Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Search results

Create the page "Problem 18" on this wiki! See also the search results found.

Page title matches

Page text matches

  • 2005 AMC 12A Problems/Problem 16
    == Problem == draw((0,0)--(18,0));
    2 KB (307 words) - 15:30, 30 March 2024
  • 2016 AMC 10A Problems/Problem 7
    == Problem == ...[[User:Integralarefun|Integralarefun]] ([[User talk:Integralarefun|talk]]) 18:19, 27 September 2023 (EDT)
    2 KB (268 words) - 18:19, 27 September 2023
  • 1953 AHSME Problems/Problem 13
    == Problem == ...in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:
    878 bytes (143 words) - 20:56, 1 April 2017
  • Simon's Favorite Factoring Trick
    ...ecause this keeps showing up in number theory problems. Let's look at this problem below: ...u through the thinking behind SFFT). Now we use factor pairs to solve this problem.
    7 KB (1,107 words) - 07:35, 26 March 2024
  • Pigeonhole Principle
    ...at <math>n>k</math>. The definition that <math>|N| > |K|</math> (as in our problem) is that there exists a surjective mapping from <math>N</math> to <math>K</ ...select a fifth sock without creating a pair. We may use this to prove the problem:
    11 KB (1,985 words) - 21:03, 5 August 2023
  • Principle of Inclusion-Exclusion
    ...d only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. ==== Problem ====
    9 KB (1,703 words) - 07:25, 24 March 2024
  • Factorial
    * <math>18! = 6402373705728000</math> ([[2007 iTest Problems/Problem 6|Source]])
    10 KB (809 words) - 16:40, 17 March 2024
  • Fermat's Little Theorem
    ...hat is the units digit of <math>k^2 + 2^k</math>? ([[2008 AMC 12A Problems/Problem 15]]) ...27^5=n^5</math>. Find the value of <math>{n}</math>. ([[1989 AIME Problems/Problem 9|1989 AIME, #9]])<br><br>
    16 KB (2,658 words) - 16:02, 8 May 2024
  • Induction
    Other, odder inductions are possible. If a problem asks you to prove something for all integers greater than 3, you can use <m ...ernational Mathematics Olympiad | IMO]]. A good example of an upper-level problem that can be solved with induction is [http://www.artofproblemsolving.com/Fo
    5 KB (768 words) - 20:45, 1 September 2022
  • Complex number
    *[[2007 AMC 12A Problems/Problem 18]] *[[1984 AIME Problems/Problem 8|1984 AIME Problem 8]]
    5 KB (860 words) - 15:36, 10 December 2023
  • Without loss of generality
    * [[2017_USAJMO_Problems/Problem_3 | 2017 USAJMO Problem 3]] * [[2016_AMC_12A_Problems/Problem_17 | 2016 AMC 12A Problem 17]] (See Solution 2)
    2 KB (280 words) - 15:30, 22 February 2024
  • 2000 AMC 12 Problems/Problem 1
    ==Problem== ...and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 3
    2 KB (276 words) - 05:25, 9 December 2023
  • Functional equation
    ...ns can significantly help in solving functional identities. Consider this problem: === Problem Examples ===
    2 KB (361 words) - 14:40, 24 August 2021
  • 2006 AMC 10B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10B Problems/Problem 1]]
    2 KB (182 words) - 21:57, 23 January 2021
  • 2006 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12B Problems/Problem 1 | Problem 1]]
    2 KB (210 words) - 00:06, 7 October 2014
  • Modular arithmetic/Introduction
    * <math>91 \not\equiv 18 \pmod{6}</math> because <math>\frac{91 - 18}{6} = \frac{73}{6}</math>, which is not an integer. === Sample Problem ===
    15 KB (2,396 words) - 20:24, 21 February 2024
  • Modular arithmetic/Intermediate
    ...ngruence: <math>x \equiv 5 \pmod{21}</math>, and <math>x \equiv -3 \equiv 18 \pmod{21}</math>. (Note that two values of <math>x</math> that are congrue ...following topics expand on the flexible nature of modular arithmetic as a problem solving tool:
    14 KB (2,317 words) - 19:01, 29 October 2021
  • 2006 AMC 10A
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 10A Problems/Problem 1]]
    2 KB (180 words) - 18:06, 6 October 2014
  • 2006 AIME I Problems
    == Problem 1 == ...<math> \overline{AC} </math> is perpendicular to <math> \overline{CD}, AB=18, BC=21, </math> and <math> CD=14. </math> Find the perimeter of <math> ABCD
    7 KB (1,173 words) - 03:31, 4 January 2023
  • 2006 AIME I Problems/Problem 5
    == Problem == <math>c=18</math>
    3 KB (439 words) - 18:24, 10 March 2015
  • 2006 AIME I Problems/Problem 1
    == Problem == ...ne{AC}</math> is [[perpendicular]] to <math>\overline{CD}</math>, <math>AB=18</math>, <math>BC=21</math>, and <math>CD=14</math>. Find the [[perimeter]]
    2 KB (217 words) - 21:43, 2 February 2014
  • 2021 JMC 10
    ==Problem 1== [[2021 JMC 10 Problems/Problem 1|Solution]]
    12 KB (1,784 words) - 16:49, 1 April 2021
  • 2006 AMC 12B Problems
    == Problem 1 == [[2006 AMC 12B Problems/Problem 1|Solution]]
    13 KB (2,058 words) - 12:36, 4 July 2023
  • 2006 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AMC 12A Problems/Problem 1]]
    1 KB (168 words) - 21:51, 6 October 2014
  • 2004 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 12A Problems/Problem 1]]
    2 KB (186 words) - 17:35, 16 December 2019
  • 2004 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 12B Problems/Problem 1]]
    2 KB (181 words) - 21:40, 6 October 2014
  • 2005 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2005 AMC 12A Problems/Problem 1|Problem 1]]
    2 KB (202 words) - 21:30, 6 October 2014
  • 2005 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 AMC 12B Problems/Problem 1|Problem 1]]
    2 KB (206 words) - 23:23, 21 June 2021
  • 2000 AMC 12
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AMC 12 Problems/Problem 1]]
    1 KB (126 words) - 13:28, 20 February 2020
  • 2001 AMC 12
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AMC 12 Problems/Problem 1]]
    1 KB (127 words) - 21:36, 6 October 2014
  • 2002 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 12A Problems/Problem 1]]
    1 KB (158 words) - 21:33, 6 October 2014
  • 2003 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 12A Problems/Problem 1]]
    1 KB (162 words) - 21:52, 6 October 2014
  • 2002 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 12B Problems/Problem 1]]
    1 KB (154 words) - 00:32, 7 October 2014
  • 2003 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 12B Problems/Problem 1]]
    1 KB (160 words) - 20:46, 1 February 2016
  • 2006 AMC 12A Problems
    == Problem 1 == [[2006 AMC 12A Problems/Problem 1|Solution]]
    15 KB (2,223 words) - 13:43, 28 December 2020
  • 2005 AMC 12A Problems
    == Problem 1 == [[2005 AMC 12A Problems/Problem 1|Solution]]
    13 KB (1,971 words) - 13:03, 19 February 2020
  • 2004 AMC 12A Problems
    == Problem 1 == [[2004 AMC 12A Problems/Problem 1|Solution]]
    13 KB (1,953 words) - 00:31, 26 January 2023
  • 2003 AMC 12A Problems
    == Problem 1 == [[2003 AMC 12A Problems/Problem 1|Solution]]
    13 KB (1,955 words) - 21:06, 19 August 2023
  • 2002 AMC 12A Problems
    == Problem 1 == [[2002 AMC 12A Problems/Problem 1|Solution]]
    12 KB (1,792 words) - 13:06, 19 February 2020
  • 2000 AMC 12 Problems
    == Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]
    13 KB (1,948 words) - 12:26, 1 April 2022
  • 2001 AMC 12 Problems
    == Problem 1 == [[2001 AMC 12 Problems/Problem 1|Solution]]
    13 KB (1,957 words) - 12:53, 24 January 2024
  • 2002 AMC 12B Problems
    == Problem 1 == [[2002 AMC 12B Problems/Problem 1|Solution]]
    10 KB (1,547 words) - 04:20, 9 October 2022
  • 2003 AMC 12B Problems
    == Problem 1 == <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>
    13 KB (1,987 words) - 18:53, 10 December 2022
  • 2004 AMC 12B Problems
    == Problem 1 == [[2004 AMC 12B Problems/Problem 1|Solution]]
    13 KB (2,049 words) - 13:03, 19 February 2020
  • 2005 AMC 12B Problems
    == Problem 1 == [[2005 AMC 12B Problems/Problem 1|Solution]]
    12 KB (1,781 words) - 12:38, 14 July 2022
  • 2006 AMC 12B Problems/Problem 15
    == Problem == // from amc10 problem series
    3 KB (458 words) - 16:40, 6 October 2019
  • 2006 AMC 12B Problems/Problem 17
    == Problem == \mathrm{(B)}\ \frac 18
    1 KB (188 words) - 22:10, 9 June 2016
  • 2006 AMC 12B Problems/Problem 19
    == Problem == {{AMC12 box|year=2006|ab=B|num-b=18|num-a=20}}
    4 KB (696 words) - 09:47, 10 August 2015
  • 2006 AMC 12B Problems/Problem 20
    == Problem == \mathrm{(A)}\ \frac 18
    3 KB (485 words) - 14:09, 21 May 2021
  • 2006 AMC 12B Problems/Problem 24
    == Problem == &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\
    3 KB (563 words) - 22:45, 24 October 2021
  • 2006 AMC 12A Problems/Problem 17
    == Problem == {{AMC12 box|year=2006|ab=A|num-b=16|num-a=18}}
    6 KB (958 words) - 23:29, 28 September 2023
  • 2006 AMC 12A Problems/Problem 19
    == Problem == {{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
    ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}} == Problem ==
    5 KB (908 words) - 19:23, 22 September 2022
  • 2005 AMC 12B Problems/Problem 4
    == Problem == ...zes 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 9
    == Problem == {{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 17
    == Problem == {{AMC12 box|year=2005|ab=B|num-b=16|num-a=18}}
    1 KB (159 words) - 21:18, 21 December 2020
  • 2005 AMC 12B Problems/Problem 19
    == Problem == {{AMC12 box|year=2005|ab=B|num-b=18|num-a=20}}
    2 KB (283 words) - 20:02, 24 December 2020
  • 2005 AMC 12B Problems/Problem 24
    == Problem == ...ad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}} </math>
    4 KB (761 words) - 09:10, 1 August 2023
  • 2006 AMC 10A Problems
    ==Problem 1== [[2006 AMC 10A Problems/Problem 1|Solution]]
    13 KB (2,028 words) - 16:32, 22 March 2022
  • 2006 AMC 12A Problems/Problem 12
    ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #12]] and [[2006 AMC 10A Problems/Problem 14|2006 AMC 10A #14]]}} == Problem ==
    2 KB (292 words) - 11:56, 17 December 2021
  • 2006 AMC 10A Problems/Problem 17
    == Problem == {{AMC10 box|year=2006|num-b=16|num-a=18|ab=A}}
    6 KB (1,066 words) - 00:21, 2 February 2023
  • 2006 AMC 10A Problems/Problem 19
    == Problem == {{AMC10 box|year=2006|ab=A|num-b=18|num-a=20}}
    2 KB (259 words) - 03:10, 22 June 2023
  • 1991 AJHSME Problems
    ==Problem 1== [[1991 AJHSME Problems/Problem 1|Solution]]
    17 KB (2,246 words) - 13:37, 19 February 2020
  • 2022 USAJMO Problems/Problem 1
    ==Problem== ...metric sequence to be <math>\{ g, gr, gr^2, \dots \}</math>. Rewriting the problem based on our new terminology, we want to find all positive integers <math>m
    4 KB (792 words) - 00:29, 13 April 2024
  • 2005 AIME II Problems/Problem 4
    == Problem == ...sitive integer]]s that are divisors of at least one of <math> 10^{10},15^7,18^{11}. </math>
    3 KB (377 words) - 18:36, 1 January 2024
  • 2005 AIME II Problems
    == Problem 1 == [[2005 AIME II Problems/Problem 1|Solution]]
    7 KB (1,119 words) - 21:12, 28 February 2020
  • 2005 AIME I Problems/Problem 4
    == Problem == 2 & 18 & no\\ \hline
    8 KB (1,248 words) - 11:43, 16 August 2022
  • 2005 AIME I Problems/Problem 9
    == Problem == ...^{34}\cdot 3^{18}}</math> and so the answer is <math>2 + 3 + 5 + 12 + 34 + 18 = \boxed{074}</math>.
