math> [[even integer | even]]. Find <math> |a-b|. </math> It is well-known that <math>\tau(n)</math> is odd if and only if <math>n</math> is a [[

4 KB (647 words) - 02:29, 4 May 2021
2005 AIME I Problems/Problem 13
...n range from <math>l = 5</math> to <math>9</math>; notice that <math>d = 10-l</math> gives the number of diagonals. Let <math>R</math> represent a move ...jacent. The <math>D</math>'s split the string into three sections (<math>-D-D-</math>): by the [[Pigeonhole principle]] all of at least one of the two l

5 KB (897 words) - 00:21, 29 July 2022
2005 AIME II Problems/Problem 15
...the [[circle]]s <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positi ...n equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>.

12 KB (2,000 words) - 13:17, 28 December 2020
2005 AIME II Problems/Problem 14
...> Point <math> D </math> is on <math> \overline{BC} </math> with <math> CD=6. </math> Point <math> E </math> is on <math> \overline{BC} </math> such tha ...from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangl

13 KB (2,129 words) - 18:56, 1 January 2024
2004 AIME I Problems/Problem 15
...o place our <math>BA^{[10]}</math> block. Thus, our answer is <math>2^9(2^9-1)-8\cdot 2^7 = 2^{18} - 2^9 - 8 \cdot 2^7 = 2^9 \times 509</math>, and the ...e note that <math>\mathcal{A}_{1,k}=\emptyset</math> for <math>k=0,2,3,4,5,6,7,8,9</math>.

9 KB (1,491 words) - 01:23, 26 December 2022
2004 AIME I Problems/Problem 14
real x = 20 - ((750)^.5)/3, CE = 8*(6^.5) - 4*(5^.5), CD = 8*(6^.5), h = 4*CE/CD; pair Cxy = 8*expi((3*pi)/2-CE/8);

4 KB (729 words) - 01:00, 27 November 2022
2004 AIME I Problems/Problem 13
The polynomial <math> P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17} </math> has <math>34</math> complex roots of the form <math> z_k = r_ ...19}+ \frac3{19}+ \frac 1{17}+ \frac2{17}= \frac6{19} + \frac 3{17} = \frac{6\cdot17 + 3\cdot19}{17\cdot19} = \frac{159}{323}</math> and so the answer is

2 KB (298 words) - 20:02, 4 July 2013
2004 AIME I Problems/Problem 12
...left(\frac{\frac{4}{5}}{1-\frac{1}{25}}\right)= \frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},</cmath> and the answer is <math>m+n = 5 + 9 = \boxed{014}</ {{AIME box|year=2004|n=I|num-b=11|num-a=13}}

2 KB (303 words) - 22:28, 11 September 2020
2004 AIME I Problems/Problem 10
A [[circle]] of [[radius]] 1 is randomly placed in a 15-by-36 [[rectangle]] <math> ABCD </math> so that the circle lies completely with [[Image:2004_I_AIME-10.png]]

5 KB (836 words) - 07:53, 15 October 2023
2004 AIME I Problems/Problem 9
...be a [[triangle]] with sides 3, 4, and 5, and <math> DEFG </math> be a 6-by-7 [[rectangle]]. A segment is drawn to divide triangle <math> ABC </math> in ...BC</math>. Thus <math>U_1</math>, and hence <math>U_2</math>, are <math>3-4-5\,\triangle</math>s.

4 KB (618 words) - 20:01, 4 July 2013
2004 AIME I Problems/Problem 1
...n + 1) + n = 3210 + 1111n</math>, for <math>n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace</math>. Adding these numbers up, we get <math>(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}</math>

2 KB (374 words) - 14:53, 27 December 2019
2004 AIME I Problems/Problem 2
...th>. Therefore, the largest element in <math>A</math> is <math>2 + \frac{m-1}{2}</math>. ...f <math>m < 3</math>, which is clearly not possible, thus <math>2 + \frac{m-1}{2} < m</math>.

8 KB (1,437 words) - 21:53, 19 May 2023
2004 AIME I Problems/Problem 5
.... Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is <math> m/n, </math> where <math \frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}

3 KB (436 words) - 18:31, 9 January 2024
2004 AIME I Problems/Problem 7
...\le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}</cmath> where the left-hand sum can be computed from:

5 KB (833 words) - 19:43, 1 October 2023
2004 AIME I Problems/Problem 8
...two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there? We use the [[Principle of Inclusion-Exclusion]] (PIE).

