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1977 Canadian MO Problems
....</math> Find the smallest positive integer <math>b</math> for which <math>N</math> is the fourth power of an integer. == Problem 4 ==

3 KB (560 words) - 19:23, 10 March 2015
1959 IMO Problems/Problem 4
<math> ab = \frac{c^2}{4}. </math> ...math> and a line parallel to <math>c</math> and of distance <math>\frac{c}{4}</math> from <math>c</math>, either point of intersection between the line

6 KB (939 words) - 17:31, 15 July 2023
1959 IMO Problems/Problem 5
...t at <math>M </math> and also at another point <math>N </math>. Let <math>N' </math> denote the point of intersection of the straight lines <math>AF </ (a) Prove that the points <math>N </math> and <math>N' </math> coincide.

2 KB (408 words) - 01:40, 2 January 2023
1970 IMO Problems/Problem 4
...th>n</math> with the property that the set <math>\{ n, n+1, n+2, n+3, n+4, n+5 \} </math> can be partitioned into two sets such that the product of the ...e factors. The only such number is 1, so the set must be <math>\{ 1, 2, 3, 4, 5, 6 \}</math>, but that does not work because only one of the numbers is

1 KB (250 words) - 03:31, 2 January 2023
1970 IMO Problems/Problem 3
<math>1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots</math>. <math>b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}</math>

3 KB (577 words) - 16:10, 26 July 2009
2006 Romanian NMO Problems
Let <math>ABC</math> be a triangle and the points <math>M</math> and <math>N</math> on the sides <math>AB</math> respectively <math>BC</math>, such that ...th> unit squares, each colored in red, yellow or green. Find minimal <math>n</math>, such that for each coloring, there exists a line and a column with

11 KB (1,779 words) - 14:57, 7 May 2012
2006 Romanian NMO Problems/Grade 9/Problem 4
...h>(n \geq 5)</math> participated at table tennis contest, which took <math>4</math> days. Every day, every student played a match. (It is possible that * no student lost all <math>4</math> matches.

2 KB (325 words) - 15:53, 11 December 2011
2005 Canadian MO Problems
...its a triangle. An example of one such path is illustrated below for <math>n=5</math>. Determine the value of <math>f(2005)</math>. ...ath>(a,b,c)</math> satisfying <math>\left(\frac{c}{a}+\frac{c}{b}\right)^2=n</math>.

2 KB (436 words) - 11:45, 26 December 2018
2005 Canadian MO Problems/Problem 3
Let <math>S</math> be a set of <math>n\ge 3</math> points in the interior of a circle. * Show that for no value of <math>n</math> can four such points in <math>S</math> (and corresponding points on

2 KB (460 words) - 13:35, 9 June 2011
2005 Canadian MO Problems/Problem 5
...\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 ...l ordered triples (if any) which are <math>n</math>-powerful for all <math>n \ge 1</math>.

984 bytes (177 words) - 18:21, 28 November 2023
Midpoint
draw((0,0)--(4,0)); label("A",(0,0),N);

4 KB (596 words) - 17:09, 9 May 2024
P versus NP
...\text{TIME}(f(n))</math> is the set of languages decidable by an <math>O(f(n))</math>-time deterministic Turing machine. *Given a list of <math>n</math> integers, is it sorted in non-decreasing order?

6 KB (1,104 words) - 15:11, 25 October 2017
2005 AMC 10A Problems/Problem 10
<math>4x^2 + ax + 8x + 9 = (mx + n)^2</math> <math>4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2</math>

3 KB (443 words) - 18:05, 29 March 2023
2005 AMC 10A Problems/Problem 11
...urth of the total number of faces of the unit cubes are red. What is <math>n</math>? <math> \textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7 </math>

883 bytes (144 words) - 12:40, 13 December 2021
2005 AMC 10A Problems/Problem 13
How many positive integers <math>n</math> satisfy the following condition: <math> (130n)^{50} > n^{100} > 2^{200}\ ?</math>

2 KB (223 words) - 07:18, 16 July 2022
2005 AMC 10A Problems/Problem 21
For how many positive integers <math>n</math> does <math> 1+2+...+n </math> evenly divide from <math>6n</math>? ...n </math> evenly [[divide]]s <math>6n</math>, then <math>\frac{6n}{1+2+...+n}</math> is an [[integer]].

1 KB (220 words) - 12:54, 14 December 2021
2005 AMC 10A Problems/Problem 25
label("$A$", A, N); label("$A$", A, N);

6 KB (854 words) - 20:25, 24 July 2022
2005 AMC 10A Problems/Problem 24
...t true that both <math> P(n) = \sqrt{n} </math> and <math> P(n+48) = \sqrt{n+48} </math>? ...bf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5 </math>

6 KB (895 words) - 13:52, 4 April 2024
Volume
...> is not divisible by the square of any prime. Find <math> \lfloor m+\sqrt{n}\rfloor. </math> (The notation <math> \lfloor x\rfloor </math> denotes the

3 KB (523 words) - 20:24, 17 August 2023
Committee forming
...ee of size <math>k</math> from a set of size <math>n</math>. We use <math>{n\choose k}</math> which is called a [[binomial coefficient]]. From here on ...ath>{n\choose 0} + {n\choose 1} + {n\choose 2} + \cdots + {n\choose n} = 2^n.</math></center>

3 KB (485 words) - 19:49, 16 July 2018
Orthocenter
dot((4,0)); draw((0,0)--(4,0)--(3,3)--cycle);

5 KB (829 words) - 13:11, 20 February 2024
1970 Canadian MO Problems
draw((2.5,0)--(2.5,7.5/4)--(5,0)--cycle,black); MP("C",(0,0),SW);MP("D",(16/5,12/5),N);MP("B",(5,0),SE);

4 KB (604 words) - 04:32, 8 October 2014
Perfect number
A [[positive integer]] <math>n</math> is called a '''perfect number''' if it is the sum of its [[proper di * <math>28 = 14 + 7 + 4 + 2 + 1</math>

1 KB (193 words) - 19:09, 3 March 2022
Stirling number
...kind''' <math>c(n, k)</math> is the number of [[permutation]]s of an <math>n</math>-[[element]] [[set]] with exactly <math>k</math> [[cycle of a permuta ...h>\{(1)(234), (1)(243), (134)(2),(143)(2),(124)(3),(142)(3),(123)(4),(132)(4), (12)(34), (13)(24), (14)(23)\}</math>.

2 KB (253 words) - 00:50, 15 May 2024
1970 IMO Problems
...the system with base <math>a</math> and <math>B_{n-1}</math> and <math>B_{n}</math> are numbers in the system with base <math>b</math>; these are relat <math>A_{n} = x_{n}x_{n-1}\cdots x_{0}, A_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}</math>,

3 KB (558 words) - 00:17, 10 December 2022
Argument
...est considered as an [[equivalence class]] <math>\mathbf r = \{r + 2\pi n, n \in \mathbb{Z}\}</math>. The advantages of this are several: most importan

1 KB (188 words) - 20:59, 31 July 2020
Proper divisor
...ger]] <math>n</math> is any [[divisor]] of <math>n</math> other than <math>n</math> itself. Thus <math>1</math> has no proper divisors, [[prime number] *[[2017 USAJMO Problems/Problem 4]]

436 bytes (63 words) - 16:27, 10 May 2021
2007 iTest Problems
<math>\mathrm{(A)}\, 4</math> ===Problem 4===

30 KB (4,794 words) - 23:00, 8 May 2024
1981 IMO Problems/Problem 2
..., n \} </math>. Each of these subsets has a smallest member. Let <math>F(n,r) </math> denote the arithmetic mean of these smallest numbers; prove that F(n,r) = \frac{n+1}{r+1}.

5 KB (879 words) - 11:18, 27 June 2020
1981 IMO Problems/Problem 3
...rs satisfying <math> m, n \in \{ 1,2, \ldots , 1981 \} </math> and <math>( n^2 - mn - m^2 )^2 = 1 </math>. ...m</math>, since if we had <math>n < m</math>, then <math>n^2 -nm -m^2 = n(n-m) - m^2 </math> would be the sum of two negative integers and therefore le

1 KB (248 words) - 10:23, 13 May 2019
1981 IMO Problems/Problem 4
...t is a [[divisor]] of the [[least common multiple]] of the remaining <math>n-1</math> numbers? (b) For which values of <math>n>2</math> is there exactly one set having the stated property?

3 KB (516 words) - 09:43, 28 March 2012
1987 IMO Problems/Problem 5
...an integer greater than or equal to 3. Prove that there is a set of <math>n </math> points in the plane such that the distance between any two points i ...the set of points <math>S = \{ (x,x^2) \mid 1 \le x \le n , x \in \mathbb{N} \}</math> in the <math>xy</math>-plane.

2 KB (290 words) - 13:37, 26 July 2009
Apothem
...</math>, the length of the apothem is <math>\frac{s}{2\tan\left(\frac{\pi}{n}\right)}</math>. ..., <math>R</math>, the length of the apothem is <math>R\cos\left(\frac{\pi}{n}\right)</math>.

