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

• 2007 AMC 8 Problems/Problem 14
  == Problem == ...)}\ 8 \qquad \mathrm{(C)}\ 13 \qquad \mathrm{(D)}\ 14 \qquad \mathrm{(E)}\ 18</math>
  1,004 bytes (162 words) - 12:22, 16 August 2021
• 2010 AMC 10B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2010 AMC 10B Problems/Problem 1|Problem 1]]
  2 KB (197 words) - 18:19, 6 October 2014
• 1994 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1994 AJHSME Problems/Problem 1]]
  2 KB (141 words) - 23:41, 8 October 2014
• 1994 AJHSME Problems
  ==Problem 1== [[1994 AJHSME Problems/Problem 1|Solution]]
  16 KB (2,292 words) - 13:36, 19 February 2020
• 2010 AIME II Problems/Problem 3
  == Problem == ...ath>(1^{19})(2^{18})\cdots(19^1)</math> (or alternatively, <math>19! \cdot 18! \cdots 1!</math>.)
  2 KB (303 words) - 15:17, 12 May 2020
• 2010 AIME II Problems/Problem 11
  == Problem == Therefore, there are <math>3 \cdot 2 \cdot 3=18</math> cases.
  6 KB (1,057 words) - 01:58, 8 January 2023
• 2010 USAJMO Problems/Problem 1
  == Problem == ...<math>S_{18}</math> with an <math>m</math> of <math>2</math> and <math>P_{18}=26</math> with an <math>m</math> of <math>26</math>, note that their produ
  12 KB (2,338 words) - 20:30, 13 February 2024
• 2015 AMC 10B Problems/Problem 20
  ==Problem== ...d\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}</math>
  4 KB (710 words) - 10:04, 17 August 2023
• 2010 AIME II Problems/Problem 15
  == Problem 15 == <cmath>NE=AN-AE=\frac 12\cdot AB - \frac{AC}{AC+BC}\cdot AB=\frac 5{18}</cmath>
  9 KB (1,523 words) - 15:24, 21 November 2023
• 1995 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1995 AJHSME Problems/Problem 1]]
  2 KB (141 words) - 23:42, 8 October 2014
• 1995 AJHSME Problems
  ==Problem 1== [[1995 AJHSME Problems/Problem 1|Solution]]
  14 KB (2,096 words) - 18:29, 2 January 2023
• 1956 AHSME Problems/Problem 1
  ==Problem #1== <math>\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 16 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 36 \qquad\textbf{(E)}\ 64 </math>
  434 bytes (57 words) - 16:54, 14 March 2023
• 2010 AMC 12B Problems/Problem 2
  == Problem 2 == ...be <math>(8 - 2) \times (5 - 2) = 18</math>. The answer is then <math>40 - 18 = 22</math> <math>(A)</math>.
  802 bytes (110 words) - 10:58, 4 July 2013
• The Ross Program
  ...umber Theory]] and consist of daily lectures, seminar groups, and rigorous problem sets. ...iversity in an effort to emphasize development of students' creativity and problem-solving skills. In 1964, the program was moved to Ohio State University. Ro
  3 KB (474 words) - 20:48, 28 May 2013
• 2010 AMC 12B Problems/Problem 7
  == Problem == <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qq
  2 KB (302 words) - 15:49, 23 November 2023
• 2010 AMC 12B Problems/Problem 8
  == Problem == {{AMC10 box|year=2010|num-b=16|num-a=18|ab=B}}
  2 KB (367 words) - 13:40, 11 July 2021
• 2010 AMC 12B Problems/Problem 11
  == Problem == .... This adds another 9 palindromes to the list, bringing our total to <math>18/90 = \boxed {\frac{1}{5} } = \boxed {E}</math>
  2 KB (354 words) - 19:54, 10 March 2024
• 2010 AMC 12B Problems/Problem 16
  ...12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}} == Problem ==
  3 KB (419 words) - 13:42, 11 July 2021
• 2010 AMC 12B Problems/Problem 17
  == Problem == <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qq
  3 KB (440 words) - 19:34, 17 October 2022
• 2010 AMC 12B Problems/Problem 19
  == Problem == {{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}}
  6 KB (990 words) - 19:14, 25 March 2024
• 2010 AMC 10A Problems/Problem 3
  == Problem 3 == \mathrm{(C)}\ 18
  2 KB (276 words) - 19:50, 5 March 2023
• 1996 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1996 AJHSME Problems/Problem 1]]
  2 KB (143 words) - 00:28, 6 November 2020
• 1996 AJHSME Problems
  ==Problem 1== [[1996 AJHSME Problems/Problem 1|Solution]]
  13 KB (1,880 words) - 13:35, 19 February 2020
• 2007 AMC 8 Problems/Problem 20
  ==Problem== ...y had won <math>x</math> of these games. From the information given in the problem, we can say that <math>\frac{x}{y}=0.45.</math> Next, the Unicorns win 6 mo
  6 KB (958 words) - 18:32, 20 January 2024
• 2007 AMC 8 Problems/Problem 19
