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  • 1987 AIME Problems/Problem 7
    == Problem == ...to 3, it would overlap with case 1). Thus, there are <math>2(3 \cdot 3) = 18</math> cases.
    3 KB (547 words) - 22:54, 4 April 2016
  • 1987 AIME Problems/Problem 2
    == Problem == ...s: <math>\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31</math>. The largest possible distance would be the sum of th
    697 bytes (99 words) - 18:46, 14 February 2014
  • 1988 AIME Problems/Problem 12
    == Problem == <cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath>
    4 KB (727 words) - 23:37, 7 March 2024
  • 1989 AIME Problems/Problem 14
    == Problem == <math>(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i</math>
    2 KB (408 words) - 17:28, 16 September 2023
  • 1989 AIME Problems/Problem 12
    == Problem == ...be a [[tetrahedron]] with <math>AB=41</math>, <math>AC=7</math>, <math>AD=18</math>, <math>BC=36</math>, <math>BD=27</math>, and <math>CD=13</math>, as
    2 KB (376 words) - 13:49, 1 August 2022
  • 1990 AIME Problems/Problem 14
    == Problem == ...e [[Pythagorean Theorem]]. Thus <math>A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}</math>.
    7 KB (1,086 words) - 08:16, 29 July 2023
  • 1990 AIME Problems/Problem 10
    == Problem == The sets <math>A = \{z : z^{18} = 1\}</math> and <math>B = \{w : w^{48} = 1\}</math> are both sets of comp
    3 KB (564 words) - 04:47, 4 August 2023
  • 1990 AIME Problems/Problem 7
    == Problem == <!-- replaced: Image:1990 AIME Problem 7.png by I_like_pie -->
    8 KB (1,319 words) - 11:34, 22 November 2023
  • 2006 AMC 10B Problems/Problem 5
    == Problem == The area of a 2x3 rectangle and a 3x4 rectangle combined is 18, so a 4x4 square is impossible without overlapping. Thus, the next smallest
    1 KB (242 words) - 18:35, 15 August 2023
  • 1991 AIME Problems/Problem 14
    == Problem == ...=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));
    2 KB (284 words) - 03:56, 23 January 2023
  • 2006 AMC 10B Problems/Problem 17
    == Problem == {{AMC10 box|year=2006|ab=B|num-b=16|num-a=18}}
    1 KB (211 words) - 04:32, 4 November 2022
  • 1991 AIME Problems/Problem 8
    == Problem == ...2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since <math>c</math> is the [[m
    2 KB (310 words) - 11:25, 13 June 2023
  • 1991 AIME Problems/Problem 6
    == Problem == ...s take a value of 7. So, <math>\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor \ge ...
    3 KB (447 words) - 17:02, 24 November 2023
  • 1992 AIME Problems/Problem 8
    == Problem == <math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math>
    5 KB (778 words) - 21:36, 3 December 2022
  • 2006 AMC 10B Problems/Problem 19
    == Problem == {{AMC10 box|year=2006|ab=B|num-b=18|num-a=20}}
    5 KB (873 words) - 15:39, 29 May 2023
  • 2006 AMC 10B Problems/Problem 23
    == Problem == ...mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math>
    5 KB (861 words) - 00:53, 25 November 2023
  • 2006 AMC 10B Problems/Problem 24
    == Problem == <math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \ma
    4 KB (558 words) - 14:38, 6 April 2024
  • 2006 AMC 10B Problems/Problem 25
    == Problem == Another way to do the problem is by the process of elimination. The only possible correct choices are the
    5 KB (878 words) - 14:39, 3 December 2023
  • 1993 AIME Problems/Problem 5
    == Problem == ...19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}</cmath>
    2 KB (355 words) - 13:25, 31 December 2018
  • 1994 AIME Problems/Problem 15
    == Problem == ...[circumcenter]]s of <math>\triangle PAB, PBC, PCA</math>. According to the problem statement, the circumcenters of the triangles cannot lie within the interio
    4 KB (717 words) - 22:20, 3 June 2021
  • 1994 AIME Problems/Problem 12
    == Problem == ...argest multiple of <math>6</math> that is <math>\le 19</math> is <math>n = 18</math>, which we can easily verify works, and the answer is <math>\frac{13}
    3 KB (473 words) - 17:06, 1 January 2024
  • 1995 AIME Problems/Problem 14
    == Problem == ...ath>78</math> intersect at a point whose distance from the center is <math>18</math>. The two chords divide the interior of the circle into four regions
    3 KB (484 words) - 13:11, 14 January 2023
  • 1996 AIME Problems/Problem 11
    == Problem == Hence, the answer is <math>\frac{18\pi}{15}+\frac{5\pi}{15}=\frac{23\pi}{15}\cdot\frac{180}{\pi}=\boxed{276}</m
    6 KB (1,022 words) - 20:23, 17 April 2021
  • 1996 AIME Problems/Problem 10
    == Problem == ...math> yields <math>x \equiv 141 \cdot 19 \equiv (140+1)(18+1) \equiv 0+140+18+1 \equiv 159 \pmod{180}</math>.
