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  • 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
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  • 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}
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  • 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]]
    33 KB (5,177 words) - 21:05, 4 February 2023
  • 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

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