1 KB (169 words) - 18:22, 9 March 2014
2005 AMC 10A Problems
<math> \mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 20 < ...-8\qquad \mathrm{(B) \ } -4\qquad \mathrm{(C) \ } -2\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 8 </math>

14 KB (2,026 words) - 11:45, 12 July 2021
Euler line
...ath>G</math>, [[circumcenter]] <math>O</math>, [[nine-point center]] <math>N</math> and [[De Longchamps point | de Longchamps point]] <math>L</math>. I ...ath>, and <math>\triangle CH_AH_B</math> [[concurrence | concur]] at <math>N</math>, the nine-point circle of <math>\triangle ABC</math>.

59 KB (10,203 words) - 04:47, 30 August 2023
2003 AMC 10A Problems
Members of the Rockham 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 memb <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18

13 KB (1,900 words) - 22:27, 6 January 2021
1997 AJHSME Problems/Problem 24
A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); label("$B$",B,N);

2 KB (394 words) - 17:05, 20 October 2023
Fermat point
pair A=(2,4), B=(1,1), C=(6,1); .../math> minimizes <math>m\cdot AP+n\cdot BP+p\cdot CP</math>, where <math>m,n,p</math> are positive reals?

4 KB (769 words) - 16:07, 29 December 2019
Balkan MO Problems and Solutions, with authors
* [[1984 BMO Problems/Problem 4 | Problem 4]] * [[1985 BMO Problems/Problem 4 | Problem 4]]

4 KB (371 words) - 16:41, 1 January 2024
Chinese Remainder Theorem
...on of each residue class mod <math>m</math> with a residue class mod <math>n</math> is a residue class mod <math>mn</math>. ...an deduce that <math>b \equiv c \pmod{m}</math> and <math>b \equiv c \pmod{n}.</math>

6 KB (1,022 words) - 14:57, 6 May 2023
1985 IMO Problems
...rs, <math>k < n</math>. Each number in the set <math>M = \{ 1,2, \ldots , n-1 \} </math> is colored either blue or white. It is given that (i) for each <math> i \in M </math>, both <math>i </math> and <math>n-i </math> have the same color;

3 KB (465 words) - 03:00, 29 March 2021
2003 AMC 10A Problems/Problem 21
The assortments are: <math>\{(0,6,0); (0,5,1); (0,4,2); (0,3,3); (0,2,4); (0,1,5); (0,0,6)\} \rightarrow 7</math> assortments. The assortments are: <math>\{(1,5,0); (1,4,1); (1,3,2); (1,2,3); (1,1,4); (1,0,5) \} \rightarrow 6</math> assortments.

7 KB (994 words) - 17:51, 11 April 2024
2003 AMC 10A Problems/Problem 22
...with <math>BH=6</math>, <math>E</math> is on <math>AD</math> with <math>DE=4</math>, line <math>EC</math> intersects line <math>AH</math> at <math>G</ma pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);

9 KB (1,446 words) - 22:48, 8 May 2024
2003 AMC 10A Problems/Problem 23
Cn[4]=centroid(Cp,Ep,Dp); label("$4$",Cn[3]);

5 KB (686 words) - 18:01, 28 January 2021
1985 IMO Problems/Problem 3
...\left( (1+x)^k \sum_{j=1}^{n}Q_{i_j - k} \right) = 2 w \left( \sum_{j=1}^{n}Q_{i_j - k} \right)</math>, which is greater than or equal to <math> w( Q_{ \sum_{j=1}^{n}Q_{i_j} = \sum_{j=0}^{k-1}a_j x^j + (1+x)^k \sum_{j=0}^{k-1} b_j x^j \equiv

2 KB (354 words) - 04:56, 11 March 2023
2003 AMC 10A Problems/Problem 25
...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>? ...and therefore <math>11|(100q+r)</math>, so <math>11|n</math>. Then, <math>n</math> can range from <math>10010</math> to <math>99990</math> for a total

6 KB (943 words) - 20:57, 29 May 2023
2004 AMC 10A Problems
...\ } -\frac{1}{4} \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } \frac{1}{4} \qquad \mathrm{(E) \ } \frac{1}{2} </math> ==Problem 4 ==

15 KB (2,092 words) - 20:32, 15 April 2024
2001 AMC 8 Problems
<math> \mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 <math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 12</math>

13 KB (1,994 words) - 13:04, 18 February 2024
1985 AIME Problems
Let <math>x_1=97</math>, and for <math>n>1</math> let <math>x_n=\frac{n}{x_{n-1}}</math>. Calculate the product <math>x_1x_2 \ldots x_8</math>. ==Problem 4==

7 KB (1,071 words) - 19:24, 23 February 2024
1985 AIME Problems/Problem 1
Let <math>x_1=97</math>, and for <math>n>1</math>, let <math>x_n=\frac{n}{x_{n-1}}</math>. Calculate the [[product]] <math>x_1x_2x_3x_4x_5x_6x_7x_8</math> ...his equation gives us respectively <math>x_1x_2 = 2</math>, <math>x_3x_4 = 4</math>, <math>x_5x_6 = 6</math> and <math>x_7x_8 = 8</math> so <cmath>x_1x_

2 KB (279 words) - 15:07, 4 September 2020
Pythagorean triple
<math>(3, 4, 5)</math><nowiki>*</nowiki> ...m^2 + n^2)</math> or <math>(2mn, m^2-n^2, m^2+n^2)</math>, where <math>m > n</math> are relatively prime positive integers of different [[parity]].

4 KB (684 words) - 16:45, 1 August 2020
2003 AMC 12A Problems/Problem 1
The first <math>2003</math> even counting numbers are <math>2,4,6,...,4006</math>. Thus, the problem is asking for the value of <math>(2+4+6+...+4006)-(1+3+5+...+4005)</math>.

3 KB (436 words) - 20:31, 28 December 2021
2003 AMC 12A Problems/Problem 8
...ac{1}{10}\qquad \mathrm{(B) \ } \frac{1}{6}\qquad \mathrm{(C) \ } \frac{1}{4}\qquad \mathrm{(D) \ } \frac{1}{3}\qquad \mathrm{(E) \ } \frac{1}{2} </math ...square, exactly half of the positive factors will be less than <math>\sqrt{n}</math>.

2 KB (326 words) - 15:40, 19 August 2023
Perfect power
...to be a ''perfect <math>k</math>th power''. For example, <math>64 = 8^2 = 4^3 = 2^6</math>, so <math>64</math> is a perfect <math>2</math>nd, <math>3</

870 bytes (148 words) - 16:52, 18 August 2013
1987 IMO Problems/Problem 3
|a_1x_1 + a_2x_2 + \cdots + a_nx_n| \le \frac{ (k-1) \sqrt{n} }{ k^n - 1 } ...ath> \frac{ (k-1)\sqrt{n} }{k^n - 1} </math>. But since there are <math>k^n </math> such sums, by the [[pigeonhole principle]], two must fall into the

2 KB (349 words) - 04:36, 28 May 2023
1987 IMO Problems
...h> be the number of permutations of the set <math>\{ 1, \ldots , n \} , \; n \ge 1 </math>, which have exactly <math>k </math> fixed points. Prove that \sum_{k=0}^{n} k \cdot p_n (k) = n!

3 KB (459 words) - 14:24, 17 September 2023
2002 IMO Shortlist Problems/N3
...w that <math> 2^{p_{1}p_{2} \cdots p_{n}} + 1 </math> has at least <math>4^n </math> divisors. ...ime; hence <math>(2^u +1)(2^v + 1)/3 </math> has at least <math>2 \cdot 4^{n-1} </math> factors, by the inductive hypothesis.

10 KB (1,739 words) - 06:38, 12 November 2019
Mock AIME 4 2006-2007 Problems/Problem 1
...accomplish this by writing 169 four-digit numbers for a total of <math>321+4(169)=997</math> digits. The last of these 169 four-digit numbers is 1168, {{Mock AIME box|year=2006-2007|n=4|before=First question|num-a=2|source=125025}}

1 KB (149 words) - 23:41, 22 April 2010
Mock AIME 4 2006-2007 Problems/Problem 7
...math> such that <math>N\equiv n\pmod{100}</math> so that <math>3^N\equiv 3^n\pmod{1000}</math>. ...th>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we have the following:

1 KB (127 words) - 00:15, 5 January 2010
Mock AIME 4 2006-2007 Problems/Problem 10
<math>\sum_{k=0}^{n} {n \choose k} =2^n</math>, and <math>\sum_{k=0}^{3n} {3n \choose k} =2^{3n}</math> ...binomial. So we divide the whole sum by 3 and we add or subtract <math>q(n)</math> to correct for the integer based on the modularity of the sum with

4 KB (595 words) - 12:14, 25 November 2023
Mock AIME 4 2006-2007 Problems/Problem 15
...ime]]. Find the greatest [[integer]] less than or equal to <math>m + \sqrt{n}</math>. ...</math>. By the [[quadratic formula]], <math>MD = \frac{2R + \sqrt{4R^2 - 4\cdot24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>.

3 KB (532 words) - 20:29, 31 August 2020
2020 AMC 10A Problems
<cmath>x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?</cmath> <cmath>\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}</cmath>

13 KB (1,968 words) - 18:32, 29 February 2024
Asymptote: Reference
...math>t</math> ranges from <math>0</math> to the number of nodes (say <math>n</math>) that determine the path. The most basic way to make paths is by jo ...tells Asymptote to form a cyclic path by joining the endpoint with <math>t=n</math> to that with <math>t=0</math>.