  == Problem == Let the smaller of the two numbers be <math>x</math>. Then, the problem states that <math> (x+1)+x<100</math>. <math> (x+1)^2-x^2=x^2+2x+1-x^2=2x+1
  1 KB (242 words) - 01:12, 16 November 2023
• 2004 AMC 8 Problems
  ==Problem 1== [[2004 AMC 8 Problems/Problem 1|Solution]]
  13 KB (1,860 words) - 19:58, 8 May 2023
• 2006 AMC 8 Problems
  ==Problem 1== ...bf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math>
  16 KB (2,215 words) - 19:18, 10 April 2024
• 1997 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1997 AJHSME Problems/Problem 1]]
  2 KB (141 words) - 23:42, 8 October 2014
• 1997 AJHSME Problems
  ==Problem 1== [[1997 AJHSME Problems/Problem 1|Solution]]
  12 KB (1,702 words) - 12:35, 6 November 2022
• 1998 AJHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1998 AJHSME Problems/Problem 1]]
  2 KB (141 words) - 23:43, 8 October 2014
• 1998 AJHSME Problems
  ==Problem 1== [[1998 AJHSME Problems/Problem 1|Solution]]
  14 KB (1,920 words) - 19:31, 31 January 2024
• 2002 AMC 8 Problems/Problem 8
  ==Problem== /* AMC8 2002 #8, 9, 10 Problem */
  2 KB (250 words) - 01:37, 6 January 2024
• 2007 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2007|num-b=16|num-a=18}}
  1 KB (166 words) - 00:28, 2 July 2023
• 2010 AMC 10B Problems/Problem 20
  == Problem == <math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \q
  3 KB (483 words) - 18:41, 4 May 2024
• 1965 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1965 AHSME Problems/Problem 1|Problem 1]]
  2 KB (217 words) - 14:15, 20 February 2020
• 2010 AMC 10B Problems/Problem 19
  == Problem == ...qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
  3 KB (443 words) - 18:34, 4 May 2024
• 2010 AMC 10B Problems/Problem 16
  == Problem== ...(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}</ma
  3 KB (544 words) - 20:54, 24 March 2024
• 2011 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2011 AMC 12A Problems/Problem 1|Problem 1]]
  2 KB (201 words) - 21:46, 6 October 2014
• 2011 AMC 12A Problems
  == Problem 1 == [[2011 AMC 12A Problems/Problem 1|Solution]]
  13 KB (1,994 words) - 13:52, 3 July 2021
• 2011 AMC 12A Problems/Problem 8
  == Problem == \textbf{(B)}\ 18 \qquad
  3 KB (540 words) - 23:12, 8 April 2024
• 2011 AMC 12A Problems/Problem 9
  == Problem == ...ach handshake was counted twice, we get a total of <math>\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}</math>
  1 KB (196 words) - 23:47, 15 July 2020
• 2011 AMC 12A Problems/Problem 10
  == Problem == \textbf{(E)}\ \frac{5}{18} </math>
  2 KB (248 words) - 14:51, 5 May 2021
• 2011 AMC 12A Problems/Problem 13
  == Problem == ...>ABC</math> has side-lengths <math>AB = 12, BC = 24,</math> and <math>AC = 18.</math> The line through the incenter of <math>\triangle ABC</math> paralle
  4 KB (683 words) - 03:12, 23 January 2023
• 2011 AMC 12A Problems/Problem 17
  == Problem == {{AMC12 box|year=2011|num-b=16|num-a=18|ab=A}}
  3 KB (462 words) - 17:49, 3 February 2024
• 2011 AMC 12A Problems/Problem 19
  == Problem == {{AMC12 box|year=2011|num-b=18|num-a=20|ab=A}}
  4 KB (563 words) - 11:12, 3 December 2023
• 2011 AMC 10A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2011 AMC 10A Problems/Problem 1|Problem 1]]
  2 KB (194 words) - 18:14, 6 October 2014
• 1993 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1993 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 00:31, 2 October 2014
• 1958 AHSME Problems
  == Problem 1 == [[1958 AHSME Problems/Problem 1|Solution]]
  25 KB (3,872 words) - 14:21, 20 February 2020
• 2011 AMC 10A Problems
  == Problem 1 == [[2011 AMC 10A Problems/Problem 1|Solution]]
  13 KB (1,903 words) - 18:09, 19 April 2021
• 1993 AHSME Problems
  == Problem 1 == [[1993 AHSME Problems/Problem 1|Solution]]
  20 KB (2,814 words) - 08:15, 27 June 2021
• 2011 AMC 10A Problems/Problem 3
  == Problem == ...\frac{5}{18} \qquad\textbf{(C)}\ \frac{1}{3} \qquad\textbf{(D)}\ \frac{7}{18} \qquad\textbf{(E)}\ \frac{2}{3} </math>
  915 bytes (132 words) - 15:48, 16 January 2021
• 2011 AMC 10A Problems/Problem 19
  == Problem 19 == Then go through the same routine as demonstrated above to finish this problem.
  3 KB (545 words) - 20:54, 21 August 2023
• 2011 AMC 10A Problems/Problem 16
  ==Problem 16== ...\ = \ &\sqrt{18+2\sqrt{81-72}}\\\\ = \ &\sqrt{18+2\sqrt{9}}\\\\ = \ &\sqrt{18+6}\\\\= \ &\sqrt{24}\\\\ = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}.
