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

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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

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