7 KB (1,205 words) - 21:38, 26 March 2024
Asymptote: Useful commands and their Output
draw((-1,0)--(4,0),EndArrow(5)); draw((0,-1)--(0,4),EndArrow(5)); label("Interest rate", (4,0), E); label("Money supply", (0,4), N);

6 KB (871 words) - 21:14, 12 June 2023
Power's of 2 in pascal's triangle
1 4 6 4 1 Remember that <math>\binom{n}{r}=\frac{n!}{k!(n-k)!}</math> where <math>n \ge r</math>.

2 KB (341 words) - 16:57, 16 June 2019
Asymptote (geometry)
...n</math>; thus the functions is undefined at <math>x=\frac{\pi}{6} + \frac{n\pi}{3}</math>. [[File:Slantasymptote.png|thumb|500px|The function <math>y=\tfrac{x^2+2x+4} {x+1}</math> has a slant asymptote at <math>y=x+1</math> ]]

4 KB (664 words) - 11:44, 8 May 2020
Pigeonhole Principle/Solutions
Also, note that it is possible to pull out 4 socks without obtaining a pair. ...re exist distinct <math>a, b \in S</math> such that <math>a \equiv b \pmod n</math>, as desired.

10 KB (1,617 words) - 01:34, 26 October 2021
2005 IMO Shortlist Problems/A1
...t for <math>|z| \ge 2 </math>, <math>n \ge 2 </math> (<math> n \in \mathbb{N} </math>), we have ...\frac{|z|^n -1}{|z|-1} = \sum_{i=0}^{n-1}|\pm z|^i \ge \left| \sum_{i=0}^{n-1} \pm z^i \right|

2 KB (346 words) - 08:56, 14 May 2024
2006 Canadian MO Problems/Problem 1
...(3,6)=1</math>, and <math>f(3,4)=6</math>. Evaluate <math>f(2006,1)+f(2006,4)+f(2006,7)+\dots+f(2006,1003)</math>.

447 bytes (64 words) - 11:32, 30 November 2023
2000 AMC 12 Problems/Problem 7
...1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 4 } </math> If <math>\log_{b} 729 = n</math>, then <math>b^n = 729</math>. Since <math>729 = 3^6</math>, <math>b</math> must be <math>3<

616 bytes (86 words) - 23:49, 29 December 2023
2007 AIME II Problems/Problem 9
...ngst the numbers given. <math>BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105</math>. Also, the length of <math>EF = \sqrt{63^2 + (448 - 2\cdot8 ==Solution 4==

5 KB (818 words) - 11:05, 7 June 2022
2007 AIME II Problems/Problem 10
...d <math>m</math> and <math>n</math> are [[relatively prime]]. Find <math>m+n+r.</math> (The set <math>S-A</math> is the set of all elements of <math>S</ ...ements. The total probability is <math>\frac{2}{64} + \frac{2}{64} = \frac{4}{64}</math>.

8 KB (1,367 words) - 11:48, 23 October 2022
2002 USAMO Problems/Problem 3
...the average of two monic polynomials of degree <math>n </math> with <math>n </math> real roots. ...ath> such that <math>P(x_i) = y_i </math> for all integers <math> i \in [1,n] </math>.

4 KB (688 words) - 13:38, 4 July 2013
2002 USAMO Problems/Problem 5
...ghtarrow n(d-1) </math>. It follows that in fact <math> n \leftrightarrow n(d-1)^k </math>, for any nonnegative integer <math>k </math>, and ...n(d-1) \leftrightarrow n(d-1)(d-2) \leftrightarrow \cdots \leftrightarrow n \prod_{i=c}^{d-1}i

7 KB (1,194 words) - 15:39, 28 March 2015
Mock AIME 1 Pre 2005 Problems
...<math>n</math> are relatively prime positive integers. Determine <math>m + n</math>. == Problem 4 ==

6 KB (1,100 words) - 22:35, 9 January 2016
Mock AIME 2 Pre 2005/Problems
...<math>n</math> are relatively prime positive integers. Compute <math>m + n</math>. ==Problem 4==

6 KB (990 words) - 15:23, 11 November 2009
Mock AIME 3 Pre 2005 Problems
...igits are in increasing order. Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. (Repeated digits are allowed.) <cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4.</cmath>

7 KB (1,135 words) - 23:53, 24 March 2019
Mock AIME 3 Pre 2005 Problems/Problem 1
..., or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(10+r)</math>. <math>84*4=14*24</math>, so <math>r=14</math>. Thus the area of the circle is <math>\b {{Mock AIME box|year=Pre 2005|n=3|before=First Question|num-a=2}}

795 bytes (129 words) - 10:22, 4 April 2012
Mock AIME 3 Pre 2005 Problems/Problem 3
<cmath>2f\left(x\right) + f\left(\frac{1}{x}\right) = 5x + 4</cmath> <cmath>2f\left(\frac 1x\right) + f\left(x\right) = \frac{5}{x} + 4</cmath>

1 KB (191 words) - 10:22, 4 April 2012
Mock AIME 3 Pre 2005 Problems/Problem 5
...math>-letter Zuminglish words. Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>. ...a vowel (<tt>CV</tt>), and let <math>c_n</math> denote the number of <math>n</math>-letter words ending in a vowel followed by a constant (<tt>VC</tt> -

5 KB (795 words) - 16:03, 17 October 2021
Mock AIME 3 Pre 2005 Problems/Problem 7
...ls of <math>ABCD</math> intersect at <math>P</math>. If <math>AB = 1, CD = 4,</math> and <math>BP : DP = 3 : 8,</math> then the area of the inscribed ci .../math>. Therefore <math>\frac{AB}{DC}=\frac{AP}{DP}=\frac{BP}{PC}=\frac{1}{4}</math>. This shows that <math>AP=2x</math> and <math>CP=12x</math>. More a

2 KB (330 words) - 10:23, 4 April 2012
Mock AIME 3 Pre 2005 Problems/Problem 8
Let <math>N</math> denote the number of <math>8</math>-tuples <math>(a_1, a_2, \dots, a Determine the remainder obtained when <math>N</math> is divided by <math>1000</math>.

3 KB (520 words) - 12:55, 11 January 2019
Mock AIME 3 Pre 2005 Problems/Problem 10
<math>\{A_n\}_{n \ge 1}</math> is a sequence of positive integers such that <math>a_{n} = 2a_{n-1} + n^2</math>

2 KB (306 words) - 10:36, 4 April 2012
Mock AIME 3 Pre 2005 Problems/Problem 14
...th>C</math> and <math>D</math> respectively. If <math>AD = 3, AP = 6, DP = 4,</math> and <math>PQ = 32</math>, then the area of triangle <math>PBC</math then <math>r_1=\frac{|AD| \times |DP| \times |AP|}{4A_1}=\frac{(3)(4)(6)}{4A_1}=\frac{18}{A_1}</math>

3 KB (563 words) - 02:05, 25 November 2023
Mock AIME 3 Pre 2005 Problems/Problem 15
<math>\sum_{k=1}^{40} \cos^{-1}\left(\frac{k^2 + k + 1}{\sqrt{k^4 + 2k^3 + 3k^2 + 2k + 2}}\right)</math> ...d <math>n</math> are relatively prime positive integers. Compute <math>m + n</math>.

2 KB (312 words) - 10:38, 4 April 2012
1960 IMO Problems/Problem 3
...f length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> an odd integer). Let <math>\alpha</math> be the acute angle subtendi \tan{\alpha}=\frac{4nh}{(n^2-1)a}.

3 KB (501 words) - 00:14, 17 May 2015
1960 IMO Problems
...ac{N}{11}</math> is equal to the sum of the squares of the digits of <math>N</math>. ...f length <math>a</math>, is divided into <math>n</math> equal parts (<math>n</math> and odd integer). Let <math>\alpha</math> be the acute angle subtend

3 KB (511 words) - 21:21, 20 August 2020
Mock AIME 5 2005-2006 Problems/Problem 1
...th>f(n)</math> be the sum of the distinct positive prime divisors of <math>n</math> less than <math>50</math> (e.g. <math>f(12) = 2+3 = 5</math> and <ma <math>\lfloor 99/23\rfloor =4</math>

2 KB (209 words) - 12:43, 10 August 2019
2007 AIME I Problems
{{AIME Problems|year=2007|n=I}} ...seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkw

7 KB (1,218 words) - 15:28, 11 July 2022
2007 AIME I Problems/Problem 2
...seconds later and strolls forward along the walkway at a constant rate of 4 feet per second. Two seconds after that, Cy reaches the start of the walkw \frac{8(s-4)+10(s-2)}{2}=6s

2 KB (301 words) - 08:09, 4 November 2022
2007 AIME I Problems/Problem 3
{{AIME box|year=2007|n=I|num-b=2|num-a=4}}

1,003 bytes (163 words) - 15:34, 18 February 2017
2007 AIME I Problems/Problem 5
...k</math>. Testing every value of <math>k</math> shows that <math>k = 0, 2, 4, 5, 7</math>, so <math>5</math> of every <math>9</math> values of <math>k</ ...> onwards, <math>995,\ 997,\ 999,\ 1000</math> work, giving us <math>535 + 4 = \boxed{539}</math> as the solution.