  4 KB (558 words) - 20:43, 21 August 2023
• 2021 JMPSC Accuracy Problems/Problem 11
  ==Problem== ...\cdot 3^2 \cdot 7</math>, so the divisors are <math>\{1,2,3,4,6,7,9,12,14,18,21,28,36,42,63,84,126,252 \}</math>. We see the set <math>\{21,42,63,84 \}<
  2 KB (296 words) - 01:17, 12 July 2021
• 2003 AMC 12A Problems/Problem 19
  ==Problem== {{AMC12 box|year=2003|ab=A|num-b=18|num-a=20}}
  2 KB (355 words) - 22:55, 4 January 2019
• 2011 AMC 10B
  **[[2011 AMC 10B Problems/Problem 1|Problem 1]] **[[2011 AMC 10B Problems/Problem 2|Problem 2]]
  2 KB (175 words) - 18:19, 6 October 2014
• 2011 AMC 12B Problems
  ==Problem 1== [[2011 AMC 12B Problems/Problem 1|Solution]]
  13 KB (1,978 words) - 16:28, 12 July 2020
• 2011 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2011 AMC 12B Problems/Problem 1|Problem 1]]
  2 KB (197 words) - 21:18, 6 October 2014
• 2012 IMO Problems/Problem 1
  == Problem == --[[User:Suli|Suli]] 18:21, 8 February 2015 (EST)
  7 KB (1,189 words) - 01:22, 19 November 2023
• 2011 AMC 12B Problems/Problem 21
  ==Problem== ...h>, <math>22</math>, <math>11</math> respectively. And since <math>a+b \le 18</math>, <math>a+b = 11</math>, <math>a-b = 4</math>, but there is no intege
  3 KB (507 words) - 19:48, 4 November 2023
• 2011 AMC 12B Problems/Problem 19
  ==Problem== ...s not affect our answer. Thus, we can ignore it and consider an equivalent problem with the line <math>y = mx</math>.
  2 KB (349 words) - 19:38, 4 November 2023
• 2001 AMC 10 Problems/Problem 5
  == Problem == draw(shift(18,8)*unitsquare);
  2 KB (266 words) - 13:51, 7 August 2023
• 2001 AMC 10 Problems/Problem 10
  == Problem == <math>\textbf{(A) }18\qquad\textbf{(B) }19\qquad\textbf{(C) }20\qquad\textbf{(D) }22\qquad\textbf
  2 KB (257 words) - 23:42, 9 April 2024
• 2001 AMC 10 Problems/Problem 19
  == Problem == ...)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 18 </math>
  2 KB (335 words) - 17:44, 11 April 2024
• 2001 AMC 10 Problems/Problem 22
  ==Problem== label("$18$",(0.5,1.5));
  5 KB (721 words) - 16:44, 9 August 2022
• 2011 AIME I Problems/Problem 1
  == Problem == <cmath>\frac{18}{5}+\frac{km}{50n}=4+\frac{m}{n}</cmath>
  4 KB (786 words) - 16:43, 5 February 2022
• 2011 AIME I Problems/Problem 6
  ==Problem== ...must be divisible by <math>16</math>. Let <math>9a=z</math>. If <math>{z-18}\equiv {0} \pmod{16}</math>, then <math>{z}\equiv {2} \pmod{16}</math>. The
  4 KB (661 words) - 01:18, 11 December 2023
• 2002 AMC 8 Problems
  ==Problem 1== [[2002 AMC 8 Problems/Problem 1 | Solution]]
  15 KB (2,102 words) - 09:58, 5 May 2024
• 2011 AIME II Problems/Problem 3
  ==Problem== The degree measures of the angles in a [[convex polygon|convex]] 18-sided polygon form an increasing [[arithmetic sequence]] with integer value
  3 KB (547 words) - 01:13, 31 January 2024
• 2011 AIME II Problems/Problem 4
  == Problem 4 == pair A = (0,0), C= (11,0), B=IP(CR(A,20),CR(C,18)), D = IP(B--C,CR(B,20/31*abs(B-C))), M = (A+D)/2, P = IP(M--2*M-B, A--C),
  6 KB (944 words) - 21:31, 14 January 2024
• 2011 AIME II Problems/Problem 6
  ==Problem 6== ...inations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is <math>(10*9*8*7/24)=2
  9 KB (1,535 words) - 01:28, 16 January 2023
• 2011 AIME II Problems/Problem 8
  == Problem == <cmath>(z^6-2^{18})(z^6+2^{18})=0</cmath>
  5 KB (805 words) - 18:46, 27 January 2024
• 2011 AIME II Problems
  == Problem 1 == [[2011 AIME II Problems/Problem 1|Solution]]
  8 KB (1,301 words) - 08:43, 11 October 2020
• 2011 AIME II Problems/Problem 12
  == Problem 12 == Finally, there are <math> 9 \times 2 = 18.</math> ways for the candidates from all the countries to sit in three bloc
  5 KB (848 words) - 19:15, 30 April 2023
• 2011 AIME II Problems/Problem 14
  ==Problem 14== ...e of these triples are repeated and all are used. By the conditions of the problem, if <math>i</math> is the same in two different triples, then the two numbe
  10 KB (1,581 words) - 22:09, 27 August 2023
• 2011 AIME II Problems/Problem 10
  == Problem 10 == <math>z = \frac{18}{7}</math>.
  11 KB (1,720 words) - 03:12, 18 December 2023
• 2011 AIME II Problems/Problem 13
  ==Problem== ...ath> and thus the equation to be <math>y=x+(12-a)</math>, we get <math>(6, 18-a)</math>. Likewise for <math>O_{2}</math> it's <math>(6, 6-a)</math>. Now
  13 KB (2,055 words) - 05:25, 9 September 2022
• 1996 AJHSME Problems/Problem 1
  ==Problem== The factors of <math>36</math> are <math>1, 2, 3, 4, 6, 9, 12, 18, </math> and <math>36</math>.