    4 KB (600 words) - 21:44, 20 November 2023
  • 2005 AIME I Problems/Problem 13
    == Problem == ...next, giving <math>2</math> ways. This totals <math>6 + 3\cdot 2\cdot 2 = 18</math> ways.
    5 KB (897 words) - 00:21, 29 July 2022
  • 2005 AIME I Problems/Problem 15
    == Problem == ...eft with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 =
    5 KB (906 words) - 23:15, 6 January 2024
  • 2005 AIME II Problems/Problem 14
    == Problem == ...ac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the
    13 KB (2,129 words) - 18:56, 1 January 2024
  • 2004 AIME I Problems/Problem 15
    == Problem == ...ions in a string format, starting with the operation that sends <math>f(x_{18}) = x_{19}</math> and so forth downwards. There are <math>2^9</math> ways t
    9 KB (1,491 words) - 01:23, 26 December 2022
  • 2004 AIME I Problems/Problem 13
    == Problem == P(x) &= \left(\frac{x^{18} - 1}{x - 1}\right)^2 - x^{17} = \frac{x^{36} - 2x^{18} + 1}{x^2 - 2x + 1} - x^{17}\\ &= \frac{x^{36} - x^{19} - x^{17} + 1}{(x -
    2 KB (298 words) - 20:02, 4 July 2013
  • 2004 AIME II Problems/Problem 15
    == Problem == Now, consider the strip of length <math>1024</math>. The problem asks for <math>s_{941, 10}</math>. We can derive some useful recurrences f
    6 KB (899 words) - 20:58, 12 May 2022
  • 2004 AIME II Problems/Problem 14
    == Problem == ...n't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that <math>n \equiv 1
    11 KB (1,857 words) - 21:55, 19 June 2023
  • 2004 AIME II Problems/Problem 10
    == Problem == <math>= 7 + 6 \cdot 3 \cdot 7 = 7 + 18 \cdot 7 = 19 \cdot 7 = 133</math>.
    8 KB (1,283 words) - 19:19, 8 May 2024
  • 2004 AIME II Problems/Problem 5
    ==Problem== From the initial problem statement, we have <math>1000w\cdot\frac{1}{4}t=\frac{1}{4}</math>.
    4 KB (592 words) - 19:02, 26 September 2020
  • 2004 AIME II Problems/Problem 2
    == Problem == ...wo red candies after Terry chooses two red candies is <math>\frac{7\cdot8}{18\cdot17} = \frac{28}{153}</math>. So the probability that they both pick tw
    2 KB (330 words) - 13:42, 1 January 2015
  • 2019 AIME I Problems/Problem 2
    ==Problem== ...possible combinations of <math>B</math> and <math>J</math>, and <math>1 + 18 \cdot 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{3
    5 KB (830 words) - 22:15, 28 December 2023
  • 1989 AIME Problems
    == Problem 1 == [[1989 AIME Problems/Problem 1|Solution]]
    7 KB (1,045 words) - 20:47, 14 December 2023
  • 1990 AIME Problems
    == Problem 1 == [[1990 AIME Problems/Problem 1|Solution]]
    6 KB (870 words) - 10:14, 19 June 2021
  • 1995 AIME Problems
    == Problem 1 == [[Image:AIME 1995 Problem 1.png]]
    6 KB (1,000 words) - 00:25, 27 March 2024
  • 2002 AIME I Problems
    == Problem 1 == [[2002 AIME I Problems/Problem 1|Solution]]
    8 KB (1,374 words) - 21:09, 27 July 2023
  • 2000 AIME II Problems
    == Problem 1 == [[2000 AIME II Problems/Problem 1|Solution]]
    6 KB (947 words) - 21:11, 19 February 2019
  • 1983 AIME Problems/Problem 3
    == Problem == ...th of the roots of this equation are real, since its discriminant is <math>18^2 - 4 \cdot 1 \cdot 20 = 244</math>, which is positive. Thus by [[Vieta's f
    3 KB (532 words) - 05:18, 21 July 2022
  • 1983 AIME Problems/Problem 9
    == Problem == ...h>9z^2 - yz + 4 = 0</math> to see that <math>z = \frac{y\pm\sqrt{y^2-144}}{18}</math>. We know that <math>y</math> must be an integer and as small as it
    4 KB (722 words) - 20:25, 14 January 2023
  • 1983 AIME Problems/Problem 12
    == Problem == ...th> and <math>y</math> are different digits, <math>1+0=1 \leq x+y \leq 9+9=18</math>, so the only possible multiple of <math>11</math> is <math>11</math>
    2 KB (412 words) - 18:23, 1 January 2024
  • 1983 AIME Problems/Problem 14
    == Problem == ...hat circle bisects the chord, so <math>QM = MP = PN = NR</math>, since the problem told us <math>QP = PR</math>.
    13 KB (2,149 words) - 18:44, 5 February 2024
  • 1984 AIME Problems/Problem 10
    == Problem == ...achieved with TWO or MORE methods. (Note: This is actually the exact same problem as the original, just reworded differently and now we are adding the score.
    7 KB (1,163 words) - 23:53, 28 March 2022
  • 1984 AIME Problems/Problem 14
    == Problem == If <math>x \ge 18</math> and is <math>0 \bmod{6}</math>, <math>x</math> can be expressed as <
    8 KB (1,346 words) - 01:16, 9 January 2024
  • 1985 AIME Problems/Problem 15
    == Problem == [[Image:AIME 1985 Problem 15.png]]
    2 KB (245 words) - 22:44, 4 March 2024
  • 1985 AIME Problems/Problem 11
    == Problem == ...= Y</math>). Finding the optimal location for <math>X</math> is a classic problem: for any path from <math>F_1</math> to <math>X</math> and then back to <mat
    5 KB (932 words) - 17:00, 1 September 2020
  • 1985 AIME Problems/Problem 10
    == Problem == *<math>1</math>: Easily possible, for example try plugging in <math>x =\frac 18</math>.
    12 KB (1,859 words) - 18:16, 28 March 2022
  • 1985 AIME Problems/Problem 9
    == Problem == D(anglemark(A2,O,A1,5)); D(anglemark(A3,O,A2,5)); D(anglemark(A2,A3,A1,18));
    5 KB (763 words) - 16:20, 28 September 2019
  • 1986 AIME Problems/Problem 11
    == Problem == ...1</math>). By the [[Binomial Theorem]], this is <math>(-1) \cdot (-1)^{15}{18 \choose 3} = \boxed{816}</math>.
    6 KB (872 words) - 16:51, 9 June 2023
  • 1986 AIME Problems/Problem 10
    == Problem == ...of the multiples of <math>222</math> from <math>15\cdot222</math> to <math>18\cdot222</math> by subtracting <math>N</math> from each <math>222(a+b+c)</ma
    3 KB (565 words) - 16:51, 1 October 2023
  • 1986 AIME Problems/Problem 9
    == Problem == ...10}\cdot450=\frac{15}{17}d</math> and <math>EC=\frac{d}{425}\cdot450=\frac{18}{17}d</math>. Since <math>FD'=BC-EE'</math>, we have <math>900-\frac{33}{17
    11 KB (1,850 words) - 18:07, 11 October 2023
  • 1986 AIME Problems/Problem 4
    == Problem == x_3-x_1&=18\\
    1 KB (212 words) - 16:25, 17 November 2019
  • 1987 AIME Problems/Problem 14
    == Problem == &= (x(x-6) + 18)(x(x+6)+18),
    7 KB (965 words) - 10:42, 12 April 2024
  • 1987 AIME Problems/Problem 12
    == Problem == ...than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>. <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.