7 KB (1,076 words) - 00:10, 29 November 2023
2007 AIME I Problems/Problem 8
...th>(x - a)(x - m) = x^2 + (k - 29)x - k</math> and that <math>2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)</math>. ...s (and we have four [[variable]]s), <math>a + m = 29 - k</math>, <math>a + n = \frac{43}{2} - k</math>, <math>am = -k</math>, and <math>an = \frac{k}{2}

4 KB (728 words) - 00:11, 29 November 2023
2007 AIME I Problems/Problem 10
...the number of shadings with this property. Find the remainder when <math>N</math> is divided by 1000. ...oose where the ball in column 3 goes. (The location of the ball in column 4 is forced.) Again, we think about what happens in the bottom half of the b

13 KB (2,328 words) - 00:12, 29 November 2023
2007 AIME I Problems/Problem 6
**There are <math>1 \cdot \left(\frac{24-15}{3} + 1\right) = 4</math> ways to reach <math>26</math>. *There are <math>4 + 25 = 29</math> ways to reach <math>26</math>.

10 KB (1,519 words) - 00:11, 29 November 2023
2007 AIME I Problems/Problem 9
...frac{1 + \frac{15}{17}}{1 - \frac{15}{17}}}</math>, and <math>x = \frac{r}{4}</math>. Therefore, <math>x + y + 2r = \frac{r}{4} + \frac{3r}{5} + 2r = \frac{57r}{20} = 34 \Longrightarrow r = \frac{680}{5

11 KB (1,851 words) - 12:31, 21 December 2021
2007 AIME I Problems/Problem 11
...the formula for the sum of the first <math>n</math> squares (<math>\frac{n(n+1)(2n+1)}{6}</math>), we get <math>\sum_{k=1}^{44}2k^2=2\frac{44(44+1)(2*44 ...m{3}{2}+\cdots +\binom{44}{2}\right)</math>. We can collapse this to <math>4\binom{45}{3}=56760</math>. Now, we have to consider <math>p</math> from <ma

3 KB (562 words) - 20:02, 30 December 2023
2007 AIME I Problems/Problem 12
...,5); MA("75^\circ",C,B,A,2); MA("30^\circ",A,C,B,4); MA("30^\circ",A,Cp,Bp,4); MA("45^\circ",A,extension(A,C,Bp,Cp),Bp,3); MA("15^\circ",C,Y,Cp,8); MA( ...ath>AF = 20\sin 75 = 20 \sin (30 + 45) = 20\left(\frac{\sqrt{2} + \sqrt{6}}4\right) = 5\sqrt{2} + 5\sqrt{6}</math></center>

10 KB (1,458 words) - 20:50, 3 November 2023
2007 AIME I
* [[2007 AIME I Problems/Problem 4]] {{AIME box|year=2007|n=I|before=[[2006 AIME I]], [[2006 AIME II|II]]|after=[[2007 AIME II]]}}

1 KB (135 words) - 12:32, 22 March 2011
2007 AIME I Problems/Problem 14
...in the following way: <math>a_{0}=a_{1}=3</math>, <math>a_{n+1}a_{n-1}=a_{n}^{2}+2007</math>. ...rac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n}}</cmath>

13 KB (2,185 words) - 23:30, 5 December 2022
2007 AIME I Problems/Problem 13
...math>ABCD</math> and vertex <math>E</math> has eight edges of length <math>4</math>. A plane passes through the midpoints of <math>AE</math>, <math>BC< currentprojection = perspective(2.5,-12,4);

7 KB (1,034 words) - 21:56, 22 September 2022
2007 AIME I Problems/Problem 15
...}</math>. This simplifies to <math>\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)</math>. Some terms will cancel out, lea ...- 4 \cdot 9 \cdot 100}</math><math> = \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}</math>. The answer is <math>\boxed{989}</ma

4 KB (673 words) - 22:14, 6 August 2022
LaTeX:Basics
...command. For example, \sqrt[n]{x} produces <math>\sqrt[n]{x}</math>. The [n] does not have to be included if no root is needed, so you can just use \sq ...have to find the mistake in your source file by line number. If you have a 4 or 5 line 'line', finding the error can be a real headache. (You can see wh

6 KB (1,112 words) - 08:03, 22 August 2023
2004 USAMO Problems
(a) For <math> i = 1, \ldots , n </math>, <math> a_i \in S </math>. <br> (b) For <math> i, j = 1, \ldots, n </math> (not necessarily distinct), <math> a_i - a_j \in S </math>. <br>

3 KB (474 words) - 09:20, 14 May 2021
2004 USAMO Problems/Problem 4
...<math>(m,n) </math> is the square in the <math>m</math>th column and <math>n</math>th row. Thus Bob can always win. ...es a number in rows 3, 4, 5, or 6, then Bob also writes a number in row 3, 4, 5, or 6.

2 KB (430 words) - 13:40, 4 July 2013
2007 AIME II
* [[2007 AIME II Problems/Problem 4]] {{AIME box|year=2007|n=II|before=[[2007 AIME I]]|after=[[2008 AIME I]], [[2008 AIME II|II]]}}

1 KB (135 words) - 12:34, 22 March 2011
2007 AIME II Problems/Problem 3
dot("$E$", E, N); dot("$A$", A, N);

6 KB (933 words) - 00:05, 8 July 2023
2007 AIME II Problems/Problem 5
...{9} \rfloor = 6</math>. Adding them up, we get <math>864 + 1 + 2 + 3 + 3 + 4 + 5 + 6 = 888</math>. === Solution 4 ===

4 KB (559 words) - 00:38, 3 January 2023
2007 AIME II Problems/Problem 6
...math> in the second column, we note that <math>3</math> is less than <math>4,6,8</math>, but greater than <math>1</math>, so there are four possible pla | 1 || 1 || 4 || 16 || 64

2 KB (338 words) - 15:30, 7 August 2022
2007 AIME II Problems/Problem 7
...th> (the one to be inclusive) integers will fit the conditions. <math>3k + 4 = 70 \Longrightarrow k = 22</math>. {{AIME box|year=2007|n=II|num-b=6|num-a=8}}

3 KB (428 words) - 18:04, 4 December 2020
2007 AIME II Problems/Problem 2
...ath>n - 1</math> ways to find pairs of <math>(x,y)</math> if <math>x + y = n</math>. Thus, there are <math>0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxe {{AIME box|year=2007|n=II|num-b=1|num-a=3}}

1 KB (228 words) - 08:41, 4 November 2022
2007 AIME II Problems/Problem 8
A [[rectangle|rectangular]] piece of paper measures 4 units by 5 units. Several [[line]]s are drawn [[parallel]] to the edges of ...e number of basic rectangles determined. Find the [[remainder]] when <math>N</math> is divided by 1000.

3 KB (399 words) - 21:17, 24 February 2021
2007 AIME II Problems/Problem 12
...{3}(x_{n}) = 308</math> and <math>56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,</math> ...can rewrite the first given equation as <math>n + n + m + n+2m + n+3m+...+n+7m = 308</math>. Simplify to get <math>2n+7m=77</math>. (1)

5 KB (829 words) - 12:22, 8 January 2024
2007 AIME II Problems/Problem 13
...equal to <math>1</math>. This gives <math>{4\choose0} + {4\choose3} = 1 + 4 = 5</math> possible combinations of numbers that work. ...>{n\choose0}x_0 + {n\choose1}x_1 + {n\choose2}x_2 + \ldots {n\choose{n}}x_{n}</math>.

3 KB (600 words) - 11:10, 22 January 2023
2007 AIME II Problems/Problem 15
...nd <math>n</math> are [[relatively prime]] positive integers. Find <math>m+n.</math> label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P));

11 KB (2,099 words) - 17:51, 4 January 2024
2007 AIME II Problems/Problem 14
...n \ge 1</math>. The condition <math>f(2) + f(3) = 125</math> implies <math>n = 2</math>, giving <math>f(5) = \boxed{676}</math>. <cmath>(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)</cmath>

7 KB (1,335 words) - 17:44, 25 January 2022
2007 AIME II Problems
{{AIME Problems|year=2007|n=II}} ...ach possible sequence appears exactly once contains N license plates. Find N/10.

9 KB (1,435 words) - 01:45, 6 December 2021
2005 USAMO Problems
...s <math>n</math> for which it is possible to arrange all divisors of <math>n</math> that are greater than 1 in a circle so that no two adjacent divisors == Problem 4 ==

4 KB (609 words) - 09:24, 14 May 2021
2002 USA TST Problems
...denote the union of all segments <math> P_1P_2, P_2P_3, \dots , P_{n-1}P_{n} </math>. Determine if it is always possible to find points <math> \display === Problem 4 ===

3 KB (544 words) - 06:58, 3 August 2017
2007 AMC 12A Problems
One ticket to a show costs <math>\$20</math> at full price. Susan buys <math>4</math> tickets using a coupon that gives her a <math>25\%</math> discount. <math>\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \

11 KB (1,750 words) - 13:35, 15 April 2022
Pyramid
...es, and <math>\displaystyle n+1</math> faces (of which <math>\displaystyle n</math> are triangular, and the remaining one is the base). ...n to that of the larger. ([[2003 AIME II Problems/Problem 4|2003 AIME II, #4]])

2 KB (329 words) - 21:05, 20 August 2021
1989 USAMO Problems/Problem 1
For each positive [[integer]] <math>n</math>, let U_n &= \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}.