  1 KB (186 words) - 11:19, 27 June 2023
• 1996 AJHSME Problems/Problem 2
  ==Problem== ...gets <math>10 - 1 = 9</math>, then <math>9 \cdot 2 = 18</math>, then <math>18 + 2 = 20</math>.
  1 KB (152 words) - 13:49, 23 October 2016
• 1999 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1999 AMC 8 Problems/Problem 1|Problem 1]]
  2 KB (163 words) - 00:24, 31 March 2015
• 1999 AMC 8 Problems
  ==Problem 1== [[1999 AMC 8 Problems/Problem 1|Solution]]
  17 KB (2,394 words) - 19:51, 8 May 2023
• 2000 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2000 AMC 8 Problems/Problem 1]]
  1 KB (138 words) - 21:51, 25 November 2013
• 2011 AMC 12B Problems/Problem 17
  ==Problem== {{AMC12 box|year=2011|num-b=16|num-a=18|ab=B}}
  2 KB (285 words) - 19:36, 7 August 2023
• 2000 AMC 8 Problems/Problem 4
  ==Problem== unitsize(18);
  3 KB (406 words) - 17:29, 22 October 2020
• 2002 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2002 AMC 8 Problems/Problem 1]]
  1 KB (138 words) - 21:51, 25 November 2013
• 2001 AMC 8 Problems/Problem 3
  ==Problem== <math>\text{(A)}\ 17 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 19 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 23</math>
  651 bytes (94 words) - 18:03, 27 March 2015
• 2001 AMC 8 Problems/Problem 8
  ==Problem== .../math> and <math> 7 </math>, so the diagonals of the large kite are <math> 18 </math> and <math> 21 </math>, and the amount of bracing Genevieve needs is
  1 KB (213 words) - 01:56, 17 January 2021
• 2001 AMC 8 Problems/Problem 9
  ==Problem== ...6)\times3(7) </math>, or <math> 18\times21 </math>, so the area is <math> (18)(21)=378 </math>. The area of the kite is half of the area of the rectangle
  1 KB (214 words) - 18:33, 2 January 2022
• 2001 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2001|num-b=16|num-a=18}}
  2 KB (241 words) - 18:16, 10 January 2024
• 2001 AMC 8 Problems/Problem 18
  ==Problem== <math>\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \
  2 KB (333 words) - 22:55, 17 October 2023
• 2011 AMC 10B Problems
  == Problem 1 == [[2011 AMC 10B Problems/Problem 1|Solution]]
  13 KB (2,090 words) - 18:05, 7 January 2021
• 2011 AMC 10B Problems/Problem 6
  == Problem== \frac{2}{3} x - 2 &= 18\\
  1 KB (220 words) - 05:34, 25 June 2022
• 2001 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2001|num-b=18|num-a=20}}
  2 KB (397 words) - 23:05, 13 August 2019
• 2001 AMC 8 Problems/Problem 21
  ==Problem== The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
  1 KB (211 words) - 00:23, 26 December 2022
• 2011 AMC 10B Problems/Problem 17
  == Problem== {{AMC10 box|year=2011|ab=B|num-b=16|num-a=18}}
  5 KB (699 words) - 04:53, 21 January 2023
• 2011 AMC 10B Problems/Problem 19
  == Problem== {{AMC10 box|year=2011|ab=B|num-b=18|num-a=20}}
  3 KB (528 words) - 11:54, 29 May 2022
• 2003 AMC 10B Problems
  ==Problem 1== <cmath> \frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath>
  15 KB (2,166 words) - 21:17, 16 February 2021
• 2007 AMC 10B Problems
  ==Problem 1== [[2007 AMC 10B Problems/Problem 1|Solution]]
  15 KB (2,297 words) - 12:57, 19 February 2020
• 2007 AMC 10B Problems/Problem 19
  ==Problem== Alternatively, we may analyze this problem a little further.