    4 KB (673 words) - 19:48, 28 December 2023
  • 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 (1,107 words) - 20:34, 27 January 2023
  • 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);
    7 KB (988 words) - 15:14, 10 April 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
  • Chicken McNugget Theorem
    ...n McNugget Theorem''' (or '''Postage Stamp Problem''' or '''Frobenius Coin Problem''') states that for any two [[relatively prime]] [[positive integer]]s <mat ...t Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest
    17 KB (2,748 words) - 19:22, 24 February 2024
  • 2007 AMC 12A Problems/Problem 10
    ...ate|[[2007 AMC 12A Problems|2007 AMC 12A #10]] and [[2007 AMC 10A Problems/Problem 14|2007 AMC 10A #14]]}} ==Problem==
    2 KB (231 words) - 14:02, 3 June 2021
  • 2007 AMC 12A Problems/Problem 12
    ...ate|[[2007 AMC 12A Problems|2007 AMC 12A #12]] and [[2007 AMC 10A Problems/Problem 16|2007 AMC 10A #16]]}} ==Problem==
    3 KB (445 words) - 08:59, 24 March 2023
  • Parity
    <b>Problem: </b> ...e add an even and odd. We can use complementary counting to help solve the problem. There are a total of <math>90</math> possibilities since Jack can choose <
    4 KB (694 words) - 22:00, 12 January 2024
  • 2007 AMC 12A Problems/Problem 13
    ==Problem== ...mouse is at <math>(4,-2)</math> and is running up the [[line]] <math>y=-5x+18</math>. At the point <math>(a,b)</math> the mouse starts getting farther fr
    2 KB (387 words) - 18:20, 27 November 2023
  • 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
  • 2007 AMC 12A Problems/Problem 17
    == Problem == {{AMC12 box|year=2007|ab=A|num-b=16|num-a=18}}
    1,022 bytes (153 words) - 14:56, 7 August 2017
  • 2007 AMC 12A Problems/Problem 19
    == Problem == ...t 2}{223} = 18</math>. Thus <math>A</math> lies on the lines <math>y = \pm 18 \quad \mathrm{(1)}</math>.
    4 KB (565 words) - 17:01, 2 April 2023
  • 2005 AMC 12A Problems/Problem 19
    == Problem == We can simply apply casework to this problem.
    4 KB (536 words) - 21:18, 22 May 2023
  • 2005 AMC 12A Problems/Problem 17
    == Problem == {{AMC12 box|year=2005|num-b=16|num-a=18|ab=A}}
    2 KB (215 words) - 13:56, 19 January 2021
  • 2006 AIME II Problems/Problem 3
    == Problem == ...t\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \fra
    4 KB (562 words) - 18:37, 30 October 2020
  • 2006 AIME II Problems/Problem 7
    == Problem == ...es. Thus, thus total possibilities for <math>(a,b)</math> is <math>576+144+18=738</math>.
    7 KB (1,114 words) - 03:41, 12 September 2021
  • 2005 PMWC Problems
    == Problem I1 == [[2005 PMWC Problems/Problem I1|Solution]]
    9 KB (1,449 words) - 20:49, 2 October 2020
  • 1998 PMWC Problems
    == Problem I1 == [[1998 PMWC Problems/Problem I1|Solution]]
    11 KB (1,738 words) - 19:25, 10 March 2015
  • 1999 PMWC Problems
    == Problem I1 == [[1999 PMWC Problems/Problem I1|Solution]]
    6 KB (703 words) - 21:21, 21 April 2014
  • 1997 PMWC Problems/Problem I10
    ==Problem== <math>4</math> | <math>18</math>
    1 KB (203 words) - 19:14, 7 April 2016
  • 2006 Cyprus MO/Lyceum/Problem 17
    ==Problem== {{CYMO box|year=2006|l=Lyceum|num-b=16|num-a=18}}
    789 bytes (123 words) - 22:00, 30 November 2015
  • 2006 Cyprus MO/Lyceum/Problem 19
    ==Problem== {{CYMO box|year=2006|l=Lyceum|num-b=18|num-a=20}}
    1 KB (214 words) - 23:44, 22 December 2016
  • 2004 AMC 10A Problems/Problem 21
    ==Problem== fill((-30,0)..(-24,18)--(0,0)--(-24,-18)..cycle,gray(0.7));
    3 KB (476 words) - 03:50, 23 January 2023
  • 2004 AMC 12A Problems/Problem 18
    ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #18]] and [[2004 AMC 10A Problems/Problem 22|2004 AMC 10A #22]]}} == Problem ==
    5 KB (738 words) - 13:11, 27 March 2023
  • 2007 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2007 AMC 12B Problems/Problem 1]]
    1 KB (168 words) - 21:22, 6 October 2014
  • 2007 AMC 12B Problems
    ==Problem 1== [[2007 AMC 12B Problems/Problem 1 | Solution]]
    12 KB (1,814 words) - 12:58, 19 February 2020
  • 1999 AHSME Problems
    == Problem 1 == [[1999 AHSME Problems/Problem 1|Solution]]
    13 KB (1,945 words) - 18:28, 19 June 2023
  • 2004 AMC 12A Problems/Problem 14
    ...ms|2004 AMC 12A #14]] and [[2004 AMC 10A Problems/Problem 18|2004 AMC 10A #18]]}} == Problem ==
    4 KB (689 words) - 03:35, 16 January 2023
  • 2004 AMC 12A Problems/Problem 17
    ...ate|[[2004 AMC 12A Problems|2004 AMC 12A #17]] and [[2004 AMC 10A Problems/Problem 24|2004 AMC 10A #24]]}} == Problem ==
    2 KB (233 words) - 08:14, 6 September 2021
  • 2004 AMC 12A Problems/Problem 20
    == Problem == ...>c < \frac 12</math>. Thus the chance is <math>\frac{\frac{1}{4}}2 = \frac 18</math>.
    3 KB (552 words) - 23:26, 28 December 2020
  • 2005 iTest
    *[[2005 iTest Problems/Problem 1|Problem 1]] *[[2005 iTest Problems/Problem 2|Problem 2]]
    3 KB (283 words) - 02:37, 24 January 2024
  • 2003 Polish Mathematical Olympiad Problems/Problem 3
    == Problem == ...2^1</math>, <math>(6-2)2! = 8 > 4 = 2^2</math>, and <math>(6-3) \cdot 3! = 18 > 8 = 2^3</math>. For the other integers less less than or equal to six, w
    5 KB (919 words) - 23:29, 20 January 2016
  • 2000 AMC 12 Problems/Problem 16
    == Problem == ...the first row is numbered <math>1,2,\ldots,17</math>, the second row <math>18,19,\ldots,34</math>, and so on down the board. If the board is renumbered s
    2 KB (310 words) - 11:28, 3 August 2021
  • 2000 AMC 12 Problems/Problem 17
    == Problem == ...trig to guess and check: the only trig facts we need to know to finish the problem is:
    6 KB (979 words) - 12:50, 17 July 2022
  • 2000 AMC 12 Problems/Problem 18
    {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #18]] and [[2000 AMC 10 Problems|2000 AMC 10 #25]]}} == Problem ==
    2 KB (382 words) - 19:20, 12 May 2023
  • 2000 AMC 12 Problems/Problem 19
    == Problem == {{AMC12 box|year=2000|num-b=18|num-a=20}}
    3 KB (547 words) - 17:37, 17 February 2024
  • 2000 AMC 12 Problems/Problem 21
    == Problem == {{AMC10 box|year=2000|num-b=18|num-a=20}}
    4 KB (666 words) - 09:22, 28 October 2022
  • 2000 AMC 12 Problems/Problem 24
    == Problem == ...s <math>36/{\pi}-r</math>. By the Pythagorean Theorem, you get <math>r^2+(18/{\pi})^2=(36/{\pi}-r)^2 \implies r=27/2{\pi}</math>. Finally, we see that t
    2 KB (263 words) - 19:59, 18 April 2024
  • 2007 AMC 10A
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2007 AMC 10A Problems/Problem 1]]
    2 KB (182 words) - 03:21, 31 December 2019
  • 2007 AMC 10A Problems/Problem 17
    == Problem == {{AMC10 box|year=2007|ab=A|num-b=16|num-a=18}}
    1 KB (201 words) - 08:04, 11 February 2023
  • 2007 AMC 10A Problems
    == Problem 1 == [[2007 AMC 10A Problems/Problem 1|Solution]]
    13 KB (2,058 words) - 17:54, 29 March 2024
  • 2007 AMC 10A Problems/Problem 11
    == Problem == <math>\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 24</math>
    1 KB (195 words) - 20:44, 21 January 2024
  • 1995 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1995 AHSME Problems/Problem 1|Problem 1]]
    2 KB (174 words) - 00:30, 2 October 2014
  • 1995 AHSME Problems
    == Problem 1 == [[1995 AHSME Problems/Problem 1|Solution]]
    17 KB (2,387 words) - 22:44, 26 May 2021
  • 1995 AHSME Problems/Problem 25
    == Problem == ...of five positive integers has [[mean]] <math>12</math> and [[range]] <math>18</math>. The [[mode]] and [[median]] are both <math>8</math>. How many diffe
    1 KB (180 words) - 20:59, 10 February 2019
  • 1995 AHSME Problems/Problem 11
    ==Problem== <math> \mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \
    2 KB (233 words) - 10:37, 30 March 2023
  • 1997 PMWC Problems
    == Problem I1 == [[1997 PMWC Problems/Problem I1|Solution]]
    15 KB (2,057 words) - 19:13, 10 March 2015
  • 1997 PMWC Problems/Problem I13
    == Problem == ...ath>70 \text{ km/h}</math>. It traveled <math>3</math> rounds within <math>18</math> hours. What is the distance between <math>A</math> and <math>B</math
    857 bytes (149 words) - 13:13, 21 January 2019
  • 1951 AHSME Problems
    == Problem 1 == [[1951 AHSME Problems/Problem 1|Solution]]
    23 KB (3,641 words) - 22:23, 3 November 2023
  • 1951 AHSME Problems/Problem 2
    == Problem == ...})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}</math>
    743 bytes (121 words) - 12:19, 5 July 2013
  • Algebraic manipulation
    x^2 + 18 &= 43 \\ ...e right side to maintain equality. The right hand side becomes <math>43 - 18 = 25</math>.