2 KB (267 words) - 19:10, 18 July 2016
1989 USAMO Problems
For each positive integer <math>n</math>, let ...>U_n = \frac{T_1}{2} + \frac{T_2}{3} + \frac{T_3}{4} + \cdots + \frac{T_n}{n+1}</math>.

2 KB (326 words) - 18:52, 18 July 2016
2003 USAMO Problems
...</math> there exists an <math>n </math>-digit number divisible by <math>5^n </math> all of whose digits are odd. Let <math> n \neq 0 </math>. For every sequence of integers

3 KB (487 words) - 09:21, 14 May 2021
2003 USAMO Problems/Problem 1
...</math> there exists an <math>n </math>-digit number divisible by <math>5^n </math> all of whose digits are odd. ...n-1} + a </math> is divisible by 5, which is true when <math> k \equiv -3^{n-1}a \pmod{5} </math>. Since there is an odd digit in each of the residue c

4 KB (736 words) - 22:17, 3 March 2023
2003 USAMO Problems/Problem 3
Let <math> n \neq 0 </math>. For every sequence of integers satisfying <math>0 \le a_i \le i</math>, for <math>i=0,\dots,n</math>, define another sequence

3 KB (636 words) - 13:39, 4 July 2013
2007 USAMO Problems
...nce is <math>9, 1, 2, 0, 3, 3, 3, \ldots</math>. Prove that for any <math>n</math> the sequence <math>a_1, a_2, a_3, \ldots</math> eventually becomes c ...consists of all points <math>(m,n)</math>, where <math>m</math> and <math>n</math> are integers. Is it possible to cover all grid points by an infinite

3 KB (539 words) - 13:42, 4 July 2013
2007 USAMO Problems/Problem 1
...nce is <math>9, 1, 2, 0, 3, 3, 3, \ldots</math>. Prove that for any <math>n</math> the sequence <math>a_1, a_2, a_3, \ldots</math> eventually becomes [ Let <math>a_1 = n</math>. Since <math>a_k\leq k - 1</math>, we have that

6 KB (1,204 words) - 20:06, 23 August 2023
2007 USAMO Problems/Problem 3
...th> are partitioned into two classes. Prove that there are at least <math>n</math> [[pairwise disjoint]] sets in the same class. Claim: If we have instead <math>k(n+1)-1</math> elements, then we must have k disjoint subsets in a class.

6 KB (1,071 words) - 08:40, 21 October 2020
2007 USAMO Problems/Problem 4
...'' with <math>n</math> ''cells'' is a connected figure consisting of <math>n</math> equal-sized [[Square (geometry)|square]] cells.<math>{}^1</math> The <div style="text-align:center;">[[Image:2007 USAMO-4.PNG|350px]]</div>

10 KB (1,878 words) - 14:56, 30 June 2021
2007 USAMO Problems/Problem 5
...or every [[nonnegative]] [[integer]] <math>n</math>, the number <math>7^{7^n}+1</math> is the [[product]] of at least <math>2n+3</math> (not necessarily The proof is by induction. The base is provided by the <math>n = 0</math> case, where <math>7^{7^0} + 1 = 7^1 + 1 = 2^3</math>. To prove t

2 KB (240 words) - 09:47, 7 August 2014
Nesbitt's Inequality
If <math> a_1, \ldots a_n </math> are positive and <math> \sum_{i=1}^{n}a_i = s </math>, then \sum_{i=1}^{n}\frac{a_i}{s-a_i} \geq \frac{n}{n-1}

7 KB (1,224 words) - 16:21, 24 October 2022
2007 BMO Problems/Problem 3
...exists a permutation <math>\sigma </math> on the set <math> \{ 1, \ldots, n \} </math> for which \sqrt{\sigma(1) + \sqrt{\sigma(2) + \sqrt{ \cdots + \sqrt{\sigma(n)}}}}

4 KB (682 words) - 10:53, 13 January 2016
Pascal Triangle Related Problems
4: 1 4 6 4 1 # The coefficient of the <math>x^{4}y^{6}</math> term.

4 KB (513 words) - 20:18, 3 January 2023
2007 Cyprus MO/Lyceum/Problems
Given the formula <math>f(x) = 4^x</math>, then <math>f(x+1)-f(x)</math> equals to ...\ 4\qquad\mathrm{(B)}\ 4^x\qquad\mathrm{(C)}\ 2\cdot4^x\qquad\mathrm{(D)}\ 4^{x+1}\qquad\mathrm{(E)}\ 3\cdot4^x</math>

13 KB (1,990 words) - 08:29, 19 December 2009
2005 IMO Shortlist Problems/A2
...hen setting <math>x=y=y_0 </math> in the original equation gives <math>f(3^n y_0) = 2 </math>, by induction. It follows that for any <math>x </math>, t ...tyle f(c_n) = 2 </math>, for all nonnegative integers <math> \displaystyle n </math>.

7 KB (1,179 words) - 20:59, 28 March 2010
Circumradius
label("$B$", B, N); <cmath> R=\frac{abc}{4\times [ABC]}\text{ or }[ABC]=\frac{abc}{4R}</cmath>

4 KB (729 words) - 16:52, 19 February 2024
Sum and difference of powers
<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math> <math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>

3 KB (450 words) - 06:25, 6 May 2024
LaTeX:Layout
Name&\#1&\#2&\#3&\#4&\#5&Total\\\hline Jane Doe&5&5&5&4&5&24\\

30 KB (5,171 words) - 10:16, 4 April 2021
LaTeX:Commands
| <math>\sqrt[n]{x}</math>||\sqrt[n]{x} | <math>\textstyle \prod_{n=1}^5\frac{n}{n-1}</math>||\prod_{n=1}^5\frac{n}{n-1}

12 KB (1,898 words) - 15:31, 22 February 2024
Chicken McNugget Theorem
...m + bn</math> for [[nonnegative]] integers <math>a, b</math> is <math>mn-m-n</math>. ...he proof is based on the fact that in each pair of the form <math>(k, mn-m-n-k)</math>, exactly one element is expressible.

17 KB (2,748 words) - 19:22, 24 February 2024
Triangular number
...the first <math>n</math> [[natural number]]s from <math>1</math> to <math>n</math>. ...th}</math> triangular number is the sum of all natural numbers from one to n.

2 KB (275 words) - 08:39, 7 July 2021
2007 AMC 12A Problems/Problem 3
<math>\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \ ...the smaller term. Then <math>n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1</math>

585 bytes (81 words) - 02:09, 16 February 2021
Parity
...integer; it can then be shown that an odd number is an integer of the form n = 2k + 1. <math>\text{(A)}\ \frac{4}{9} \qquad \text{(B)}\ \frac{9}{19} \qquad \text{(C)}\ \frac{1}{2} \qquad \

4 KB (694 words) - 22:00, 12 January 2024
Sophie Germain Identity
<div style="text-align:center;"><math>a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)</math></div> a^4 + 4b^4 &= a^4 + 4a^2b^2 + 4b^4 - 4a^2b^2 \\ &= (a^2 + 2b^2)^2 - (2ab)^2 \\ &= (a^2 + 2b^2 - 2ab) (a^2 + 2

2 KB (222 words) - 15:04, 30 December 2023
2007 AMC 12A Problems/Problem 22
...th>n.</math> For how many values of <math>n</math> is <math>n + S(n) + S(S(n)) = 2007?</math> ...(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5</math>

15 KB (2,558 words) - 19:33, 4 February 2024
2007 AMC 12A Problems/Problem 24
...h> on the interval <math>[0,\pi]</math>. What is <math>\sum_{n=2}^{2007} F(n)</math>? <math>F(n) = n + 1</math>

4 KB (588 words) - 14:40, 23 August 2023
2007 AMC 12A Problems/Problem 25
...math>S_{n}</math> denote the number of spacy subsets of <math>\{ 1, 2, ... n \}</math>. We have <math>S_{0} = 1, S_{1} = 2, S_{2} = 3</math>. The spacy subsets of <math>S_{n + 1}</math> can be divided into two groups:

9 KB (1,461 words) - 23:07, 27 January 2024
2005 AMC 12A Problems/Problem 10
...urth of the total number of faces of the unit cubes are red. What is <math>n</math>? (\mathrm {A}) \ 3 \qquad (\mathrm {B}) \ 4 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 6 \qquad (\mathrm {E})\ 7

693 bytes (107 words) - 21:19, 3 July 2013
2005 AMC 12A Problems/Problem 15
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad ===Solution 4===

14 KB (1,970 words) - 17:02, 18 August 2023
2009 AIME II
** [[2009 AIME II Problems/Problem 4|Problem 4]] {{AIME box|year=2009|n=II|before=[[2009 AIME I]]|after=[[2010 AIME I]], [[2010 AIME II|II]]}}

1 KB (156 words) - 17:59, 20 June 2020
2006 AIME II Problems/Problem 3
...uence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is just 27, 81, 135, 189, so 4 multiples here. Finally, we have 33+11+4+1=49.