4 KB (620 words) - 21:26, 5 June 2021
2004 AIME I Problems
...h are quadrilaterals. A space diagonal is a line segment connecting two non-adjacent vertices that do not belong to the same face. How many space diagon ...io was <math>\frac{300}{500} = \frac{3}{5}</math>. The largest possible two-day success ratio that Beta could achieve is <math> m/n, </math> where <math

9 KB (1,434 words) - 13:34, 29 December 2021
2004 AIME II Problems/Problem 15
<cmath>s_{82, 9} = 2s_{82, 8} = 4s_{82, 7} = 8s_{127 - 82, 6} = 8s_{45, 6}</cmath> <cmath>s_{45, 6} = 2s_{63 - 45, 5} + 1 = 2s_{18, 5} + 1 = 4s_{31 - 18, 4} + 1 = 4s_{13, 4}

6 KB (899 words) - 20:58, 12 May 2022
2004 AIME II Problems/Problem 14
...set <math>(b,c): n=1000-9(b+12c)</math>, it is possible that <math>a = n-2b-3c < 0</math> (for example, suppose we counted the solution set <math>(b,c) ...th>(b,c): n \equiv b \pmod{12}, 0 \le b \le 11</math> and <math>c = \frac{n-b}{12} \le 7</math> (in other words, we take the greatest possible value of

11 KB (1,857 words) - 21:55, 19 June 2023
2004 AIME II Problems/Problem 12
...CD </math> be an [[isosceles trapezoid]], whose dimensions are <math> AB = 6, BC=5=DA, </math>and <math> CD=4. </math> Draw [[circle]]s of [[radius]] 3 pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5);

3 KB (431 words) - 23:21, 4 July 2013
2004 AIME II Problems/Problem 10
...}</math>. For example, with the binary string 0001001000 <math>y</math> is 6 and <math>x</math> is 3 (note that it is zero indexed). ...math> is clearly never divisible by 9, so we can just focus on <math>1+2^{y-x}</math>.

8 KB (1,283 words) - 19:19, 8 May 2024
2004 AIME II Problems/Problem 8
...ays and likewise we can partition the 167 in three ways. So we have <math>6\cdot 3\cdot 3 = \boxed{54}</math> as our answer. 6, 2, 167

2 KB (353 words) - 18:08, 25 November 2023
2004 AIME II Problems/Problem 7
...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 \frac{14}{l}x &= y-8 = \frac{9}{2}\\

9 KB (1,501 words) - 05:34, 30 October 2023
2004 AIME II Problems/Problem 6
...two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given t ...amount of bananas each monkey has by 2, 3, 6. Now, the ratio is <math>6:6:6</math>, so, <math>12x+6y+22z = 3x+6y+33z = 6x+18y+12z</math>. Subtracting <

6 KB (950 words) - 14:18, 15 January 2024
2004 AIME II Problems/Problem 5
...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 co ...math> mph to make it on time. Therefore the company needs to add <math>1566-800 = \boxed{766}</math> more workers.

4 KB (592 words) - 19:02, 26 September 2020
2004 AIME II Problems
...the smaller can be expressed in the form <math> \frac{a\pi+b\sqrt{c}}{d\pi-e\sqrt{f}}, </math> where <math> a, b, c, d, e, </math> and <math> f </math> ...lock is viewed so that three of its faces are visible, exactly 231 of the 1-cm cubes cannot be seen. Find the smallest possible value of <math> N. </mat

9 KB (1,410 words) - 05:05, 20 February 2019
1983 AIME Problems
Let <math>f(x)=|x-p|+|x-15|+|x-p-15|</math>, where <math>0 . Determine the [[minimum]] value t ...ircle is <math>\sqrt{50}</math> cm, the length of <math>AB</math> is <math>6</math> cm and that of <math>BC</math> is <math>2</math> cm. The angle <math

7 KB (1,104 words) - 12:53, 6 July 2022
2019 AIME I Problems/Problem 2
...\ne 0</math> because <math>B \ne J</math>, so the probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry. The probability that <math>B-J = 1</math> is <math>\frac{19}{20 \times 19} = \frac{1}{20}</math> because

5 KB (830 words) - 22:15, 28 December 2023
1984 AIME Problems
...A+(B-A)*3/4--A+(C-A)*3/4); D(B+(C-B)*5/6--B+(A-B)*5/6);D(C+(B-C)*5/12--C+(A-C)*5/12); MP("t_3",(A+B+(B-A)*3/4+(A-B)*5/6)/2+(-1,0.8),N);

6 KB (933 words) - 01:15, 19 June 2022
1986 AIME Problems
...sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).</cmath> <center><math>2x_1+x_2+x_3+x_4+x_5=6</math></center>

5 KB (847 words) - 15:48, 21 August 2023
1987 AIME Problems
.../math> requires no carrying. Find the number of simple ordered pairs of non-negative integers that sum to <math>1492</math>. Find the area of the region enclosed by the graph of <math>|x-60|+|y|=\left|\frac{x}{4}\right|.</math>