    4 KB (562 words) - 18:49, 8 November 2020
  • 1959 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1959 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:19, 20 February 2020
  • 1959 AHSME Problems
    == Problem 1== [[1959 AHSME Problems/Problem 1|Solution]]
    22 KB (3,345 words) - 20:12, 15 February 2023
  • 1966 AHSME Problems
    == Problem 1 == <math>\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)}
    19 KB (3,159 words) - 22:10, 11 March 2024
  • 1966 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1966 AHSME Problems/Problem 1|Problem 1]]
    2 KB (217 words) - 14:15, 20 February 2020
  • 1966 AHSME Problems/Problem 1
    == Problem == <math>\text{(A)} \ \frac 18 \qquad \text{(B)} \ \frac 73 \qquad \text{(C)} \ \frac78 \qquad \text{(D)}
    712 bytes (99 words) - 12:39, 5 July 2013
  • 2002 AMC 12B Problems/Problem 19
    == Problem == {{AMC12 box|year=2002|ab=B|num-b=18|num-a=20}}
    852 bytes (119 words) - 10:22, 4 July 2013
  • 2002 AMC 12B Problems/Problem 14
    ...12B Problems|2002 AMC 12B #14]] and [[2002 AMC 10B Problems|2002 AMC 10B #18]]}} == Problem ==
    2 KB (282 words) - 14:04, 12 July 2021
  • 2002 AMC 12B Problems/Problem 18
    == Problem == [[Image:2002_12B_AMC-18.png]]
    3 KB (376 words) - 19:16, 20 August 2019
  • 2002 AMC 12B Problems/Problem 13
    == Problem == The sum of <math>18</math> consecutive positive integers is a [[perfect square]]. The smallest
    2 KB (261 words) - 23:34, 18 March 2023
  • 2002 AMC 12B Problems/Problem 12
    == Problem == ...thrm{(D)}\ 4}</math> solutions (respectively yielding <math>n = 0, 10, 16, 18</math>).
    4 KB (579 words) - 05:54, 17 October 2023
  • 2007 AMC 8 Problems
    ==Problem 1== [[2007 AMC 8 Problems/Problem 1|Solution]]
    12 KB (1,800 words) - 20:01, 8 May 2023
  • 2003 AMC 12B Problems/Problem 17
    == Problem == We rewrite the logarithms in the problem. <cmath>\log(x) + 3\log(y) = 1</cmath> <cmath>2\log(x) + \log(y) = 1</cmath
    2 KB (329 words) - 13:49, 4 April 2024
  • 2003 AMC 12B Problems/Problem 1
    ==Problem== <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>
    970 bytes (134 words) - 00:09, 14 September 2015
  • 2004 AMC 10A Problems/Problem 19
    ==Problem== {{AMC10 box|year=2004|ab=A|num-b=18|num-a=20}}
    2 KB (220 words) - 14:19, 21 April 2021
  • 2007 AMC 10A Problems/Problem 19
    == Problem == {{AMC10 box|year=2007|ab=A|num-b=18|num-a=20}}
    4 KB (560 words) - 14:57, 3 June 2021
  • 2003 AMC 12B Problems/Problem 19
    == Problem == ...elements. Hence the answer is <math>\frac{3 \cdot 3!}{4 \cdot 4!} = \frac {18}{96} = \frac{3}{16}</math>, and <math>a+b=19 \Rightarrow \mathrm{(E)}</math
    2 KB (320 words) - 22:59, 5 May 2024
  • 1998 AHSME Problems/Problem 17
    == Problem == {{AHSME box|year=1998|num-b=16|num-a=18}}
    656 bytes (94 words) - 22:16, 28 March 2024
  • 2004 AMC 12B Problems/Problem 24
    == Problem == \qquad\mathrm{(E)}\ 18</math>
    2 KB (302 words) - 19:59, 3 July 2013
  • 2004 AMC 12B Problems/Problem 19
    == Problem == A truncated cone has horizontal bases with radii <math>18</math> and <math>2</math>. A sphere is tangent to the top, bottom, and late
    3 KB (520 words) - 19:12, 20 November 2023
  • 2004 AMC 12B Problems/Problem 17
    == Problem == {{AMC12 box|year=2004|ab=B|num-b=16|num-a=18}}
    936 bytes (144 words) - 02:47, 13 July 2021
  • 2008 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 AMC 12A Problems/Problem 1|Problem 1]]
    2 KB (193 words) - 21:49, 6 October 2014
  • 2008 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 AMC 12B Problems/Problem 1|Problem 1]]
    2 KB (193 words) - 00:11, 7 October 2014
  • 2008 AMC 10A
    ...The second link contains the answer key. The rest contain each individual problem and its solution. **[[2008 AMC 10A Problems/Problem 1|Problem 1]]
    2 KB (195 words) - 18:08, 28 June 2021
  • 2008 AMC 10B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2008 AMC 10B Problems/Problem 1|Problem 1]]
    2 KB (188 words) - 18:44, 6 October 2014
  • 2002 AMC 10A Problems/Problem 22
    == Problem == <math>\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>
    3 KB (430 words) - 23:13, 13 September 2023
  • 2008 AMC 10A Problems
    ==Problem 1== [[2008 AMC 10A Problems/Problem 1|Solution]]
    14 KB (2,138 words) - 15:08, 18 February 2023
  • 2008 AMC 12A Problems
    ==Problem 1== [[2008 AMC 12A Problems/Problem 1|Solution]]
    13 KB (2,025 words) - 13:56, 2 February 2021
  • 2008 AMC 12A Problems/Problem 19
    ==Problem== {{AMC12 box|year=2008|ab=A|num-b=18|num-a=20}}
    5 KB (895 words) - 22:54, 9 January 2021
  • 2008 AMC 12A Problems/Problem 17
    ==Problem== {{AMC12 box|year=2008|ab=A|num-b=16|num-a=18}}
    2 KB (386 words) - 13:52, 21 December 2020
  • 2002 AMC 12B Problems/Problem 17
    == Problem == {{AMC12 box|year=2002|ab=B|num-b=16|num-a=18}}
    2 KB (380 words) - 15:00, 17 August 2023
  • 2008 AMC 10A Problems/Problem 20
    ==Problem== ...KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</ma
    2 KB (294 words) - 21:53, 17 October 2023
  • 2008 AMC 12A Problems/Problem 20
    ==Problem== <cmath>[ACD] = [ABC] - [BCD] = \frac 12 (3)(4) - \frac{24}{7} = \frac{18}{7}</cmath>-->
    6 KB (951 words) - 16:31, 2 August 2019
  • 2007 AMC 12B Problems/Problem 24
    ...dex.php/2007_AMC_10B_Problems/Problem_25#Solution| 2007 AMC 10B #25] (same problem) == Problem ==
    9 KB (1,522 words) - 22:46, 12 May 2022
  • 2008 AMC 12A Problems/Problem 14
    == Problem == What is the area of the region defined by the [[inequality]] <math>|3x-18|+|2y+7|\le3</math>?
    902 bytes (125 words) - 10:28, 13 March 2016
  • 2007 AMC 12B Problems/Problem 18
    ==Problem== ...}\ 10 \qquad \mathrm{(B)}\ 13 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 18 \qquad \mathrm{(E)}\ 21</math>
    5 KB (758 words) - 16:35, 15 February 2021
  • 2007 AMC 12B Problems/Problem 14
    ==Problem== ...three smaller triangles. The altitudes of these triangles are given in the problem as <math>PQ</math>, <math>PR</math>, and <math>PS</math>.
    3 KB (401 words) - 22:58, 8 May 2023
  • 2007 AMC 12B Problems/Problem 17
    ==Problem== {{AMC12 box|year=2007|ab=B|num-b=16|num-a=18}}
    1 KB (242 words) - 12:41, 19 June 2023
  • 2008 AMC 12B Problems/Problem 13
    ==Problem== \textbf{(C)}\ \frac{\sqrt3}{18} \qquad
    2 KB (328 words) - 10:54, 4 July 2013
  • 2007 AMC 12B Problems/Problem 19
    ==Problem== {{AMC12 box|year=2007|ab=B|num-b=18|num-a=20}}
    1 KB (166 words) - 16:35, 15 February 2021
  • 2007 AMC 12B Problems/Problem 23
    ==Problem== <math>(13, 84), (14, 48), (15, 36), (16, 30), (18, 24), (20, 21).</math>
    4 KB (725 words) - 19:59, 4 January 2024
  • 2007 AMC 12B Problems/Problem 16
    ==Problem== <math>\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ 18 \qquad \mathrm{(C)}\ 27 \qquad \mathrm{(D)}\ 54 \qquad \mathrm{(E)}\ 81</ma
    3 KB (463 words) - 16:35, 15 February 2021
  • 2004 AMC 12A Problems/Problem 19
    == Problem 19 == {{AMC12 box|year=2004|ab=A|num-b=18|num-a=20}}
    5 KB (785 words) - 00:29, 31 July 2023
  • 2008 AMC 10B Problems
    ==Problem 1== [[2008 AMC 10B Problems/Problem 1|Solution]]
    12 KB (1,838 words) - 16:52, 7 October 2022
  • 2008 AMC 12B Problems
    ==Problem 1== [[2008 AMC 12B Problems/Problem 1|Solution]]
    14 KB (2,199 words) - 13:43, 28 August 2020
  • 2008 AMC 12B Problems/Problem 2
    ...cate|[[2008 AMC 12B Problems|2008 AMC 12B #2]] and [[2008 AMC 10B Problems/Problem 2|2008 AMC 10B #2]]}} ==Problem==
    2 KB (219 words) - 23:53, 31 July 2023
  • 2008 AMC 12B Problems/Problem 4
    ==Problem== ...ac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}</mat
    1 KB (183 words) - 22:35, 10 June 2017
  • 2008 AIME I
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2008 AIME I Problems/Problem 1]]
    1 KB (140 words) - 16:18, 17 October 2020
  • 2008 AMC 12B Problems/Problem 19
    ==Problem== {{AMC12 box|year=2008|ab=B|num-b=18|num-a=20}}
    2 KB (307 words) - 19:28, 26 September 2023
  • 1985 AJHSME Problems
    == Problem 1 == [[1985 AJHSME Problem 1 | Solution]]
    12 KB (1,670 words) - 17:42, 24 November 2021
  • 1985 AJHSME
    ...he second link contains the answer key. The rest contain each individual problem and its solution. ** [[1985 AJHSME Problems/Problem 1|Problem 1]]
    2 KB (175 words) - 16:47, 12 April 2012
  • 2008 AIME I Problems
    == Problem 1 == [[2008 AIME I Problems/Problem 1|Solution]]
    9 KB (1,536 words) - 00:46, 26 August 2023
  • 2008 AIME I Problems/Problem 4
    == Problem == Indeed, by solving, we find <math>(x,y) = (18,62)</math> is the unique solution.