4 KB (562 words) - 18:37, 30 October 2020
2006 AIME II Problems/Problem 6
label("$1$", A--D, N); ...= \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}</math>, so <math>AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}

7 KB (1,067 words) - 12:23, 8 April 2024
2006 AIME II Problems/Problem 8
...ee rotation. This is known as the identity rotation, and there are <math>6^4=1296</math> choices because we don't have any restrictions. Case 4: symmetry about lines. We multiply by 3 for these because the amount of col

4 KB (695 words) - 10:37, 4 November 2023
2006 AIME II Problems/Problem 12
D(MP("A",A,N,s)--MP("B",B,W,s)--MP("C",C,E,s)--cycle);D(O); ...\angle AEF = \frac12\cdot11\cdot13\cdot\sin{120^\circ} = \frac {143\sqrt3}4.</cmath>

6 KB (1,033 words) - 02:36, 19 March 2022
2006 AIME II Problems/Problem 5
...<math> n </math> are [[relatively prime]] positive integers, find <math> m+n. </math> ...<math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <math>4 \cdot \frac{1}{36} = \frac{1}{9}</math>. Subtracting all these probabilitie

5 KB (712 words) - 12:10, 5 November 2023
2006 AIME II Problems/Problem 9
...enter]]s at (0,0), (12,0), and (24,0), and have [[radius|radii]] 1, 2, and 4, respectively. Line <math> t_1 </math> is a common internal [[tangent]] to ..., we can do the same thing to get <math>O_3s_1 = 4</math> and <math>s_1s = 4\sqrt{3}</math>.

3 KB (553 words) - 10:45, 26 August 2015
2006 AIME II Problems/Problem 13
How many integers <math> N </math> less than <math>1000</math> can be written as the sum of <math> j < ...-1)}{2}\right) = j(2n+j)</math>. Thus, <math>j</math> is a factor of <math>N</math>.

4 KB (675 words) - 10:40, 14 July 2022
2006 AIME II Problems/Problem 15
...math> n </math> is not divisible by the square of any prime, find <math> m+n.</math> ...must have that <math>h_x^2 = \frac1{16}</math> and so <math>h_x = \frac{1}4</math> and similarly <math>h_y = \frac15</math> and <math>h_z = \frac16</ma

4 KB (725 words) - 17:18, 27 June 2021
2006 AIME II Problems/Problem 11
...a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> ...e the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be:

3 KB (417 words) - 10:07, 12 October 2023
2006 AIME II Problems/Problem 4
An example of such a permutation is <math> (6,5,4,3,2,1,7,8,9,10,11,12). </math> Find the number of such permutations. ...spaces between <math>a_6</math> and <math>a_{12}</math>. <math>\binom{10}{4}</math> is 210. This splits the remaining 10 numbers into two distinct sets

2 KB (384 words) - 14:12, 20 April 2024
2006 AIME II Problems/Problem 7
=== Solution 4 (Similar to Solution 2)=== {{AIME box|year=2006|n=II|num-b=6|num-a=8}}

7 KB (1,114 words) - 03:41, 12 September 2021
2006 AIME II Problems/Problem 10
...and <math> n </math> are relatively prime positive integers. Find <math> m+n. </math> ...th>\frac{126}{512} + \frac{193}{512} = \frac{319}{512}</math>, and <math>m+n = \boxed{831}</math>.

6 KB (983 words) - 13:42, 8 December 2021
2006 AIME II Problems
{{AIME Problems|year=2006|n=II}} .../math> is a positive integer. Find the number of possible values for <math>n</math>.

8 KB (1,350 words) - 12:00, 4 December 2022
2005 PMWC Problems
4. <math>101a + 53 \ge 2332</math> ...such a way that each man gets <math>6</math> apples, each woman gets <math>4</math> apples and each child gets <math>1</math> apple. In how many possibl

9 KB (1,449 words) - 20:49, 2 October 2020
1998 PMWC Problems
Calculate: <math>\frac{1*2*3+2*4*6+3*6*9+4*8*12+5*10*15}{1*3*5+2*6*10+3*9*15+4*12*20+5*15*25}</math> int triangle(pair z, int n){

11 KB (1,738 words) - 19:25, 10 March 2015
1999 PMWC Problems
label("$5$",(P+Q)/2,N); import graph; size(4.62cm);

6 KB (703 words) - 21:21, 21 April 2014
2000 PMWC Problems
Given that <math>A^4=75600\times B</math>. If <math>A</math> and <math>B</math> are positive int ...oduct of two <math>4</math>-digit numbers formed by the digits <math>1,2,3,4,5,6,7,8</math> without any repetition. Find the largest value of <math>p</m

11 KB (1,713 words) - 22:47, 13 July 2023
2001 PMWC Problems
O=(A+B+C+D)/4; E=(O.x+cos(-eta/4),O.y+sin(-eta/4));

4 KB (641 words) - 21:24, 21 April 2014
2002 PMWC Problems
A=(0,4); C=(4,0);

5 KB (725 words) - 16:07, 23 April 2014
2003 PMWC Problems
pair M=(-1,0), N=(1,0),a=4/5*expi(pi/10),b=expi(37pi/100); draw((M--N)^^(origin--a)^^(origin--b));

7 KB (918 words) - 16:15, 22 April 2014
2005 PMWC Problems/Problem T6
4+5+6 &=& 7+8 \\ ...use the fact that the sum of the first <math>n</math> odd numbers is <math>n^2</math>):

882 bytes (140 words) - 19:07, 10 March 2015
1989 USAMO Problems/Problem 3
...</math> such that <math>P(a + bi) = 0</math> and <math>(a^2 + b^2 + 1)^2 < 4 b^2 + 1</math>. <cmath> P(z) = \prod_{j=1}^n (z- z_j) . </cmath>

2 KB (340 words) - 19:11, 18 July 2016
1971 Canadian MO Problems
== Problem 4 == [[1971 Canadian MO Problems/Problem 4 | Solution]]

3 KB (519 words) - 08:58, 13 September 2012
1997 PMWC Problems/Problem T2
&& 1 \left(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1} &+& 3\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}

1 KB (139 words) - 19:07, 10 March 2015
1961 IMO Problems
<math>\cos^n{x} - \sin^n{x} = 1</math> where ''n'' is a given positive integer.

3 KB (425 words) - 21:18, 20 August 2020
2004 AMC 10A Problems/Problem 20
label("$F$", F, N); <math> \mathrm{(A) \ } \frac{4}{3} \qquad \mathrm{(B) \ } \frac{3}{2} \qquad \mathrm{(C) \ } \sqrt{3} \qqu

4 KB (710 words) - 02:47, 18 April 2024
1990 USAMO Problems
f_{n + 1}(x) &= \sqrt {x^2 + 6f_n(x)} \quad \text{for } n \geq 1. ...od to represent the positive square root.) For each positive integer <math>n</math>, find all real solutions of the equation <math>\, f_n(x) = 2x \,</ma

3 KB (386 words) - 20:47, 3 July 2013
2007 AMC 12B Problems
==Problem 4== ...t Market, 3 bananas cost as much as 2 apples, and 6 apples cost as much as 4 oranges. How many oranges cost as much as 18 bananas?

12 KB (1,814 words) - 12:58, 19 February 2020
1990 USAMO Problems/Problem 2
f_{n + 1}(x) &= \sqrt {x^2 + 6f_n(x)} \quad \text{for } n \geq 1. ...o represent the positive [[square root]].) For each positive integer <math>n</math>, find all real solutions of the equation <math>\, f_n(x) = 2x \,</ma

2 KB (322 words) - 19:14, 18 July 2016
1999 AHSME Problems
<math>1 - 2 + 3 -4 + \cdots - 98 + 99 = </math> == Problem 4 ==

13 KB (1,945 words) - 18:28, 19 June 2023
1960 IMO Problems/Problem 1
...ac{N}{11}</math> is equal to the sum of the squares of the digits of <math>N</math>. Let <math>N = 100a + 10b+c</math> for some digits <math>a,b,</math> and <math>c</math>.

4 KB (751 words) - 05:01, 17 August 2022
Asymptote: Useful functions
Label ''L''="", int ''n''=1, real ''radius''=0, real ''space''=0,<br/> fill(x[0]--x[1]--x[2]--x[3]--x[4]--x[5]--cycle,black);

4 KB (646 words) - 21:18, 26 March 2024
2001 USAMO Problems/Problem 2
..."F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D ...}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0.</cmath>

11 KB (2,091 words) - 08:35, 16 November 2017
Summation
As an example, <math>\sum_{i=3}^6 i^3 = 3^3 + 4^3 + 5^3 + 6^3</math>. Note that if <math>a>b</math>, then the sum is <math> ...ere the sum only includes the terms <math>i</math> which divide into <math>n</math>.

3 KB (482 words) - 16:39, 8 October 2023
Isaac Newton
Isaac Newton was born on January 4, 1643 in Lincolnshire, England. Newton was born very shortly after the deat ...th> places a force on matter with the same mass <math>n</math>, then <math>n</math> will put an equivalent force in the opposite direction.

2 KB (391 words) - 13:36, 22 November 2020
Root (operation)
...h>n</math>th '''root''' of a number <math>x</math>, denoted by <math>\sqrt[n]{x}</math>, is a common operation on numbers and a partial [[inverse]] to [ ...wish to take both the positive and negative roots, we write <math>\pm\sqrt[n]{x}</math>.

3 KB (532 words) - 16:52, 20 May 2020
1962 IMO Problems
Find the smallest natural number <math>n</math> which has the following properties: ..., the resulting number is four times as large as the original number <math>n</math>.