    4 KB (503 words) - 15:46, 3 August 2022
  • 1997 AIME Problems/Problem 7
    == Problem == \frac{5}{18}t^2 - 110t + 110^2 - 51^2 &\le& 0\\
    4 KB (617 words) - 18:47, 17 July 2022
  • 1998 AIME Problems/Problem 8
    == Problem == ...\frac{F_{7}}{F_{8}} \cdot 1000 = \frac{13}{21} \cdot 1000 = 618.\overline{18}</math>
    2 KB (354 words) - 19:37, 24 September 2023
  • 1998 AIME Problems/Problem 5
    == Problem == Though the problem may appear to be quite daunting, it is actually not that difficult. <math>\
    1 KB (225 words) - 02:20, 16 September 2017
  • 1999 AIME Problems/Problem 15
    == Problem == ...itude from <math>P</math> to <math>\triangle ABC</math>. The crux of this problem is the following lemma.
    7 KB (1,169 words) - 15:28, 13 May 2024
  • 1999 AIME Problems/Problem 3
    == Problem == ...s <math>n=1, 9, 10,</math> or <math>18</math>, and the sum is <math>1+9+10+18=\boxed{38}</math>.
    2 KB (296 words) - 01:18, 29 January 2021
  • 1999 AIME Problems/Problem 2
    == Problem == ...h congruent area is <math>621.</math> Therefore, since the height is <math>18,</math> the sum of the bases of each trapezoid must be <math>69.</math>
    3 KB (423 words) - 11:06, 27 April 2023
  • 1999 AIME Problems/Problem 8
    == Problem == This problem just requires a good diagram and strong 3D visualization.
    3 KB (445 words) - 19:40, 4 July 2013
  • 2000 AIME I Problems/Problem 5
    == Problem == If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way
    7 KB (1,011 words) - 20:09, 4 January 2024
  • 2001 AIME I Problems/Problem 14
    == Problem == ...ive mail and <math>1</math> represent a house that does receive mail. This problem is now asking for the number of <math>19</math>-digit strings of <math>0</m
    13 KB (2,298 words) - 19:46, 9 July 2020
  • 2001 AIME I Problems/Problem 13
    == Problem == ...th>-degree arcs can be represented as <math>x + 20</math>, as given in the problem.
    3 KB (561 words) - 19:25, 27 November 2022
  • 2005 AMC 10B
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2005 AMC 10B Problems/Problem 1]]
    1 KB (165 words) - 12:40, 14 August 2020
  • 2001 AIME I Problems/Problem 11
    == Problem == <cmath>18 c_2 \equiv 2 \pmod{31}</cmath>
    3 KB (493 words) - 13:51, 22 July 2020
  • 2001 AIME I Problems/Problem 10
    == Problem == <math>0,2,2</math> 18
    8 KB (1,187 words) - 02:40, 28 November 2020
  • 2002 AIME I Problems/Problem 13
    == Problem == ...ath>\overline{AD}</math> and <math>\overline{CE}</math> have lengths <math>18</math> and <math>27</math>, respectively, and <math>AB=24</math>. Extend <m
    6 KB (974 words) - 13:01, 29 September 2023
  • 2003 AIME I Problems/Problem 9
    == Problem == ...ath>19 - n</math> choices for both pairs. This gives <math>\sum_{n = 10}^{18} (19 - n)^2 = \sum_{n = 1}^9 n^2 = 285</math> balanced numbers. Thus, ther
    4 KB (696 words) - 11:55, 10 September 2023
  • 2003 AIME I Problems/Problem 8
    == Problem 8== ...d</math>, so only <math>9</math> may work. Hence, the four terms are <math>18,\ 27,\ 36,\ 48</math>, which indeed fits the given conditions. Their sum is
    5 KB (921 words) - 23:21, 22 January 2023
  • 2003 AIME II Problems/Problem 14
    == Problem == ...other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
    9 KB (1,461 words) - 15:09, 18 August 2023
  • 2002 AIME II Problems/Problem 15
    == Problem == <cmath>\frac{1}{m_1^2}a^2-[(18/m_1)+12]a+117=0</cmath>
    7 KB (1,182 words) - 09:56, 7 February 2022
  • 2002 AIME II Problems/Problem 7
    == Problem == ...23, 14, 5, 21, 12, 3, 19, 10, 1, 17, 8, 24, 15, 6, 22, 13, 4, 20, 11, 27, 18, 9, 0,</math> and then it loops.