    4 KB (732 words) - 22:17, 28 November 2023
  • 2008 AIME I Problems/Problem 14
    == Problem == ...from <math>A</math> to line <math>CT</math>. Suppose <math>\overline{AB} = 18</math>, and let <math>m</math> denote the maximum possible length of segmen
    8 KB (1,333 words) - 00:18, 1 February 2024
  • Mock AIME 1 2007-2008 Problems
    == Problem 1 == [[Mock AIME 1 2007-2008 Problems/Problem 1|Solution]]
    6 KB (992 words) - 14:15, 13 February 2018
  • 2008 AIME II Problems/Problem 5
    == Problem== *[http://usamts.org/Solutions/Solution2_2_18.pdf USAMTS Year 18 Problem 2]
    8 KB (1,338 words) - 23:15, 28 November 2023
  • 2008 AIME II Problems/Problem 10
    == Problem == <cmath>\sqrt{18},\ \sqrt{13},\ \sqrt{10},\ \sqrt{9},\ \sqrt{8},\ \sqrt{5},\ \sqrt{4},\ \sqr
    4 KB (569 words) - 09:44, 25 November 2019
  • 2008 AIME II Problems/Problem 12
    == Problem == The well known problem of ordering <math>x</math> elements of a string of <math>y</math> elements
    10 KB (1,550 words) - 12:58, 15 July 2023
  • 1995 AHSME Problems/Problem 19
    ==Problem== {{AHSME box|year=1995|num-b=18|num-a=20}}
    1 KB (200 words) - 20:58, 10 February 2019
  • 1952 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1952 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:25, 20 February 2020
  • 1953 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1953 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:24, 20 February 2020
  • 1954 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1954 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:23, 20 February 2020
  • 1955 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1955 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:23, 20 February 2020
  • 1956 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1956 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:22, 20 February 2020
  • 1957 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1957 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:21, 20 February 2020
  • 1958 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1958 AHSME Problems/Problem 1|Problem 1]]
    3 KB (257 words) - 14:20, 20 February 2020
  • 1960 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1960 AHSME Problems/Problem 1|Problem 1]]
    2 KB (217 words) - 14:19, 20 February 2020
  • 2008 AMC 10B Problems/Problem 11
    ==Problem== ...quation by 2 and substituting into the first equation gives <math>128=5u_4+18 \Longrightarrow u_4=22</math>, therefore <math>u_5=\frac{128-22}{2}=53 \lon
    913 bytes (135 words) - 19:42, 24 October 2022
  • 2008 AMC 10B Problems/Problem 17
    ==Problem== In more concise terms, this problem is an extension of the binomial distribution. We find the number of ways on
    1 KB (202 words) - 12:42, 30 January 2024
  • 2008 AMC 10B Problems/Problem 19
    ==Problem== {{AMC10 box|year=2008|ab=B|num-b=18|num-a=20}}
    2 KB (373 words) - 13:21, 7 June 2021
  • 2008 AMC 10B Problems/Problem 20
    ==Problem== <math>\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9</math>
    4 KB (572 words) - 17:44, 14 June 2023
  • 2008 AMC 10A Problems/Problem 17
    ==Problem== ...athrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}
    2 KB (348 words) - 19:59, 4 June 2021
  • 2008 AMC 10A Problems/Problem 19
    ==Problem== {{AMC10 box|year=2008|ab=A|num-b=18|num-a=20}}
    2 KB (321 words) - 17:52, 7 November 2021
  • 1961 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1961 AHSME Problems/Problem 1|Problem 1]]
    2 KB (218 words) - 14:18, 20 February 2020
  • 1962 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1962 AHSME Problems/Problem 1|Problem 1]]
    2 KB (218 words) - 14:17, 20 February 2020
  • 2003 AMC 12A Problems/Problem 17
    == Problem == {{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}}
    6 KB (1,026 words) - 22:35, 29 March 2023
  • 1995 AHSME Problems/Problem 9
    ==Problem== ...} \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 } </math>
    760 bytes (100 words) - 13:59, 5 July 2013
  • How should I prepare?
    ...e also many books and online handouts/lectures you can use to improve your problem-solving skills. Depending on your current abilities, you will want to star The '''Art of Problem Solving books''' are an excellent resource to help prepare for math contest
    13 KB (1,926 words) - 11:22, 30 November 2023
  • 1967 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1967 AHSME Problems/Problem 1|Problem 1]]
    2 KB (216 words) - 14:14, 20 February 2020
  • 1999 AHSME Problems/Problem 19
    == Problem == {{AHSME box|year=1999|num-b=18|num-a=20}}
    1 KB (173 words) - 14:35, 5 July 2013
  • 1999 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1999 AHSME Problems/Problem 1|Problem 1]]
    2 KB (178 words) - 13:28, 20 February 2020
  • 1998 AHSME Problems
    == Problem 1 == [[1998 AHSME Problems/Problem 1|Solution]]
    15 KB (2,222 words) - 10:40, 11 August 2020
  • 1998 AHSME Problems/Problem 28
    == Problem == <math> \mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26</math>
    4 KB (662 words) - 00:51, 3 October 2023
  • 1998 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1998 AHSME Problems/Problem 1|Problem 1]]
    2 KB (173 words) - 03:44, 29 September 2014
  • 2003 USAMO Problems/Problem 5
    == Problem == ...\left(\frac {1}{3} + \frac {8a + 6}{6}\right) = 1 + \frac {8(a + b + c) + 18}{6} = 8</math>, as desired.
    6 KB (1,063 words) - 02:36, 9 August 2023
  • 1963 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1963 AHSME Problems/Problem 1|Problem 1]]
    2 KB (217 words) - 14:17, 20 February 2020
  • 1989 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1989 AHSME Problems/Problem 1|Problem 1]]
    2 KB (174 words) - 00:32, 2 October 2014
  • 1990 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1990 AHSME Problems/Problem 1|Problem 1]]
    2 KB (174 words) - 00:32, 2 October 2014
  • 1990 AHSME Problems/Problem 29
    == Problem == <cmath>\begin{array}{ccccc}1&3&9&27&81\\2&6&18&54\\4&12&36\\5&15&45\\7&21&63\\8&24&72\\10&30&90\\11&33&99\\13&39\\\vdots&\
    2 KB (285 words) - 19:25, 25 September 2020
  • 1990 AHSME Problems/Problem 26
    ==Problem== a_8 + a_{10} & = 18, &&(4) \\
    3 KB (402 words) - 23:17, 23 September 2023
  • 2001 USAMO Problems/Problem 1
    == Problem == 4 & 9 & 14 & 17 & 17 & 17 & 18 & 19 \\
    5 KB (841 words) - 17:19, 5 May 2022
  • 2008 iTest
    *[[2008 iTest Problems/Problem 1|Problem 1]] *[[2008 iTest Problems/Problem 2|Problem 2]]
    5 KB (424 words) - 15:18, 24 January 2020
  • 2008 iTest Problems
    ==Problem 1== [[2008 iTest Problems/Problem 1|Solution]]
    71 KB (11,749 words) - 01:31, 2 November 2023
  • 2008 iTest Problems/Problem 99
    == Problem == Then <math>xy = 2^{15} \cdot 3^{17} \cdot 7</math> has <math>16 \cdot 18 \cdot 2 = \boxed{576}</math> factors.