3 KB (431 words) - 21:17, 20 August 2020
2004 AMC 12A Problems/Problem 11
...xt {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4</math> ...1}=21</math>. Substituting yields: <math>20n+25=21(n+1),</math> so <math>n=4</math>, <math>v = 80.</math> Then, we see that the only way Paula can sati

2 KB (310 words) - 17:39, 31 December 2022
2004 AMC 12A Problems/Problem 17
(ii) <math>f(2n) = n \cdot f(n)</math> for any positive integer <math>n</math>. f\left(2^4\right) &= 2^3\cdot f\left(2^3\right) &&= 2^{3+2+1},

2 KB (233 words) - 08:14, 6 September 2021
2004 AMC 12A Problems/Problem 20
...we want <math>c < \frac 12</math>. Thus the chance is <math>\frac{\frac{1}{4}}2 = \frac 18</math>. Then the area producing the desired result is 3/4. Since the area of the unit square is 1, the probability is <math>\frac 34<

3 KB (552 words) - 23:26, 28 December 2020
2004 AMC 12A Problems/Problem 21
If <math>\sum_{n = 0}^{\infty}{\cos^{2n}}\theta = 5</math>, what is the value of <math>\cos{ ...ongrightarrow 1 = 5 - 5\cos^2 \theta \Longrightarrow \cos^2 \theta = \frac{4}{5}</math>. Using the formula <math>\cos 2\theta = 2\cos^2 \theta - 1 = 2\l

1 KB (176 words) - 13:19, 15 July 2022
2004 AMC 12A Problems/Problem 25
...>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is as small as possible. What is <math>m</math>? a_x &= \frac{1}{x}+\frac{3}{x^2}+\frac{3}{x^3}+\frac{1}{x^4}+\frac{3}{x^5}+\frac{3}{x^6}+\cdots \\

3 KB (480 words) - 14:50, 17 August 2020
Binet's Formula
'''Binet's formula''' is an explicit formula used to find the <math>n</math>th term of the Fibonacci sequence. If <math>F_n</math> is the <math>n</math>th [[Fibonacci number]], then

3 KB (550 words) - 16:12, 24 February 2024
1992 USAMO Problems/Problem 2
...can derive the identity <math>\tan(n+1)-\tan(n)=\frac{\sin 1}{\cos(n)\cos(n+1)}</math>. Then the left hand side of the equation simplifies to <math>\t === Solution 4 ===

4 KB (628 words) - 07:41, 19 July 2016
1995 AHSME Problems/Problem 4
...ath>N</math> is <math>50 \%</math> of <math>P</math>, then <math>\frac {M}{N} =</math> ...{(C) \ 1 } \qquad \mathrm{(D) \ \frac {6}{5} } \qquad \mathrm{(E) \ \frac {4}{3} } </math>

2 KB (232 words) - 22:30, 16 February 2018
1999 USAMO Problems/Problem 4
Let <math>a_{1}, a_{2}, \dots, a_{n}</math> (<math>n > 3</math>) be real numbers such that a_{1} + a_{2} + \cdots + a_{n} \geq n \qquad \mbox{and} \qquad

3 KB (499 words) - 09:47, 20 July 2016
1999 IMO Problems/Problem 2
Let <math>n \geq 2</math> be a fixed integer. \sum_{1\leq i<j \leq n} x_ix_j (x_i^2 + x_j^2) \leq C \left(

4 KB (838 words) - 01:04, 17 November 2023
Proofs of AM-GM
...reals <math>\lambda_1, \dotsc, \lambda_n</math> such that <math>\sum_{i=1}^n \lambda_i = 1</math>, then <cmath> \sum_{i=1}^n \lambda_i a_i \geq \prod_{i=1}^n a_i^{\lambda_i}, </cmath>

12 KB (2,171 words) - 07:55, 11 May 2023
1976 USAMO Problems/Problem 5
<cmath> P(x^5) + xQ(x^5) + x^2 R(x^5) = (x^4 + x^3 + x^2 + x +1) S(x), </cmath> In general we will show that if <math>m</math> is an integer less than <math>n</math> and <math>P_0, \dotsc, P_{m-1}</math> and <math>S</math> are polynom

3 KB (572 words) - 17:14, 16 August 2015
2003 Polish Mathematical Olympiad Problems/Problem 3
...atural number]] <math>n</math>, <math>2^n-1</math> is divisible by <math>W(n)</math>. ...q}</math>. Since <math>q</math> is an odd prime, we may divide by <math>2^n</math> to obtain <math>2^q \equiv 1 \pmod{q}</math>. But by [[Fermat's Lit

5 KB (919 words) - 23:29, 20 January 2016
2008 AMC 12B Problems/Problem 23
...ms of the divisors of <math>10^n</math> is <math>792</math>. What is <math>n</math>? ...ath>. <math>\log 10^n=n</math> so the number of factors divided by 2 times n equals the sum of all the factors, 792.

5 KB (814 words) - 18:02, 17 January 2023
2000 AMC 12 Problems/Problem 19
...B)}\ 2.5 \qquad \text {(C)}\ 3 \qquad \text {(D)}\ 3.5 \qquad \text {(E)}\ 4</math> label("$A$",A,N);

3 KB (547 words) - 17:37, 17 February 2024
2000 AMC 12 Problems/Problem 21
label("$2m$",(4,0),S); ...rac{bc}{a}</math> and <cmath>[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]</cmath>

5 KB (804 words) - 01:22, 13 May 2024
2000 AMC 12 Problems/Problem 23
<math>\textbf {(A)}\ 1/5 \qquad \textbf {(B)}\ 1/4 \qquad \textbf {(C)}\ 1/3 \qquad \textbf {(D)}\ 1/2 \qquad \textbf {(E)}\ 1 ...>5</math> from the power of <math>2</math>. This yields <math>0,1,2,-1,3,0,4,1,-2,5,2</math>, and indeed the only non-positive terms are <math>0,0,-1,-2

2 KB (348 words) - 22:26, 24 October 2021
2007 AMC 10A Problems/Problem 23
...ered pair]]s <math>(m,n)</math> of positive [[integer]]s, with <math>m \ge n</math>, have the property that their squares differ by <math>96</math>? <math>\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12</math>

2 KB (352 words) - 14:20, 3 July 2023
2003 JBMO Problems/Problem 1
...nsists of <math>n</math> digits, each of which is 8. Prove that <math>A+2B+4</math> is a perfect square. A &= 4(10^{2n-1} + 10^{2n-2} \cdots 10^1 + 10^0) \\

1,002 bytes (145 words) - 00:14, 28 August 2018
2007 AMC 10A Problems
One ticket to a show costs <math>\$20</math> at full price. Susan buys 4 tickets using a coupon that gives her a <math>25\%</math> discount. Pam buy ...- \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}</math>

13 KB (2,058 words) - 17:54, 29 March 2024
2007 AMC 10A Problems/Problem 10
<math>\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6</math> ...48 + 16(n+1)</math>, and the total number of people in the family is <math>n+2</math>. So

2 KB (411 words) - 20:54, 19 August 2023
2007 AMC 10A Problems/Problem 5
The school store sells 7 pencils and 8 notebooks for <math>\mathdollar 4.15</math>. It also sells 5 pencils and 3 notebooks for <math>\mathdollar 1. We let <math>p =</math> cost of one pencil in dollars, <math>n = </math> cost of one notebook in dollars. Then

1 KB (133 words) - 12:10, 3 June 2021
1995 AHSME Problems
...mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} } </math> == Problem 4 ==

17 KB (2,387 words) - 22:44, 26 May 2021
1995 AHSME Problems/Problem 11
How many base 10 four-digit numbers, <math>N = \underline{a} \underline{b} \underline{c} \underline{d}</math>, satisfy a (i) <math>4,000 \leq N < 6,000;</math> (ii) <math>N</math> is a multiple of 5; (iii) <math>3 \leq b < c \leq 6</math>.

2 KB (233 words) - 10:37, 30 March 2023
1997 PMWC Problems
real xmin = -0.43, xmax = 4.69, ymin = -0.49, ymax = 2.22; /* image dimensions */ draw((3.94,4.18)--(6,5.09)--(6,6)--(3.14,6)--cycle);

15 KB (2,057 words) - 19:13, 10 March 2015
1951 AHSME Problems
The percent that <math>M</math> is greater than <math>N</math> is: ...\textbf{(D) \ } \frac {M - N}{M} \qquad \textbf{(E) \ } \frac {100(M + N)}{N} </math>

23 KB (3,641 words) - 22:23, 3 November 2023
1990 USAMO Problems/Problem 3
19. Prove that for any odd integer <math>n \geq 1</math>, there is a way to number <cmath> \{ n, n+1, n+2, \dots, n+32 \} </cmath>

3 KB (460 words) - 19:14, 18 July 2016
1959 AHSME Problems
== Problem 4== [[1959 AHSME Problems/Problem 4|Solution]]

22 KB (3,345 words) - 20:12, 15 February 2023
1966 AHSME Problems
Given that the ratio of <math>3x - 4</math> to <math>y + 15</math> is constant, and <math>y = 3</math> when <mat == Problem 4 ==

19 KB (3,159 words) - 22:10, 11 March 2024
1991 USAMO Problems/Problem 5
dot(A,linewidth(4)); dot(B,linewidth(4));

5 KB (953 words) - 12:18, 4 September 2018
1991 USAMO Problems/Problem 3
Show that, for any fixed integer <math>\,n \geq 1,\,</math> the sequence <cmath> 2, \; 2^2, \; 2^{2^2}, \; 2^{2^{2^2}}, \ldots \pmod{n} </cmath>

3 KB (559 words) - 19:18, 5 June 2023
1991 USAMO Problems/Problem 1
label("$\mathsf{C}$", C, N); label("$\mathsf{x}$", A--D, N);

5 KB (755 words) - 08:58, 6 May 2023
1991 USAMO Problems
...n) - \left( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \right) (n+1), </cmath> ...sum involving all nonempty subsets <math>S</math> of <math>\{1,2,3, \ldots,n\}</math>.