    3 KB (403 words) - 12:10, 9 September 2023
  • 2002 AIME II Problems/Problem 5
    == Problem == <math>18 > 8 \cdot 3^m</math> which is true for m=0 but fails for higher integers <m
    3 KB (515 words) - 14:46, 14 February 2021
  • 2002 AIME II Problems/Problem 4
    == Problem == [[Image:AIME 2002 II Problem 4.gif]]
    2 KB (268 words) - 07:28, 13 September 2020
  • 2001 AIME II Problems/Problem 12
    == Problem == ...ath>. The total volume added here is then <math>\Delta P_1 = 4 \cdot \frac 18 = \frac 12</math>.
    2 KB (380 words) - 00:28, 5 June 2020
  • 2000 AIME II Problems/Problem 7
    == Problem == ...14!}+\frac 1{6!13!}+\frac 1{7!12!}+\frac 1{8!11!}+\frac 1{9!10!}=\frac N{1!18!}</math></center> find the [[floor function|greatest integer]] that is less
    2 KB (281 words) - 12:09, 5 April 2024
  • 2000 AIME II Problems/Problem 4
    == Problem == ...is <math>(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)</math>. If a number has <math>18 = 2 \cdot 3 \cdot 3</math> factors, then it can have at most <math>3</math>
    2 KB (397 words) - 15:55, 11 May 2022
  • 2005 AMC 10B Problems/Problem 17
    == Problem == {{AMC10 box|year=2005|ab=B|num-b=16|num-a=18}}
    2 KB (324 words) - 15:30, 16 December 2021
  • 2006 AMC 12A Problems/Problem 3
    ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #3]] and [[2006 AMC 10A Problems/Problem 3|2006 AMC 10A #3]]}} == Problem ==
    1 KB (155 words) - 17:30, 16 December 2021
  • 2006 AMC 12A Problems/Problem 6
    ...cate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}} == Problem ==
    3 KB (528 words) - 18:29, 7 May 2024
  • 2006 AMC 12A Problems/Problem 9
    == Problem == ...rm{(A)}\ 10\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 15\qquad\mathrm{(D)}\ 18\qquad\mathrm{(E)}\ 20</math>
    3 KB (429 words) - 18:14, 26 September 2020
  • 2007 iTest
    **[[2007 iTest Problems/Problem 1|Problem 1]] **[[2007 iTest Problems/Problem 2|Problem 2]]
    3 KB (305 words) - 15:10, 5 November 2023
  • 2006 AMC 10B Problems
    == Problem 1 == [[2006 AMC 10B Problems/Problem 1|Solution]]
    14 KB (2,059 words) - 01:17, 30 January 2024
  • 2005 AMC 10B Problems
    == Problem 1 == [[2005 AMC 10B Problems/Problem 1|Solution]]
    12 KB (1,874 words) - 21:20, 23 December 2020
  • 2005 AMC 10A
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AMC 10A Problems/Problem 1]]
    2 KB (182 words) - 18:09, 6 October 2014
  • 2000 AMC 12 Problems/Problem 8
    ==Problem== draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);
    8 KB (1,202 words) - 16:17, 10 May 2024
  • 2000 AMC 12 Problems/Problem 6
    ==Problem== Two different prime numbers between <math>4</math> and <math>18</math> are chosen. When their sum is subtracted from their product, which o
    1 KB (228 words) - 19:31, 29 April 2024
  • 1977 Canadian MO Problems/Problem 3
    == Problem == ...th>a=7,</math> <math>b^2+b+1=7^3.</math> Solving this quadratic, <math>b = 18 </math>.