  5 KB (723 words) - 10:58, 27 October 2021
• 2001 AMC 8 Problems/Problem 11
  == Problem == <math>\text{(A)}\ 12 \qquad \text{(B)}\ 15 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 21 \qquad \text{(E)}\ 24</math>
  2 KB (383 words) - 19:36, 24 December 2023
• One Root Equations Document
  <math>\frac{-18 \pm \sqrt{(18)^2-4*9*9}}{2*9}</math> <math>\frac{-18 \pm \sqrt{(9^2*2^2)-(9^2*2^2)}}{2*9}</math>
  6 KB (1,119 words) - 19:50, 2 June 2011
• 1992 AJHSME Problems/Problem 6
  ==Problem== unitsize(18);
  1 KB (174 words) - 00:09, 5 July 2013
• 2011 AMC 10B Problems/Problem 21
  == Problem == ...a difference of <math>8,</math> which isn't given as a possibility in the problem. This means <math>1</math> must be the difference between <math>y</math> an
  8 KB (1,303 words) - 20:29, 5 September 2022
• 1998 AHSME Problems/Problem 3
  == Problem 3 == ...\qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18 </math>
  1 KB (230 words) - 19:46, 10 March 2015
• 1998 AJHSME Problems/Problem 4
  ==Problem== draw((14,21)--(18,0)--(30,9));
  709 bytes (94 words) - 00:28, 5 July 2013
• 1998 AJHSME Problems/Problem 11
  ==Problem== ...xt{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 15 \qquad \text{(E)}\ 18</math>
  708 bytes (101 words) - 00:29, 5 July 2013
• 1998 AJHSME Problems/Problem 17
  ==Problem 17== {{AJHSME box|year=1998|num-b=16|num-a=18}}
  2 KB (354 words) - 14:30, 29 May 2021
• 2003 AMC 10B Problems/Problem 17
  ==Problem== {{AMC10 box|year=2003|ab=B|num-b=16|num-a=18}}
  1 KB (252 words) - 16:43, 29 June 2021
• 2002 AMC 10B Problems/Problem 19
  == Problem == {{AMC10 box|year=2002|ab=B|num-b=18|num-a=20}}
  3 KB (472 words) - 14:56, 17 August 2023
• 1984 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[1984 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 03:48, 29 September 2014
• 1984 AHSME Problems
  ==Problem 1== [[1984 AHSME Problems/Problem 1|Solution]]
  13 KB (1,879 words) - 14:00, 19 February 2020
• 1984 AHSME Problems/Problem 7
  ==Problem== ...ext{minutes} \qquad \mathrm{(B) \ }15 \text{minutes}\qquad \mathrm{(C) \ } 18 \text{minutes}\qquad \mathrm{(D) \ }20 \text{minutes} \qquad \mathrm{(E) \
  1 KB (208 words) - 12:49, 5 July 2013
• 1999 AMC 8 Problems/Problem 21
  ==Problem== draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle);
  2 KB (332 words) - 22:27, 24 January 2024
• 1984 AHSME Problems/Problem 13
  ==Problem== <math> =\frac{2\sqrt{12}-2\sqrt{18}-2\sqrt{30}}{2-(\sqrt{3}+\sqrt{5})^2} </math>
  2 KB (258 words) - 16:43, 1 March 2015
• 1984 AHSME Problems/Problem 17
  ==Problem== {{AHSME box|year=1984|num-b=16|num-a=18}}
  1 KB (202 words) - 18:13, 12 March 2018
• 1984 AHSME Problems/Problem 19
  ==Problem== {{AHSME box|year=1984|num-b=18|num-a=20}}
  1 KB (209 words) - 12:51, 5 July 2013
• 2010 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2010 AMC 8 Problems/Problem 1|Problem 1]]
  2 KB (152 words) - 18:46, 21 November 2018
• 1998 AJHSME Problems/Problem 22
  ==Problem== A sample sequence: <math>23, 18, 9, 81, 76, \ldots .</math>
  1 KB (191 words) - 00:32, 5 July 2013
• 2000 AMC 8 Problems/Problem 14
  ==Problem== ...o <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\b
  2 KB (249 words) - 05:23, 31 December 2022
• 2000 AMC 8 Problems/Problem 11
  ==Problem== \textbf{(D)}\ 18 \qquad
  2 KB (382 words) - 15:59, 1 January 2024
• 1984 AHSME Problems/Problem 29
  ==Problem== ...ome point <math>X</math>. We have <math>XA=\sqrt{6}</math>, <math>OA=\sqrt{18}</math>, <math>\angle OXA = 90</math> because the line is tangent to the ci
  4 KB (614 words) - 20:09, 12 September 2022
• 1967 AHSME Problems
  ==Problem 1== [[1967 AHSME Problems/Problem 1|Solution]]
  20 KB (3,108 words) - 14:14, 20 February 2020
• 2003 AMC 8 Problems
  ==Problem 1== [[2003 AMC 8 Problems/Problem 1|Solution]]
  16 KB (2,236 words) - 12:02, 19 February 2024
• 1998 AJHSME Problems/Problem 23
  ==Problem== fill((21,0)--(18,3sqrt(3))--(24,3sqrt(3))--cycle,black);
  2 KB (279 words) - 15:23, 29 May 2021
• 2003 AMC 10B Problems/Problem 19
  ==Problem== {{AMC10 box|year=2003|ab=B|num-b=18|num-a=20}}
  4 KB (606 words) - 13:19, 9 July 2021
• 1984 AHSME Problems/Problem 15
  ==Problem 15== <math> \mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad
  3 KB (493 words) - 18:16, 4 June 2021
• 1996 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1996 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 00:30, 2 October 2014
• 2005 CEMC Gauss (Grade 7)
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2005 CEMC Gauss (Grade 7) Problems/Problem 1]]
  2 KB (179 words) - 03:53, 24 October 2014
• 2005 CEMC Gauss (Grade 7) Problems
  == Problem 1 == [[2005 CEMC Gauss (Grade 7) Problems/Problem 1|Solution]]
  16 KB (2,317 words) - 03:54, 24 October 2014
• 1999 AMC 8 Problems/Problem 11
  ==Problem== ...7 + 4 = \boxed{\text{(D)}\ 24}</math>. Note: Two other common sums, <math>18</math> and <math>21</math>, are also possible.
  1 KB (181 words) - 18:23, 7 November 2020
• 2003 AMC 10B Problems/Problem 25
  ==Problem== <math> 5+18 = 23 </math>.