    4 KB (685 words) - 14:39, 7 October 2017
  • 1995 AHSME Problems/Problem 17
    ==Problem== draw(rotate(18)*polygon(5));
    1 KB (201 words) - 16:29, 12 March 2024
  • 1995 AHSME Problems/Problem 30
    ==Problem== <math> \mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 } </math>
    5 KB (828 words) - 05:52, 30 September 2023
  • 2007 AMC 12B Problems/Problem 20
    ==Problem== ...y=ax+d</math>, <math>y=bx+c</math>, and <math>y=bx+d</math> has area <math>18</math>. The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y
    7 KB (1,143 words) - 21:25, 20 December 2020
  • 1998 AJHSME Problems/Problem 19
    ==Problem== <math>8+10=18</math>
    2 KB (242 words) - 19:53, 31 October 2016
  • 2007 AMC 10B
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2007 AMC 10B Problems/Problem 1]]
    2 KB (182 words) - 18:52, 6 October 2014
  • 2007 AMC 8
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2007 AMC 8 Problems/Problem 1]]
    1 KB (127 words) - 21:46, 25 November 2013
  • 1980 AHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1980 AHSME Problems/Problem 1|Problem 1]]
    2 KB (174 words) - 22:29, 1 October 2014
  • 2005 AMC 8 Problems
    ==Problem 1== [[2005 AMC 8 Problems/Problem 1|Solution]]
    13 KB (1,821 words) - 22:18, 5 December 2023
  • 2005 AMC 8
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2005 AMC 8 Problems/Problem 1]]
    1 KB (127 words) - 21:47, 25 November 2013
  • 2003 AMC 12A Problems/Problem 10
    == Problem == ...rac{1}{6}\qquad \mathrm{(C) \ } \frac{2}{9}\qquad \mathrm{(D) \ } \frac{5}{18}\qquad \mathrm{(E) \ } \frac{5}{12} </math>
    1 KB (213 words) - 10:16, 4 July 2013
  • 1999 AHSME Problems/Problem 3
    == Problem== ...1{80} \qquad \mathrm{(B) \ } \frac 1{40} \qquad \mathrm{(C) \ } \frac 1{18} \qquad \mathrm{(D) \ } \frac 1{9} \qquad \mathrm{(E) \ } \frac 9{80} </m
    1 KB (201 words) - 14:36, 13 February 2019
  • 2002 AMC 10A
    ...ink contains the answers to each problem. The rest contain each individual problem and its solution. * [[2002 AMC 10A Problems/Problem 1]]
    1 KB (169 words) - 19:05, 26 November 2019
  • 2002 AMC 10A Problems
    ==Problem 1== [[2002 AMC 10A Problems/Problem 1|Solution]]
    11 KB (1,733 words) - 11:04, 12 October 2021
  • 2002 AMC 10B Problems
    ==Problem 1== [[2002 AMC 10B Problems/Problem 1|Solution]]
    10 KB (1,540 words) - 22:53, 19 December 2023
  • 2002 AMC 10A Problems/Problem 19
    ==Problem== {{AMC10 box|ab=A|year=2002|num-b=18|num-a=20}}
    1 KB (204 words) - 20:45, 28 May 2023
  • 2002 AMC 10B
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2002 AMC 10B Problems/Problem 1]]
    1 KB (157 words) - 00:51, 3 February 2015
  • 2004 AMC 10B
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2004 AMC 10B Problems/Problem 1]]
    2 KB (180 words) - 01:30, 7 October 2014
  • 2004 AMC 10B Problems
    ==Problem 1== [[2004 AMC 10B Problems/Problem 1|Solution]]
    13 KB (1,988 words) - 23:06, 7 March 2024
  • 1972 USAMO Problems/Problem 1
    ==Problem== ...the positive integers <math>a,b,\ldots, g</math>. For example, <math>(3,6,18)=3</math> and <math>[6,15]=30</math>. Prove that
    5 KB (1,018 words) - 11:14, 6 October 2023
  • 1972 USAMO Problems
    ==Problem 1== ...the positive integers <math>a,b,\ldots, g</math>. For example, <math>(3,6,18)=3</math> and <math>[6,15]=30</math>. Prove that
    2 KB (350 words) - 15:28, 2 June 2018
  • 2001 AMC 10 Problems
    ==Problem 1== [[2001 AMC 10 Problems/Problem 1|Solution]]
    14 KB (1,983 words) - 16:25, 2 June 2022
  • 2000 AMC 10
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2000 AMC 10 Problems/Problem 1]]
    1 KB (125 words) - 14:27, 20 February 2020
  • 2000 AMC 10 Problems
    ==Problem 1== [[2000 AMC 10 Problems/Problem 1|Solution]]
    14 KB (2,035 words) - 21:57, 2 May 2024
  • 2021 GMC 10B
    ==Problem 1== [[2021 GMC 10B Problems/Problem 1|Solution]]
    11 KB (1,695 words) - 14:33, 7 March 2022
  • 2003 AMC 10B
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2003 AMC 10B Problems/Problem 1]]
    1 KB (165 words) - 18:53, 6 October 2014
  • 2001 AMC 10
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. * [[2001 AMC 10 Problems/Problem 1]]
    2 KB (153 words) - 18:59, 6 October 2014
  • 1990 AJHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1990 AJHSME Problems/Problem 1]]
    2 KB (141 words) - 23:39, 8 October 2014
  • 2004 AMC 12B Problems/Problem 4
    == Problem == ...e digit equal to <math>7</math>. Therefore the probability is <math>\dfrac{18}{90} = \boxed{\dfrac{1}{5}} \Longrightarrow \mathrm{(B)}</math>.
    1 KB (210 words) - 19:22, 22 October 2022
  • 1998 AHSME Problems/Problem 19
    == Problem == {{AHSME box|year=1998|num-b=18|num-a=20}}
    2 KB (368 words) - 11:08, 4 February 2017
  • 1998 AHSME Problems/Problem 13
    == Problem == ...15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18 </math>
    1 KB (195 words) - 14:29, 5 July 2013
  • Areteem Institute Summer Camp
    ...help improve their problem solving skills. The focus of this program is on problem solving, exploration of mathematical knowledge, and improvement of analytic ...ornia State University, Fullerton (Residential and Day Camp options)- June 18-July 7
    3 KB (428 words) - 16:59, 9 March 2017
  • 1999 AHSME Problems/Problem 17
    == Problem == According to the problem statement, there are polynomials <math>Q(x)</math> and <math>R(x)</math> su
    2 KB (361 words) - 05:01, 7 February 2016
  • 2008 AMC 12B Problems/Problem 17
    ==Problem== <math>\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20</math>
    2 KB (313 words) - 23:51, 5 October 2023
  • 2000 AMC 10 Problems/Problem 17
    ==Problem== {{AMC10 box|year=2000|num-b=16|num-a=18}}
    1 KB (211 words) - 03:52, 20 July 2023
  • 2000 AMC 10 Problems/Problem 25
    #REDIRECT [[2000 AMC 12 Problems/Problem 18]]
    45 bytes (4 words) - 00:10, 27 November 2011
  • 1983 AHSME
    ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1983 AHSME Problems/Problem 1|Problem 1]]
    2 KB (174 words) - 00:34, 2 October 2014
  • 1999 AHSME Problems/Problem 26
    ==Problem== <math> \mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24</math>
    3 KB (451 words) - 10:19, 23 January 2024
  • 1985 AJHSME Problems/Problem 19
    ==Problem== {{AJHSME box|year=1985|num-b=18|num-a=20}}
    1 KB (219 words) - 09:04, 22 January 2023
  • 1985 AJHSME Problems/Problem 17
    ==Problem== {{AJHSME box|year=1985|num-b=16|num-a=18}}
    1 KB (234 words) - 08:14, 13 January 2023
  • 1986 AJHSME Problems
    == Problem 1 == [[1986 AJHSME Problems/Problem 1|Solution]]
    14 KB (2,054 words) - 15:41, 8 August 2020
  • 1999 AHSME Problems/Problem 22
    == Problem == ...\qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18</math>
    3 KB (538 words) - 12:02, 17 October 2020
  • 1986 AJHSME
    ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1986 AJHSME Problems/Problem 1|Problem 1]]
    2 KB (173 words) - 10:54, 15 April 2012
  • 1986 AJHSME Problems/Problem 17
    ==Problem== We can solve this problem using logic.
    3 KB (457 words) - 15:02, 4 April 2021
  • 1986 AJHSME Problems/Problem 19
    ==Problem== {{AJHSME box|year=1986|num-b=18|num-a=20}}
    1 KB (198 words) - 11:43, 24 February 2023
  • 2004 AMC 10B Problems/Problem 2
    ==Problem== <math> \mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 18\qquad \mathrm{(C) \ } 19 \qquad \mathrm{(D) \ } 20\qquad \mathrm{(E) \ } 30
    1 KB (208 words) - 19:10, 24 January 2015
  • 2004 AMC 12B Problems/Problem 15
    ==Problem== <math> \mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 4
    2 KB (326 words) - 09:47, 17 October 2020
  • 1987 AJHSME
    ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1987 AJHSME Problems/Problem 1|Problem 1]]
    2 KB (173 words) - 06:52, 16 April 2012
  • 1987 AJHSME Problems
    == Problem 1 == [[1987 AJHSME Problems/Problem 1|Solution]]
    12 KB (1,568 words) - 09:35, 31 October 2021
  • 2007 Alabama ARML TST Problems/Problem 13
    ==Problem== When in the situation <math>H</math>, we have probability <math>\frac 18</math> of winning the game right away, by throwing three more heads in a ro
    2 KB (374 words) - 11:48, 9 December 2022
  • 2002 AMC 10B Problems/Problem 17
    == Problem == {{AMC10 box|year=2002|ab=B|num-b=16|num-a=18}}
    2 KB (273 words) - 13:27, 21 May 2021
  • 2004 AMC 10B Problems/Problem 18
    == Problem == ...> and <math>\sin(\angle E) = \frac{12}{20}</math>, so <math>96 = 18 + 18 + 18 + x</math>.
    6 KB (904 words) - 12:54, 22 October 2023
  • 2002 AMC 12A Problems/Problem 10
    == Problem == {{AMC10 box|year=2002|ab=A|num-b=16|num-a=18}}
    3 KB (425 words) - 00:37, 1 February 2024
  • 2001 AMC 12 Problems/Problem 19
    == Problem == {{AMC12 box|year=2001|num-b=18|num-a=20}}
    1 KB (179 words) - 22:48, 18 August 2023
  • 2001 AMC 12 Problems/Problem 9
    == Problem == ...ad (\mathrm{C})\ \frac52 \qquad (\mathrm{D})\ 3 \qquad (\mathrm{E})\ \frac{18}5</math>
    1 KB (202 words) - 00:33, 30 December 2023
  • 1991 AJHSME Problems/Problem 10
    == Problem 10 == ...ext{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18</math>
    976 bytes (145 words) - 00:07, 5 July 2013
  • 2009 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 12A Problems/Problem 1|Problem 1]]
    2 KB (203 words) - 21:48, 6 October 2014
  • 2009 AMC 10A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 10A Problems/Problem 1|Problem 1]]
    2 KB (184 words) - 01:50, 4 March 2020
  • 2004 AMC 12B Problems/Problem 12
    ==Problem== From the values given in the problem statement we see that <math>a_3=a_1+2</math>.
    2 KB (364 words) - 11:41, 13 October 2021
  • 2009 AMC 12A Problems
    == Problem 1 == Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took <math>h</math
    13 KB (2,105 words) - 13:13, 12 August 2020
  • 2009 AMC 10A Problems
    == Problem 1 == [[2009 AMC 10A Problems/Problem 1|Solution]]
    14 KB (2,130 words) - 11:32, 7 November 2021
  • 2009 AMC 12A Problems/Problem 25
    == Problem == <cmath>\theta_{18}=165</cmath>
    7 KB (990 words) - 07:23, 24 October 2022
  • 2009 AMC 12A Problems/Problem 18
    {{duplicate|[[2009 AMC 12A Problems|2009 AMC 12A #18]] and [[2009 AMC 10A Problems|2009 AMC 10A #25]]}} == Problem ==
    6 KB (1,012 words) - 19:16, 14 September 2022
  • 2009 AMC 12A Problems/Problem 19
    == Problem == For simplicity, we solve the same problem with triangle and square instead of pentagon and heptagon. For the triangle
    4 KB (630 words) - 21:27, 30 December 2023
  • 2009 AMC 12A Problems/Problem 17
    == Problem == ...check that <math>r_1-r_2\ne 0</math>, as you cannot divide by zero. As the problem states that the series are different, <math>r_1 \ne r_2</math>, and so ther
    3 KB (453 words) - 10:22, 6 October 2023
  • 2009 AMC 12A Problems/Problem 11
    == Problem == ...4</math> congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to <math>4</m
    5 KB (759 words) - 16:48, 19 April 2022
  • 2009 AMC 12A Problems/Problem 12
    == Problem == ...sum is <math>9+9=18</math>. Hence the sum of digits will be at most <math>18</math>.