3 KB (512 words) - 19:17, 18 July 2016
2002 AMC 12B Problems/Problem 20
...] with <math>m\angle XOY = 90^{\circ}</math>. Let <math>M</math> and <math>N</math> be the [[midpoint]]s of legs <math>OX</math> and <math>OY</math>, re ...th>(2x)^2 + (2y)^2 = XY^2 \Longrightarrow XY = \sqrt{4(x^2 + y^2)} = \sqrt{4(169)} = \sqrt{676} = \boxed{\mathrm{(B)}\ 26}</cmath>

3 KB (447 words) - 15:02, 17 August 2023
2002 AMC 12B Problems/Problem 23
label("$A$",A,N); x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3)

4 KB (598 words) - 09:45, 23 January 2024
2002 AMC 12B Problems/Problem 22
For all [[integer]]s <math>n</math> greater than <math>1</math>, define <math>a_n = \frac{1}{\log_n 2002 ...\frac{1}{\frac{\log 2002}{\log n}} = \left(\frac{1}{\log 2002}\right) \log n </math>. Thus

2 KB (312 words) - 18:26, 13 January 2022
2002 AMC 12B Problems/Problem 13
...he smallest possible such perfect square is <math>25</math> when <math>a = 4</math>, and the sum is <math>225 \Rightarrow \mathrm{(B)}</math>. the normal sequence can be described as N^2+N divided by 2.

2 KB (261 words) - 23:34, 18 March 2023
2002 AMC 12B Problems/Problem 12
For how many integers <math>n</math> is <math>\dfrac n{20-n}</math> the [[perfect square|square]] of an integer? \qquad\mathrm{(D)}\ 4

4 KB (579 words) - 05:54, 17 October 2023
2002 AMC 12B Problems/Problem 15
...digit number obtained by removing the leftmost digit is one ninth of <math>N</math>? <math>\mathrm{(A)}\ 4

2 KB (395 words) - 15:50, 3 April 2022
2002 AMC 12B Problems/Problem 3
For how many positive integers <math>n</math> is <math>n^2 - 3n + 2</math> a [[prime]] number? ...umber. The only prime that is even is <math>2</math>, which is when <math>n</math> is <math>3</math> or <math>0</math>. Since <math>0</math> is not a p

1 KB (247 words) - 21:30, 24 August 2022
2002 AMC 12B Problems/Problem 4
{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #4]] and [[2002 AMC 10B Problems|2002 AMC 10B #7]]}} Let <math>n</math> be a positive [[integer]] such that <math>\frac 12 + \frac 13 + \fra

2 KB (321 words) - 10:54, 9 August 2022
Mock AIME 1 Pre 2005 Problems/Problem 1
...ds digits, then tens digits, then units digits. Every one of <math>\{1,2,3,4,5,6,7,8,9\}</math> may appear as the hundreds digit, and there are <math>9 Every one of <math>\{0,1,2,3,4,5,6,7,8,9\}</math> may appear as the tens digit; however, since <math>0</ma

1 KB (194 words) - 13:44, 5 September 2012
Construction
4. Construct a perpendicular bisector. 7. Partition a line segment into <math>n</math> different parts.

3 KB (443 words) - 20:52, 28 August 2014
2007 AMC 8 Problems
...} \frac{2}{5} \qquad \textbf{(B)} \frac{1}{2} \qquad \textbf{(C)} \frac{5}{4} \qquad \textbf{(D)} \frac{5}{3} \qquad \textbf{(E)} \frac{5}{2}</math> ==Problem 4==

12 KB (1,800 words) - 20:01, 8 May 2023
2008 iTest Problems/Problem 97
..., and there are <math>L_{n-1}</math> ways to rearrange the remaining <math>n-1</math> students. Thus, we have the recursion <cmath>L_{n+1} = L_n + L_{n-1}</cmath>

3 KB (497 words) - 13:29, 20 October 2019
2002 AMC 12B Problems/Problem 9
...<math>c=3</math>, and <math>d=4</math>. <math>\frac{a}{d}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}</math> .../math>, and <math>\frac{a}{d}=\frac{a}{k^2a}=\frac{1}{k^2}=\boxed{\frac{1}{4}} \Longrightarrow \mathrm{(C)}</math>.

2 KB (288 words) - 21:42, 11 December 2017
2003 AMC 12B Problems/Problem 22
...h>P</math> and <math>Q</math> be the feet of the perpendiculars from <math>N</math> to <math>\overline{AC}</math> and <math>\overline{BD}</math>, respec pair P = (A+O)/2, Q=(B+O)/2, N=(A+B)/2;

4 KB (551 words) - 14:17, 23 June 2022
2007 AMC 10A Problems/Problem 18
...on <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <ma dotfactor=4;

6 KB (867 words) - 00:17, 20 May 2023
2003 AMC 12B Problems/Problem 20
<math>\mathrm{(A)}\ -4 \qquad\mathrm{(E)}\ 4</math>

2 KB (382 words) - 18:01, 23 November 2020
2004 AMC 10A Problems/Problem 19
draw((-3,11)--(3, 5)^^(-3,10)--(3, 4)); draw((3,-2)--(3,-4)^^(-3,-2)--(-3,-4));

2 KB (220 words) - 14:19, 21 April 2021
2007 AMC 10A Problems/Problem 19
Each of the <math>4</math> unpainted triangles has area <math>\frac{1}8</math>. It can be obser ...c{\frac{\sqrt{2}}{2}+\frac{1}{2}}{\frac{2}{4}-\frac{1}{4}}</math> or <math>4\left(\frac{\sqrt{2}}{2}+\frac{1}{2}\right)</math> which is <math>\boxed{2\s

4 KB (560 words) - 14:57, 3 June 2021
2003 AMC 12B Problems/Problem 24
return 1(x-4)+3; draw(graph(f2,2,4));

7 KB (1,183 words) - 11:47, 15 February 2016
1974 USAMO Problems/Problem 1
...P(a_i) = a_{i+1}</math> and <math>P(a_n) = 1</math>, for <math>1 \le i \le n-1</math>. Then we have <cmath>a_1 - a_2 | a_2 - a_3 | ... | a_n - a_1 | a_1 ...d_1x^n+d_2x^{n-1}+\cdots+d_n.</math> Note that <cmath>b=P(a)=d_1a^n+d_2a^{n-1}+\cdots+d_n</cmath>

7 KB (1,291 words) - 20:30, 27 April 2020
1995 USAMO Problems
...,</math> <math>\, a_n \,</math> is the number obtained by writing <math>\, n \,</math> in base <math>\, p-1 \,</math> and reading the result in base <ma ==Problem 4==

3 KB (540 words) - 13:31, 4 July 2013
2004 AMC 12B Problems/Problem 25
...th>S = \{2^0,2^1,2^2,\ldots ,2^{2003}\}</math> have a first digit of <math>4</math>? ...\mathrm{(B)}</math> elements of <math>S</math> with a first digit of <math>4</math>.

3 KB (383 words) - 20:12, 15 October 2016
2004 AMC 12B Problems/Problem 24
label("$B$",B,N); ...alpha = 0</math>. Since <math>\tan\alpha \neq 0</math>, we have <math>\tan^4 DBE = 1 \Longrightarrow \angle DBE = 45^{\circ}</math>.

2 KB (302 words) - 19:59, 3 July 2013
2004 AMC 12B Problems/Problem 19
\qquad\mathrm{(B)}\ 4\sqrt{5} label("$G$",G,N);

3 KB (520 words) - 19:12, 20 November 2023
1990 USAMO Problems/Problem 5
...\, P \,</math> and <math>\, Q \,</math>. Prove that the points <math>\, M, N, P, Q \,</math> lie on a common circle. ...</math>, <math>Q</math>, <math>A'</math> are cyclic, <math>M</math>, <math>N</math>, <math>P</math>, <math>Q</math> are cyclic by the radical axis theor

3 KB (604 words) - 20:52, 24 October 2018
2007 AMC 10B Problems/Problem 11
label("$A$",A,N); label("$r$",(O+B)/2,N);

5 KB (851 words) - 22:02, 26 July 2021
1994 USAMO Problems
...each positive integer <math>\, n, \,</math> the interval <math>\, [s_n, s_{n + 1}) \,</math> contains at least one perfect square. ==Problem 4==

2 KB (391 words) - 07:58, 19 July 2016
2002 AMC 10A Problems/Problem 25
label("$C$",C,N); label("$D$",D,N);

8 KB (1,308 words) - 07:05, 19 December 2022
2002 AMC 10A Problems/Problem 22
4&8&64\\ 12&4&16\\

3 KB (430 words) - 23:13, 13 September 2023

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