  6 KB (882 words) - 11:45, 12 November 2023
• 2009 AMC 8 Problems
  ==Problem 1== [[2009 AMC 8 Problems/Problem 1|Solution]]
  18 KB (2,551 words) - 18:46, 27 February 2024
• 1953 AHSME Problems/Problem 34
  ==Problem 34== <math>\textbf{(A)}\ 18\text{ inches} \qquad
  655 bytes (96 words) - 01:24, 4 January 2019
• 2004 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2004 AMC 8 Problems/Problem 1]]
  1 KB (138 words) - 21:47, 25 November 2013
• 2002 AMC 8 Problems/Problem 7
  == Problem == label("SWEET TOOTH", (9.5,18), N);
  1 KB (196 words) - 00:41, 5 July 2013
• 2002 AMC 8 Problems/Problem 6
  ==Problem== ...lows in at the rate of 20 milliliters per minute and drains at the rate of 18 milliliters per minute. Which one of these graphs shows the volume of water
  1 KB (201 words) - 17:51, 19 December 2023
• 2002 AMC 8 Problems/Problem 9
  ==Problem== /* AMC8 2002 #8, 9, 10 Problem */
  2 KB (240 words) - 03:56, 6 January 2024
• 2002 AMC 8 Problems/Problem 10
  ==Problem== /* AMC8 2002 #8, 9, 10 Problem */
  2 KB (253 words) - 22:32, 17 February 2024
• 2002 AMC 8 Problems/Problem 22
  ==Problem== /* AMC8 2002 #22 Problem */
  2 KB (241 words) - 20:03, 15 April 2023
• 2002 AMC 8 Problems/Problem 17
  ==Problem== ...nswer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
  2 KB (288 words) - 18:31, 20 January 2024
• 2002 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2002|num-b=18|num-a=20}}
  665 bytes (98 words) - 20:01, 15 April 2023
• 2002 AMC 8 Problems/Problem 15
  == Problem == draw(shift(18,0)*((0,1)--(2,1)--(3,0)--(3,3)--(2,2)--(1,3)--(1,2)--(0,2)--cycle));
  1 KB (197 words) - 04:47, 25 November 2019
• 2000 AMC 8 Problems/Problem 15
  ==Problem== ...h> \text{(A)}\ 12\qquad\text{(B)}\ 13\qquad\text{(C)}\ 15\qquad\text{(D)}\ 18\qquad\text{(E)}\ 21 </math>
  2 KB (325 words) - 20:56, 6 November 2013
• 2000 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2000|num-b=16|num-a=18}}
  1 KB (140 words) - 20:05, 15 April 2023
• 2000 AMC 8 Problems/Problem 19
  == Problem == {{AMC8 box|year=2000|num-b=18|num-a=20}}
  2 KB (383 words) - 16:58, 12 January 2024
• 2000 AMC 8 Problems/Problem 25
  ==Problem== To quickly solve this multiple choice problem, make the (not necessarily valid, but very convenient) assumption that <mat
  3 KB (484 words) - 13:59, 22 October 2023
• 2005 AMC 8 Problems/Problem 5
  ==Problem== ...th>90-3(24)=18</math> cans are left. After one <math>12</math>-pack, <math>18-12=6</math> cans are left. Then buy one more <math>6</math>-pack. The total
  625 bytes (96 words) - 01:08, 5 July 2013
• 1993 AJHSME Problems/Problem 11
  == Problem == label("$80$",(18,-.25),S);
  2 KB (238 words) - 00:11, 5 July 2013
• 1993 AJHSME Problems/Problem 10
  == Problem == unitsize(18);
  1 KB (176 words) - 12:27, 27 June 2023
• 2009 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2009 AMC 8 Problems/Problem 1|Problem 1]]
  2 KB (152 words) - 18:45, 21 November 2018
• 2003 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2003 AMC 8 Problems/Problem 1]]
  1 KB (146 words) - 18:36, 6 November 2020
• 1999 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=1999|num-b=16|num-a=18}}
  2 KB (241 words) - 15:00, 21 December 2022
• 1999 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=1999|num-b=18|num-a=20}}
  2 KB (266 words) - 22:26, 24 January 2024
• 1999 AMC 8 Problems/Problem 20
  ==Problem== unitsize(18);
  3 KB (422 words) - 22:24, 24 January 2024
• 1997 AJHSME Problems/Problem 17
  ==Problem== {{AJHSME box|year=1997|num-b=16|num-a=18}}
  3 KB (427 words) - 20:50, 26 May 2021
• 1997 AJHSME Problems/Problem 19
  ==Problem== ...f the preceeding number.) Thus, <math>\frac{a}{2} = 9</math>, and <math>a=18</math>.
  2 KB (286 words) - 22:00, 21 April 2017
• 1996 AJHSME Problems/Problem 4
  ==Problem== ...number is <math>18</math>. Adding all the numbers gives <math>8\cdot 36 + 18 = 306</math>.
  2 KB (348 words) - 00:24, 5 July 2013
• 1996 AJHSME Problems/Problem 14
  ==Problem== unitsize(18);
  2 KB (254 words) - 00:24, 5 July 2013
• 1996 AJHSME Problems/Problem 17
  ==Problem== {{AJHSME box|year=1996|num-b=16|num-a=18}}
  2 KB (257 words) - 11:20, 22 March 2015
• 1996 AJHSME Problems/Problem 19
  ==Problem== unitsize(18);
  2 KB (223 words) - 00:25, 5 July 2013
• 1995 AJHSME Problems/Problem 2
  == Problem== ...lder than Inez, who is <math>15</math>. Therefore, Zack is <math>15 + 3 = 18</math> years old.