    3 KB (481 words) - 20:06, 17 December 2017
  • 2009 AMC 10A Problems/Problem 18
    == Problem == Hence the answer is <math>\frac{36}{70}=\frac{18}{35}</math>. We know this is a little bit larger than <math>\frac 12</math>
    3 KB (402 words) - 10:29, 2 August 2021
  • 2009 AMC 10A Problems/Problem 16
    == Problem == \mathrm{(D)}\ 18
    3 KB (524 words) - 16:26, 23 June 2023
  • 2009 AMC 12A Problems/Problem 1
    == Problem == Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took <math>h</math
    834 bytes (126 words) - 21:41, 3 July 2013
  • 2009 AMC 10A Problems/Problem 19
    == Problem == ...en be a factor of <math>100</math>, excluding <math>100</math> because the problem says that <math>r<100</math>. <math>100\: =\: 2^2\; \cdot \; 5^2</math>. Th
    2 KB (276 words) - 09:57, 8 June 2021
  • 2009 AMC 10A Problems/Problem 17
    == Problem == {{AMC10 box|year=2009|ab=A|num-b=16|num-a=18}}
    4 KB (684 words) - 21:14, 23 October 2023
  • 2001 AMC 12 Problems/Problem 8
    == Problem == {{AMC10 box|year=2001|num-b=16|num-a=18}}
    2 KB (279 words) - 00:32, 30 December 2023
  • 2001 AMC 12 Problems/Problem 10
    ...AMC 12 Problems|2001 AMC 12 #10]] and [[2001 AMC 10 Problems|2001 AMC 10 #18]]}} == Problem ==
    2 KB (252 words) - 00:11, 15 August 2022
  • 2001 AMC 12 Problems/Problem 12
    == Problem == ...precisely <math>30-6=24</math> numbers that satisfy all criteria from the problem statement.
    7 KB (1,110 words) - 15:20, 30 May 2022
  • 1987 AJHSME Problems/Problem 12
    ==Problem== What [[fraction]] of the large <math>12</math> by <math>18</math> [[rectangle|rectangular]] region is shaded?
    996 bytes (130 words) - 23:53, 4 July 2013
  • 2001 AMC 12 Problems/Problem 17
    == Problem == {{AMC12 box|year=2001|num-b=16|num-a=18}}
    5 KB (792 words) - 15:23, 30 November 2021
  • 2002 AMC 12A Problems/Problem 17
    == Problem == {{AMC12 box|year=2002|ab=A|num-b=16|num-a=18}}
    1 KB (161 words) - 15:01, 8 September 2022
  • 2009 AMC 10A Problems/Problem 5
    ==Problem== <math>\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mat
    2 KB (231 words) - 15:59, 9 February 2023
  • 2002 AMC 12A Problems/Problem 19
    == Problem == {{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}
    2 KB (350 words) - 11:09, 18 July 2023
  • 2002 AMC 12A Problems/Problem 18
    == Problem == \text{(B) }18
    3 KB (485 words) - 03:13, 1 September 2023
  • 2007 AMC 10B Problems/Problem 25
    ==Problem== <math>9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}</math>
    8 KB (1,425 words) - 23:05, 21 June 2023
  • 2009 AMC 10B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 10B Problems/Problem 1|Problem 1]]
    2 KB (189 words) - 17:22, 28 June 2021
  • 2009 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 12B Problems/Problem 1|Problem 1]]
    2 KB (197 words) - 21:14, 6 October 2014
  • 2009 AMC 10B Problems/Problem 3
    == Problem == \mathrm{(D)}\ 18\qquad
    791 bytes (119 words) - 15:58, 9 June 2021
  • 2009 AMC 10B Problems/Problem 23
    ...10B Problems|2009 AMC 10B #23]] and [[2009 AMC 12B Problems|2009 AMC 12B #18]]}} == Problem ==
    2 KB (304 words) - 13:23, 2 July 2021
  • 2009 AMC 10B Problems/Problem 25
    == Problem == <math>\mathrm{(A)}\frac 18\qquad
    4 KB (637 words) - 04:52, 2 July 2022
  • 2009 AMC 12B Problems
    == Problem 1 == [[2009 AMC 12B Problems/Problem 1|Solution]]
    13 KB (2,030 words) - 03:04, 5 September 2021
  • 2009 AMC 10B Problems/Problem 6
    == Problem == \mathrm{(D)}\ 18\qquad
    1 KB (159 words) - 08:13, 4 November 2022
  • 2009 AMC 12B Problems/Problem 14
    == Problem == {{AMC10 box|year=2009|ab=B|num-b=16|num-a=18}}
    4 KB (695 words) - 21:33, 7 October 2023
  • 2009 AMC 12B Problems/Problem 19
    == Problem == ...math> becomes negative for <math>n</math> between <math>2</math> and <math>18</math>, and then <math>f(19)=761</math> is again a prime number. And as <ma
    4 KB (696 words) - 13:27, 23 December 2020
  • 2009 AMC 12B Problems/Problem 10
    == Problem == {{AMC10 box|year=2009|ab=B|num-b=18|num-a=20}}
    2 KB (372 words) - 17:36, 28 June 2021
  • 2009 AMC 12B Problems/Problem 25
    == Problem == ...sum_{a=1}^{4} a(5-a)^2 = 1\cdot 16 + 2\cdot 9 + 3 \cdot 4 + 4 \cdot 1 = 16+18+12+4=50</cmath>
    15 KB (2,229 words) - 03:36, 4 September 2021
  • 2009 AMC 12B Problems/Problem 13
    == Problem == \mathrm{(D)}\ 18\qquad
    883 bytes (137 words) - 18:42, 23 February 2017
  • 2009 AMC 10B Problems
    == Problem 1 == [[2009 AMC 10B Problems/Problem 1|Solution]]
    15 KB (2,262 words) - 00:53, 18 June 2021
  • 2009 AMC 10B Problems/Problem 24
    == Problem == ...ath>\frac{180*(18-2)}{18} = 160^\circ</math> so each interior angle of the 18-gon is <math>160^\circ</math>. Let <math>x</math> be the degree measure of
    3 KB (546 words) - 04:26, 16 January 2023
  • 1987 AJHSME Problems/Problem 7
    ==Problem== ~Shadow-18
    2 KB (282 words) - 14:47, 8 June 2021
  • 1987 AJHSME Problems/Problem 19
    ==Problem== {{AJHSME box|year=1987|num-b=18|num-a=20}}
    980 bytes (136 words) - 11:36, 20 May 2017
  • 1987 AJHSME Problems/Problem 17
    ==Problem== {{AJHSME box|year=1987|num-b=16|num-a=18}}
    934 bytes (156 words) - 23:53, 4 July 2013
  • 1988 AJHSME
    ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1988 AJHSME Problems/Problem 1|Problem 1]]
    2 KB (173 words) - 17:28, 6 January 2024
  • 1988 AJHSME Problems
    == Problem 1 == [[1988 AJHSME Problems/Problem 1|Solution]]
    14 KB (1,872 words) - 15:23, 17 January 2023
  • 2009 AIME I Problems/Problem 14
    == Problem == ...egers, the two solutions to this are <math>(y,z) = (39,0)</math> or <math>(18,5)</math>.
    2 KB (326 words) - 14:12, 16 August 2020
  • 2009 AIME I Problems/Problem 5
    == Problem == Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem:
    8 KB (1,224 words) - 19:52, 7 March 2024
  • 2009 AIME I Problems/Problem 12
    == Problem == (If it's not then the answer to the problem would be irrational which can't be in the form of <math>\frac {m}{n}</math>
    12 KB (1,970 words) - 22:53, 22 January 2024
  • 1989 AJHSME Problems
    == Problem 1 == [[1989 AJHSME Problems/Problem 1|Solution]]
    13 KB (1,765 words) - 11:55, 22 November 2023
  • 2000 AMC 8 Problems
    == Problem 1 == [[2000 AMC 8 Problems/Problem 1|Solution]]
    15 KB (2,165 words) - 03:32, 13 April 2024
  • 1988 AJHSME Problems/Problem 24
    ==Problem== ...75688772935274463415059)--(18,1.7320508075688772935274463415059)--(17,0)--(18,-1.7320508075688772935274463415059)--(20,-1.7320508075688772935274463415059
    6 KB (470 words) - 23:56, 4 July 2013
  • 2009 AIME II Problems/Problem 10
    ==Problem== ...\dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\
    9 KB (1,415 words) - 13:56, 18 December 2022
  • 1988 AJHSME Problems/Problem 14
    ==Problem== 18 & 2 & 20 \\ \hline
    884 bytes (110 words) - 19:37, 10 March 2015
  • 1988 AJHSME Problems/Problem 17
    ==Problem== {{AJHSME box|year=1988|num-b=16|num-a=18}}
    1 KB (202 words) - 23:56, 4 July 2013
  • 1988 AJHSME Problems/Problem 19
    ==Problem== {{AJHSME box|year=1988|num-b=18|num-a=20}}
    946 bytes (133 words) - 10:51, 28 June 2023
  • 1989 AJHSME
    ...The second link contains the answer key. The rest contain each individual problem and its solution. ** [[1989 AJHSME Problems/Problem 1|Problem 1]]
    2 KB (173 words) - 17:29, 6 January 2024
  • Olimpiada de Mayo
    ===Problem 1=== ===Problem 2===
    4 KB (668 words) - 14:52, 17 August 2020
  • 1989 AJHSME Problems/Problem 5
    ==Problem== &= -15+18 \\
    476 bytes (64 words) - 23:57, 4 July 2013
  • 1990 AJHSME Problems
    ==Problem 1== ...x digits <math>4,5,6,7,8,9</math> in one of the six boxes in this addition problem?