    713 bytes (114 words) - 01:45, 19 August 2012
  • University of South Carolina High School Math Contest/1993 Exam/Problems
    == Problem 1 == [[University of South Carolina High School Math Contest/1993 Exam/Problem 1|Solution]]
    14 KB (2,102 words) - 22:03, 26 October 2018
  • Mock AIME 1 2006-2007 Problems/Problem 2
    ...an 9 balls. There are <math>{12 + 7 - 1 \choose 7 - 1} = {18 \choose 6} = 18,564</math> ways to place 12 objects into 7 boxes. Of these, 7 place all 12 *[[Mock AIME 1 2006-2007 Problems/Problem 1 | Previous Problem]]
    1 KB (188 words) - 15:53, 3 April 2012
  • Mock AIME 2 2006-2007 Problems/Problem 8
    == Problem == .../math>. The [[divisor | factors]] of 144 are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72 and 144 itself. As both <math>x_1</math> and <math>x_2</ma
    3 KB (470 words) - 00:33, 10 August 2019
  • Mock AIME 2 2006-2007 Problems/Problem 12
    == Problem == Thus <math>[ABCD]=\frac{25\sqrt{143}}{18}\rightarrow\boxed{186}</math>
    2 KB (311 words) - 10:53, 4 April 2012
  • 2005 AMC 10A Problems/Problem 17
    ==Problem== {{AMC10 box|year=2005|ab=A|num-b=16|num-a=18}}
    2 KB (266 words) - 03:36, 16 January 2023
  • Cyclic function
    * [[2006_AMC_12A_Problems/Problem_18 | 2006 AMC 12A Problem 18]]
    363 bytes (53 words) - 14:47, 20 July 2016
  • 2007 iTest Problems
    ===Problem 1=== [[2007 iTest Problems/Problem 1|Solution]]
    30 KB (4,794 words) - 23:00, 8 May 2024
  • 2006 iTest
    *[[2006 iTest Problems/Problem 1|Problem 1]] *[[2006 iTest Problems/Problem 2|Problem 2]]
    3 KB (320 words) - 09:56, 23 April 2024
  • 2005 AMC 10A Problems
    == Problem 1 == [[2005 AMC 10A Problems/Problem 1|Solution]]
    14 KB (2,026 words) - 11:45, 12 July 2021
  • 2004 AMC 10A Problems/Problem 7
    ==Problem== ...op of the <math>2^{\text{nd}}</math>, it will be a layer of <math>3\times6=18</math> oranges, etc.
    1 KB (180 words) - 01:14, 12 April 2022
  • 2004 AMC 10A Problems/Problem 10
    ==Problem== ...h>\left(\frac12\right)^3{3\choose2}\left(\frac12\right)^4{4\choose2}=\frac{18}{128}</math>
    2 KB (221 words) - 21:49, 15 April 2024
  • 2004 AMC 10A Problems/Problem 13
    ==Problem== ...\qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 16 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ } 24 </math>
    1 KB (167 words) - 12:10, 17 August 2021
  • 2003 AMC 10A
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 10A Problems/Problem 1]]
    1 KB (165 words) - 18:48, 6 October 2014
  • 2003 AMC 10A Problems
    ==Problem 1== [[2003 AMC 10A Problems/Problem 1|Solution]]
    13 KB (1,900 words) - 22:27, 6 January 2021
  • 2003 AMC 10A Problems/Problem 14
    == Problem == <math> \mathrm{(A) \ } 12\qquad \mathrm{(B) \ } 15\qquad \mathrm{(C) \ } 18\qquad \mathrm{(D) \ } 21\qquad \mathrm{(E) \ } 24 </math>
    2 KB (336 words) - 15:49, 19 August 2023
  • 2003 AMC 10A Problems/Problem 15
    == Problem == ...ac{17}{50}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{18}{25} </math>
    2 KB (261 words) - 14:34, 17 August 2023
  • 2003 AMC 10A Problems/Problem 17
    == Problem == {{AMC10 box|year=2003|ab=A|num-b=16|num-a=18}}
    2 KB (414 words) - 13:48, 4 April 2024
  • 2004 AMC 10A
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 10A Problems/Problem 1]]
    2 KB (182 words) - 01:29, 7 October 2014
  • 2003 AMC 10A Problems/Problem 23
    == Problem == When there are 5 equilateral triangles in the base, you need <math>18</math> toothpicks in all.