  749 bytes (108 words) - 22:08, 19 June 2020
• 1995 AJHSME Problems/Problem 4
  ==Problem== <math>\text{(A)}\ 18 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 27 \qquad \t
  986 bytes (149 words) - 00:15, 5 July 2013
• 1995 AJHSME Problems/Problem 5
  ==Problem== ...xt{(B)}\ 15 \qquad \text{(C)}\ 16 \qquad \text{(D)}\ 17 \qquad \text{(E)}\ 18</math>
  2 KB (225 words) - 13:06, 1 April 2022
• 1995 AJHSME Problems/Problem 6
  ==Problem== <math>\text{(A)}\ 9 \qquad \text{(B)}\ 18 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 72 \qquad \text{(D)}\ 81</math>
  2 KB (277 words) - 11:37, 27 June 2023
• 1995 AJHSME Problems/Problem 11
  ==Problem== ...e distance. They set off in opposite directions around the outside of the 18-block area as shown. When they meet for the first time, they will be close
  1 KB (214 words) - 00:15, 5 July 2013
• 1992 AJHSME Problems/Problem 11
  == Problem == draw((15,0)--(15,16)--(18,16)--(18,0));
  2 KB (226 words) - 00:09, 5 July 2013
• 1997 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1997 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 00:30, 2 October 2014
• 1997 AHSME Problems
  == Problem 1 == [[1997 AHSME Problems/Problem 1|Solution]]
  17 KB (2,590 words) - 13:38, 19 February 2020
• 1997 AHSME Problems/Problem 10
  ==Problem== ...}\ \frac{1}{2}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{11}{18} </math>
  1 KB (213 words) - 14:12, 5 July 2013
• 1997 AHSME Problems/Problem 11
  ==Problem== ...he first five games. If her average after ten games was greater than <math>18</math>, what is the least number of points she could have scored in the ten
  1 KB (208 words) - 14:12, 5 July 2013
• 1997 AHSME Problems/Problem 17
  ==Problem== {{AHSME box|year=1997|num-b=16|num-a=18}}
  1 KB (200 words) - 10:31, 25 September 2016
• 1997 AHSME Problems/Problem 18
  ==Problem== <math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20 </math>
  2 KB (364 words) - 14:13, 5 July 2013
• 1997 AHSME Problems/Problem 19
  ==Problem== <math> \textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qqu
  3 KB (422 words) - 19:00, 9 August 2015
• 1997 AHSME Problems/Problem 22
  ==Problem== ...1 + 8 = 38</math> dollars. This means that Dick must have <math>56 - 38 = 18</math> dollars. However, the difference between Carlos and Dick is not <ma
  2 KB (343 words) - 17:46, 9 January 2015
• 2019 AMC 10A
  **[[2019 AMC 10A Problems/Problem 1|Problem 1]] **[[2019 AMC 10A Problems/Problem 2|Problem 2]]
  2 KB (169 words) - 15:51, 9 February 2019
• 2019 AMC 10B
  **[[2019 AMC 10B Problems/Problem 1|Problem 1]] **[[2019 AMC 10B Problems/Problem 2|Problem 2]]
  2 KB (169 words) - 13:06, 14 February 2019
• 1997 AHSME Problems/Problem 25
  ==Problem== ...th>AA^{\prime} = 10</math>, <math>BB^{\prime}= 8</math>, <math>CC^\prime = 18</math>, and <math>DD^\prime = 22</math> and <math>M</math> and <math>N</mat
  2 KB (307 words) - 22:13, 7 December 2023
• 1997 AHSME Problems/Problem 27
  ==Problem== To confirm that our original period works, we may see that <math>f(18) = -6</math>, <math>f(19) = -7</math>, <math>f(20) = -8</math>, <math>f(21)
  3 KB (456 words) - 14:14, 5 July 2013
• 2008 AMC 8
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2008 AMC 8 Problems/Problem 1]]
  1 KB (135 words) - 18:42, 6 November 2020
• 2008 AMC 8 Problems/Problem 5
  ==Problem== \textbf{(C)}\ 18\qquad
  712 bytes (103 words) - 19:20, 8 August 2021
• 2008 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2008|num-b=16|num-a=18}}
  984 bytes (149 words) - 00:26, 2 May 2022
• 2008 AMC 8 Problems/Problem 18
  == Problem == draw((-18,1)--(-12, 1), EndArrow);
  2 KB (210 words) - 13:37, 19 October 2020
• 2008 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2008|num-b=18|num-a=20}}
  1 KB (179 words) - 00:41, 2 July 2023
• 2010 AMC 8 Problems/Problem 21
  ==Problem== ...the third day she read <math>1/3</math> of the remaining pages plus <math>18</math> pages. She then realized that there were only <math>62</math> pages
  2 KB (350 words) - 22:35, 21 April 2024
• 1950 AHSME Problems/Problem 19
  == Problem == {{AHSME 50p box|year=1950|num-b=18|num-a=20}}
  827 bytes (150 words) - 00:53, 12 October 2020
• 2009 AMC 8 Problems/Problem 4
  ==Problem== ...={(17,0), (17,1), (17,2), (17,3), (17,4), (18,0), (18,1), (18,2), (18,3), (18,4), (19,0), (19,1), (19,2), (19,3), (19,4)};