6 KB (869 words) - 15:34, 22 August 2023
1988 AIME Problems
...r -- the correct five buttons. The sample shown below has <math>\{1, 2, 3, 6, 9\}</math> as its combination. Suppose that these locks are redesigned so <center>[[Image:1988-1.png]]</center>

6 KB (902 words) - 08:57, 19 June 2021
1989 AIME Problems
== Problem 6 == pair A=(0,0),B=(10,0),C=6*expi(pi/3);

7 KB (1,045 words) - 20:47, 14 December 2023
1990 AIME Problems
The [[increasing sequence]] <math>2,3,5,6,7,10,11,\ldots</math> consists of all [[positive integer]]s that are neithe Find the value of <math>(52+6\sqrt{43})^{3/2}-(52-6\sqrt{43})^{3/2}</math>.

6 KB (870 words) - 10:14, 19 June 2021
1991 AIME Problems
== Problem 6 == [[1991 AIME Problems/Problem 6|Solution]]

7 KB (1,106 words) - 22:05, 7 June 2021
1992 AIME Problems
\text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1

8 KB (1,117 words) - 05:32, 11 November 2023
1993 AIME Problems
:(b) those who caught <math>3</math> or more fish averaged <math>6</math> fish each; How many ordered four-tuples of integers <math>(a,b,c,d)\,</math> with <math>0 < a < b < c < d < 5

8 KB (1,275 words) - 06:55, 2 September 2021
1994 AIME Problems
<center><math>f(x)+f(x-1) = x^2\,</math></center>. ...integer <math>n\,</math>, let <math>p(n)\,</math> be the product of the non-zero digits of <math>n\,</math>. (If <math>n\,</math> has only one digit, t

7 KB (1,141 words) - 07:37, 7 September 2018
1995 AIME Problems
...h> and <math>n</math> are relatively prime positive integers. Find <math>m-n.</math> Circles of radius <math>3</math> and <math>6</math> are externally tangent to each other and are internally tangent to a

6 KB (1,000 words) - 00:25, 27 March 2024
1996 AIME Problems
...positive integer <math>n</math> for which the expansion of <math>(xy-3x+7y-21)^n</math>, after like terms have been collected, has at least 1996 terms. Suppose that the roots of <math>x^3+3x^2+4x-11=0</math> are <math>a</math>, <math>b</math>, and <math>c</math>, and that

6 KB (931 words) - 17:49, 21 December 2018
1997 AIME Problems
...hould have obtained. What is the sum of the two-digit number and the three-digit number? The number <math>r</math> can be expressed as a four-place decimal <math>0.abcd,</math> where <math>a, b, c,</math> and <math>d</

7 KB (1,098 words) - 17:08, 25 June 2020
1998 AIME Problems
...{12}</math> the [[least common multiple]] of the positive integers <math>6^6</math> and <math>8^8</math>, and <math>k</math>? == Problem 6 ==

7 KB (1,084 words) - 02:01, 28 November 2023
1999 AIME Problems
Find the sum of all positive integers <math>n</math> for which <math>n^2-19n+99</math> is a perfect square. ...e <math>|S(x+2)-S(x)|.</math> For example, <math>T(199)=|S(201)-S(199)|=|3-19|=16.</math> How many values of <math>T(x)</math> do not exceed 1999?

7 KB (1,094 words) - 13:39, 16 August 2020
2000 AIME I Problems
...is, and let <math>E</math> be the reflection of <math>D</math> across the y-axis. The area of pentagon <math>ABCDE</math> is <math>451</math>. Find <mat The diagram shows a rectangle that has been dissected into nine non-overlapping squares. Given that the width and the height of the rectangle ar

7 KB (1,204 words) - 03:40, 4 January 2023
2001 AIME I Problems
Find the sum of all positive two-digit integers that are divisible by each of their digits. ...the roots, real and non-real, of the equation <math>x^{2001}+\left(\frac 12-x\right)^{2001}=0</math>, given that there are no multiple roots.

7 KB (1,212 words) - 22:16, 17 December 2023
2002 AIME I Problems
...ree-digit arrangement that reads the same left-to-right as it does right-to-left) is <math>m/n</math>, where <math>m</math> and <math>n</math> are relat for(i=0; i<6; i=i+1) {

8 KB (1,374 words) - 21:09, 27 July 2023
2003 AIME I Problems
...1, 13, 34, 3, 21, 2\}. </math> Susan makes a list as follows: for each two-element subset of <math> \mathcal{S}, </math> she writes on her list the gre == Problem 6 ==

6 KB (965 words) - 16:36, 8 September 2019
2000 AIME II Problems
<center><math>\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}</math></center> ...ht distinguishable rings, let <math>n</math> be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order

6 KB (947 words) - 21:11, 19 February 2019

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