    5 KB (686 words) - 18:01, 28 January 2021
  • 2004 AMC 10A Problems
    ==Problem 1 == [[2004 AMC 10A Problems/Problem 1|Solution]]
    15 KB (2,092 words) - 20:32, 15 April 2024
  • 2004 AMC 12A Problems/Problem 15
    ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #15]] and [[2004 AMC 10A Problems/Problem 17|2004 AMC 10A #17]]}} ==Problem==
    2 KB (309 words) - 22:27, 15 August 2023
  • 2001 AMC 8
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2001 AMC 8 Problems/Problem 1]]
    1 KB (138 words) - 10:26, 22 August 2013
  • 2001 AMC 8 Problems
    ==Problem 1== [[2001 AMC 8 Problems/Problem 1 | Solution]]
    13 KB (1,994 words) - 13:04, 18 February 2024
  • 2003 AMC 12A Problems/Problem 3
    == Problem == ...qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\% </math>
    1 KB (183 words) - 15:36, 19 August 2023
  • 2003 AMC 12A Problems/Problem 15
    == Problem == ...{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}</math>. We have thus solved the problem.
    3 KB (380 words) - 21:53, 19 March 2022
  • 2002 AMC 12A Problems/Problem 1
    == Problem == We expand to get <math>2x^2-8x+3x-12+2x^2-12x+3x-18=0</math> which is <math>4x^2-14x-30=0</math> after combining like terms. Us
    979 bytes (148 words) - 13:06, 8 November 2021
  • Mock AIME 4 2006-2007 Problems/Problem 15
    ==Problem== ...+ \sqrt{4R^2 - 4\cdot24^2}}{2} = \frac{43}{\sqrt 3} + \frac{11}{\sqrt3} = 18\sqrt{3}</math>. (We only take the positive sign because [[angle]] <math>B<
    3 KB (532 words) - 20:29, 31 August 2020
  • 2020 AMC 10A Problems
    ==Problem 1== [[2020 AMC 10A Problems/Problem 1|Solution]]
    13 KB (1,968 words) - 18:32, 29 February 2024
  • 2021 Fall AMC 10B
    **[[2021 Fall AMC 10B Problems/Problem 1|Problem 1]] **[[2021 Fall AMC 10B Problems/Problem 2|Problem 2]]
    2 KB (205 words) - 10:53, 1 December 2021
  • Mock AIME 3 Pre 2005 Problems/Problem 14
    ==Problem== ...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
  • 2007 AIME I Problems/Problem 3
    == Problem == ...he result, we find that <math>(81 + 18bi - b^2)(9 + bi) = \ldots + (9\cdot 18 + 81)bi - b^3i</math>.
    1,003 bytes (163 words) - 15:34, 18 February 2017
  • 2007 AIME I Problems/Problem 5
    == Problem == ...9</math>. Therefore, it must round to <math>c</math> because <math>\frac 5{18}<\frac 12</math> so <math>c</math> is the closest integer. Therefore there
    7 KB (1,076 words) - 00:10, 29 November 2023
  • 2007 AIME I Problems/Problem 10
    == Problem == ...Exclusion]] based on the stipulation that <math>j\ne k</math> to solve the problem:
    13 KB (2,328 words) - 00:12, 29 November 2023
  • 2007 AIME I Problems/Problem 6
    == Problem == ..., <math>(15,18,21,26)</math>, <math>(15,18,21,24,26)</math>, and <math>(15,18,21,24)</math>. That is a total of 6.
    10 KB (1,519 words) - 00:11, 29 November 2023
  • 2007 AIME I Problems/Problem 13
    == Problem == ...sqrt{28}</math>. The area of <math>\triangle{PHI}=\frac{1}{2}\cdot\sqrt{28-18}\cdot6\sqrt{2}=6\sqrt{5}</math>
    7 KB (1,034 words) - 21:56, 22 September 2022
  • 2007 AIME II Problems/Problem 2
    == Problem == ...math> if <math>x + y = n</math>. Thus, there are <math>0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}</math> solutions of <math>(a,b,c)</math>.
    1 KB (228 words) - 08:41, 4 November 2022
  • 2007 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2007 AMC 12A Problems/Problem 1]]
    1 KB (168 words) - 21:50, 6 October 2014
  • 2007 AMC 12A Problems
    ==Problem 1== [[2007 AMC 12A Problems/Problem 1 | Solution]]
    11 KB (1,750 words) - 13:35, 15 April 2022
  • 2007 AMC 12B Problems/Problem 4
    == Problem == ..., and 6 apples cost as much as 4 oranges. How many oranges cost as much as 18 bananas?
    597 bytes (89 words) - 16:33, 15 February 2021
  • 2005 IMO Shortlist Problems/A3
    == Problem == <math> pq + pr + ps = p(9-p) \ge 3(9-3) = 18 </math>.
    3 KB (542 words) - 17:09, 19 December 2018
  • 1951 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1951 AHSME Problems/Problem 1|Problem 1]]
    3 KB (258 words) - 14:25, 20 February 2020

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