    15 KB (2,059 words) - 15:03, 6 October 2021
  • 2016 AMC 12A Problems/Problem 17
    ==Problem 17== Note that we can also use coordinates to solve this problem. WLOG, set the side length of square <math>ABCD</math> equal to <math>6</ma
    2 KB (321 words) - 16:54, 27 November 2023
  • 1991 AJHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1991 AJHSME Problems/Problem 1]]
    2 KB (141 words) - 23:39, 8 October 2014
  • 1989 AJHSME Problems/Problem 19
    ==Problem== unitsize(18);
    2 KB (255 words) - 03:28, 28 November 2019
  • 1989 AJHSME Problems/Problem 22
    ==Problem== ...ext{(A)}\ 6 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 24</math>
    1 KB (177 words) - 16:03, 19 April 2021
  • 1989 AJHSME Problems/Problem 24
    ==Problem== ...the cut rectangles to the whole uncutted rectangle will be <math>\frac{15}{18} = \frac{5}{6}.</math>
    2 KB (333 words) - 20:05, 12 December 2021
  • 1989 AJHSME Problems/Problem 25
    ==Problem== With this knowledge, we can break down the problem into smaller problems, first, the probability that the first spinner lands
    3 KB (437 words) - 12:57, 16 October 2023
  • 1990 AJHSME Problems/Problem 14
    ==Problem== <math>\text{(A)}\ 12 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 36</math>
    833 bytes (136 words) - 20:58, 15 September 2019
  • 1990 AJHSME Problems/Problem 19
    ==Problem== {{AJHSME box|year=1990|num-b=18|num-a=20}}
    727 bytes (119 words) - 01:59, 25 November 2020
  • 1990 AJHSME Problems/Problem 25
    ==Problem== fill((18,1)--(19,1)--(19,3)--(18,3)--cycle,gray);
    2 KB (306 words) - 22:17, 11 July 2009
  • 1991 AJHSME Problems/Problem 9
    ==Problem== <math>\text{(A)}\ 18 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 25 \qquad \t
    1 KB (149 words) - 00:07, 5 July 2013
  • 1984 USAMO Problems/Problem 1
    ==Problem== <cmath>\begin{align*}a+b+c+d &= 18,\\ ab+ac+ad+bc+bd+cd &= k,\\ abc+abd+acd+bcd &=-200,\\ abcd &=-1984.\\ \end
    4 KB (704 words) - 19:25, 28 March 2024
  • 1992 AJHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1992 AJHSME Problems/Problem 1]]
    2 KB (141 words) - 23:40, 8 October 2014
  • 1991 AJHSME Problems/Problem 16
    ==Problem== unitsize(18);
    2 KB (314 words) - 00:07, 5 July 2013
  • 1991 AJHSME Problems/Problem 17
    ==Problem== {{AJHSME box|year=1991|num-b=16|num-a=18}}
    2 KB (266 words) - 00:07, 5 July 2013
  • 1991 AJHSME Problems/Problem 19
    ==Problem== {{AJHSME box|year=1991|num-b=18|num-a=20}}
    895 bytes (135 words) - 12:16, 8 October 2015
  • 1991 AJHSME Problems/Problem 20
    ==Problem== In the addition problem, each digit has been replaced by a letter. If different letters represent
    1 KB (203 words) - 13:06, 16 August 2020
  • 1992 AJHSME Problems
    ==Problem 1== [[1992 AJHSME Problems/Problem 1|Solution]]
    17 KB (2,346 words) - 13:36, 19 February 2020
  • 1961 AHSME Problems/Problem 10
    == Problem== <math>\textbf{(A)}\ \sqrt{18} \qquad
    1 KB (176 words) - 23:27, 26 May 2018
  • 1993 AJHSME
    ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1993 AJHSME Problems/Problem 1]]
    2 KB (141 words) - 23:41, 8 October 2014
  • 1993 AJHSME Problems
    ==Problem 1== [[1993 AJHSME Problems/Problem 1|Solution]]
    14 KB (1,797 words) - 11:13, 28 December 2022
  • 2006 AMC 8
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2006 AMC 8 Problems/Problem 1|Problem 1]]
    2 KB (154 words) - 21:46, 25 November 2013
  • 2003 AMC 12B Problems/Problem 25
    == Problem == \qquad\mathrm{(C)}\ \dfrac{1}{18}
    4 KB (635 words) - 11:46, 1 September 2022
  • 2005 AMC 10A Problems/Problem 19
    == Problem == {{AMC10 box|year=2005|ab=A|num-b=18|num-a=20}}
    3 KB (315 words) - 12:46, 14 December 2021
  • 2010 AMC 12A
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2010 AMC 12A Problems/Problem 1|Problem 1]]
    2 KB (203 words) - 21:47, 6 October 2014
  • 2010 AMC 10A
    **[[2010 AMC 10A Problems/Problem 1|Problem 1]] **[[2010 AMC 10A Problems/Problem 2|Problem 2]]
    2 KB (175 words) - 21:24, 15 November 2022
  • 2010 AMC 12A Problems
    == Problem 1 == [[2010 AMC 12A Problems/Problem 1|Solution]]
    12 KB (1,817 words) - 15:00, 12 August 2020
  • 2010 AMC 12A Problems/Problem 19
    == Problem == {{AMC12 box|year=2010|num-b=18|num-a=20|ab=A}}
    2 KB (379 words) - 18:30, 10 July 2022
  • 2010 AMC 12A Problems/Problem 9
    == Problem == {{AMC10 box|year=2010|ab=A|num-b=16|num-a=18}}
    2 KB (383 words) - 17:42, 28 June 2021
  • 2010 AMC 12A Problems/Problem 16
    ...12A Problems|2010 AMC 12A #16]] and [[2010 AMC 10A Problems|2010 AMC 10A #18]]}} == Problem ==
    5 KB (796 words) - 22:51, 26 November 2023
  • 2010 AMC 12A Problems/Problem 17
    == Problem == {{AMC10 box|year=2010|ab=A|num-b=18|num-a=20}}
    4 KB (690 words) - 10:13, 14 October 2022
  • 2010 AMC 12A Problems/Problem 23
    == Problem == There might be a problem when you cancel out the <math>10</math>s from <math>90!</math>. One method
    10 KB (1,525 words) - 09:44, 24 April 2024
  • 2010 AMC 12A Problems/Problem 24
    == Problem == <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 36</math>
    9 KB (1,434 words) - 17:54, 17 August 2022
  • 2007 AMC 8 Problems/Problem 4
    == Problem == <math>\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 15 \qquad\mathrm{(C)}\ 18 \qquad\mathrm{(D)}\ 30 \qquad\mathrm{(E)}\ 36</math>
    679 bytes (109 words) - 12:22, 16 August 2021
  • 2007 AMC 8 Problems/Problem 7
    == Problem == ...age of <math>5</math> people in a room is <math>30</math> years. An <math>18</math>-year-old person leaves
    1 KB (192 words) - 12:22, 16 August 2021
  • 2007 AMC 8 Problems/Problem 8
    == Problem == ...)}\ 4.5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 18</math>
    2 KB (237 words) - 04:26, 23 July 2021
  • 2010 AMC 12B
    ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2010 AMC 12B Problems/Problem 1|Problem 1]]
    2 KB (197 words) - 21:42, 6 October 2014
  • 2010 AMC 10A Problems
    == Problem 1 == [[2010 AMC 10A Problems/Problem 1|Solution]]
    13 KB (1,902 words) - 11:20, 5 March 2023
  • 2010 AMC 10B Problems
    == Problem 1 == [[2010 AMC 10B Problems/Problem 1|Solution]]
    12 KB (1,817 words) - 22:44, 22 December 2020
  • 2010 AMC 12B Problems
    == Problem 1 == [[2010 AMC 12B Problems/Problem 1|Solution]]
    12 KB (1,845 words) - 13:00, 19 February 2020
  • Proofs without words
    Fermat point problem Fagnano problem
    55 KB (7,986 words) - 17:04, 20 December 2018
  • 2010 AIME I Problems/Problem 6
    == Problem == ...and <math>c = \frac{14}{5}</math>, so <math>P(x) = \frac{9}{5}x^2 - \frac{18}{5}x + \frac {14}{5}</math>. Therefore, <math>P(16) = \boxed{406}</math>.
    6 KB (1,019 words) - 20:39, 20 November 2023
  • 2010 AIME I Problems/Problem 9
    == Problem == ...th>c = \sqrt [3]{k + 18}</math>, so we have <math>28 + 3\sqrt [3]{k(k+4)(k+18)} = 28+3abc=a^3+b^3+c^3=3k+22</math>. Move 28 over, divide both sides by 3,
    3 KB (483 words) - 07:35, 4 November 2022
  • 2010 AIME I Problems/Problem 11
    == Problem == ...\frac{33}{4}\right)</math> is given by <math>h_1 + h_2 = \sqrt{\left(\frac{18}{4} - \frac{39}{4}\right)^2 + \left(\frac{26}{4} - \frac{33}{4}\right)^2} =
    4 KB (636 words) - 16:46, 25 November 2023
  • 2010 AIME I Problems/Problem 13
    == Problem == ...rac {1}{2}\left(8 - \frac {\sqrt {3}}{9}h\right)h = 4h - \frac {\sqrt {3}}{18}h^2</math>
    10 KB (1,418 words) - 23:05, 20 October 2021
  • 2010 AIME I Problems/Problem 14
    == Problem == ...or = 3</math>. So we go up one more power of <math>10</math> and get <math>18 + 270 + 4 = 292</math>, which is very close to what we are looking for.
    4 KB (739 words) - 22:09, 25 November 2023
  • 2010 AIME I Problems/Problem 15
    == Problem == ...+BM)r}=\frac{12+AM+BM}{13+CM+BM}= \frac{12+BM}{13+BM} = \frac{17+15(AM-7)}{18+15(AM-7)}</math>
    14 KB (2,210 words) - 13:14, 11 January 2024

View (previous 500 | next 500) (20 | 50 | 100 | 250 | 500)

Retrieved from "https://artofproblemsolving.com/wiki/index.php/Special:Search"