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1998 AIME Problems/Problem 9
== Problem == label(rotate(45)*"$M_1-M_2\le m$",((m+60)/2,(60-m)/2),NW,fontsize(9));

4 KB (624 words) - 18:34, 18 February 2018
1998 AIME Problems/Problem 8
== Problem == | 0 || 1 || 2 || 3 || 4 || 5 || 6

2 KB (354 words) - 19:37, 24 September 2023
1998 AIME Problems/Problem 7
== Problem == Note that this is an algebraic bijection, we have simplified the problem and essentially removed the odd condition, so now we can finish with plain

5 KB (684 words) - 11:41, 13 August 2023
1998 AIME Problems/Problem 6
== Problem == <math>0 = RC^2 - 112\cdot847</math>

2 KB (254 words) - 19:38, 4 July 2013
1998 AIME Problems/Problem 5
== Problem == Given that <math>A_k = \frac {k(k - 1)}2\cos\frac {k(k - 1)\pi}2,</math> find <math>|A_{19} + A_{20} + \cdots + A_{98}|.</math>

1 KB (225 words) - 02:20, 16 September 2017
1998 AIME Problems/Problem 4
== Problem == Nine tiles are numbered <math>1, 2, 3, \cdots, 9,</math> respectively. Each of three players randomly selects

5 KB (917 words) - 02:37, 12 December 2022
1998 AIME Problems/Problem 3
== Problem == The graph of <math> y^2 + 2xy + 40|x|= 400</math> partitions the plane into several regions. What

1 KB (198 words) - 20:13, 23 February 2018
1998 AIME Problems/Problem 2
== Problem == [[Image:AIME_1998-2.png|thumb|300px|right]]

6 KB (913 words) - 16:34, 6 August 2020
1998 AIME Problems/Problem 1
== Problem == ...<math>k</math> has only 2s and 3s in its prime factorization, or <math>k = 2^a3^b</math>.

2 KB (289 words) - 22:50, 23 April 2024
2000 AIME I Problems/Problem 3
== Problem == ...e [[relatively prime]] positive integers, the [[coefficient]]s of <math>x^{2}</math> and <math>x^{3}</math> are equal. Find <math>a + b</math>.

679 bytes (98 words) - 00:51, 2 November 2023
2000 AIME II Problems/Problem 14
== Problem == ...on <math>(f_1,f_2,f_3,\ldots,f_m)</math>, meaning that <math>k=1!\cdot f_1+2!\cdot f_2+3!\cdot f_3+\cdots+m!\cdot f_m</math>, where each <math>f_i</math

7 KB (1,131 words) - 14:49, 6 April 2023
IMO Problems and Solutions
* [[1959 IMO Problems/Problem 1 | Problem 1]] proposed by Poland * [[1959 IMO Problems/Problem 2 | Problem 2]] proposed by Constantin Ionescu-Tiu, Romania

35 KB (4,009 words) - 20:25, 21 February 2024
Balkan Mathematical Olympiad
...u>Problem 1: 6 Problem 2: 6.5 Problem 3: 7.5 Problem 4: 8}} *In every competition or test there are four problem usually covering [[geometry]], [[number theory]], [[algebra]], and [[combin

4 KB (578 words) - 00:12, 29 August 2018
Junior Balkan Mathematical Olympiad
...ficulty=4 - 6|breakdown=Problem 1: 4 Problem 2/3: 5 Problem 4: 6}} *In every competition or test there are four problem usually covering [[geometry]], [[number theory]], [[algebra]], and [[combin

3 KB (472 words) - 18:49, 2 February 2019
JBMO Problems and Solutions, with authors
* [[1997 JBMO Problems/Problem 1 | Problem 1]] proposed by Bulgaria * [[1997 JBMO Problems/Problem 2 | Problem 2]] proposed by Cyprus

6 KB (524 words) - 17:17, 17 August 2018
Balkan MO Problems and Solutions, with authors
* [[1984 BMO Problems/Problem 1 | Problem 1]] * [[1984 BMO Problems/Problem 2 | Problem 2]]

4 KB (371 words) - 16:41, 1 January 2024
2007 AMC 12A Problems/Problem 22
...ate|[[2007 AMC 12A Problems|2007 AMC 12A #22]] and [[2007 AMC 10A Problems/Problem 25|2007 AMC 10A #25]]}} == Problem ==

15 KB (2,558 words) - 19:33, 4 February 2024
1998 PMWC Problems
== Problem I1 == 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>

11 KB (1,738 words) - 19:25, 10 March 2015
1997 PMWC Problems/Problem T3
==Problem== ...0 thousands and multiply it by 2 and subtract 5, since we are leaving out 1998 and 1999.

1 KB (202 words) - 14:39, 20 April 2014
1999 AHSME Problems
== Problem 1 == <math>1 - 2 + 3 -4 + \cdots - 98 + 99 = </math>

13 KB (1,945 words) - 18:28, 19 June 2023

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