  5 KB (633 words) - 01:56, 26 November 2023
• 2009 AMC 8 Problems/Problem 16
  ==Problem== <math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24</math>
  673 bytes (90 words) - 20:22, 28 August 2016
• 2009 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2009|num-b=16|num-a=18}}
  2 KB (307 words) - 07:08, 28 December 2023
• 2009 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2009|num-b=18|num-a=20}}
  2 KB (274 words) - 19:38, 15 April 2023
• 1970 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1970 AHSME Problems/Problem 1|Problem 1]]
  2 KB (196 words) - 14:11, 20 February 2020
• 1994 AJHSME Problems/Problem 10
  ==Problem== ...rror, the divisors of <math>36</math> are found to be <math>1,2,3,4,6,9,12,18,36</math>, for a total of <math>9</math>. However, <math>1</math> and <math
  1 KB (195 words) - 08:42, 26 May 2021
• 1994 AJHSME Problems/Problem 14
  ==Problem== <math>\text{(A)}\ 9 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 36</math>
  624 bytes (93 words) - 01:06, 7 November 2018
• 1994 AJHSME Problems/Problem 17
  ==Problem== ...ards for the first hour, <math>19</math> cubic yards for the second, <math>18</math> for the third, etc., always shoveling one cubic yard less per hour t
  1 KB (175 words) - 00:13, 5 July 2013
• 1994 AJHSME Problems/Problem 18
  ==Problem== draw((17,2)--(18,8)--(22,8)--(23,2));
  2 KB (367 words) - 13:30, 30 October 2016
• 1994 AJHSME Problems/Problem 19
  ==Problem== {{AJHSME box|year=1994|num-b=18|num-a=20}}
  1 KB (209 words) - 20:12, 21 February 2021
• 1994 AJHSME Problems/Problem 21
  ==Problem== ...ext{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 12 \qquad \text{(E)}\ 18</math>
  684 bytes (108 words) - 01:10, 7 November 2018
• 1994 AJHSME Problems/Problem 25
  ==Problem== <math>99 \cdot 44 = 4356</math> and <math>4+5+3+6 = 18 = 9 \cdot 2</math>
  1 KB (187 words) - 09:30, 12 January 2024
• 1996 AHSME Problems
  ==Problem 1== [[1996 AHSME Problems/Problem 1|Solution]]
  15 KB (2,343 words) - 13:39, 19 February 2020
• 1996 AHSME Problems/Problem 17
  ==Problem== ...math>x-2 = 6\sqrt{3} - 2</math>, and <math>y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}</math>.
  3 KB (379 words) - 16:24, 12 May 2022
• 1996 AHSME Problems/Problem 19
  ==Problem== {{AHSME box|year=1996|num-b=18|num-a=20}}
  3 KB (540 words) - 21:32, 10 July 2017
• 1996 AHSME Problems/Problem 24
  ==Problem== ...2 = 18</math>. Therefore, the sum of the first 1234 terms is <math>2401 + 18 = \boxed{2419}</math>.
  3 KB (448 words) - 20:39, 2 November 2023
• 1992 AJHSME Problems/Problem 17
  ==Problem== {{AJHSME box|year=1992|num-b=16|num-a=18}}
  768 bytes (114 words) - 00:09, 5 July 2013
• 1992 AJHSME Problems/Problem 19
  ==Problem== <math>\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98</math>
  885 bytes (138 words) - 22:26, 25 May 2021
• 1992 AJHSME Problems/Problem 20
  ==Problem== draw((14,-1)--(15,-1)--(15,0)--(16,0)--(16,-1)--(18,-1)--(18,-2)--(17,-2)--(17,-3)--(16,-3)--(16,-2)--(14,-2)--cycle);
  1 KB (239 words) - 00:10, 5 July 2013
• 1992 AJHSME Problems/Problem 22
  ==Problem== <math>\text{(A)}\ 15 \qquad \text{(B)}\ 17 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20</math>
  1 KB (160 words) - 00:10, 5 July 2013
• 1992 AJHSME Problems/Problem 24
  ==Problem== ...)}\ 7.7 \qquad \text{(B)}\ 12.1 \qquad \text{(C)}\ 17.2 \qquad \text{(D)}\ 18 \qquad \text{(E)}\ 27</math>
  1 KB (191 words) - 00:10, 5 July 2013
• 2010 PUMaC Problems
  [[2011 PuMAC Problems/Algebra Problem A1|Solution]] [[2011 PuMAC Problems/Algebra Problem A2|Solution]]
  22 KB (3,694 words) - 23:58, 3 June 2022
• 2016 AMC 10A Problems/Problem 19
  == Problem == This problem breaks down into finding <math>QP:PB</math> and <math>DQ:QB</math>. We can
  8 KB (1,386 words) - 15:10, 8 October 2023
• 1993 AJHSME Problems/Problem 17
  ==Problem== {{AJHSME box|year=1993|num-b=16|num-a=18}}
  1 KB (187 words) - 09:35, 1 December 2020
• 1993 AJHSME Problems/Problem 19
  ==Problem== {{AJHSME box|year=1993|num-b=18|num-a=20}}
  768 bytes (91 words) - 12:31, 27 June 2023
• 1995 AJHSME Problems/Problem 17
  ==Problem== {{AJHSME box|year=1995|num-b=16|num-a=18}}
  1 KB (125 words) - 03:15, 23 December 2012
• 1995 AJHSME Problems/Problem 19
  ==Problem== {{AJHSME box|year=1995|num-b=18|num-a=20}}
  2 KB (227 words) - 03:17, 23 December 2012
• 1995 AJHSME Problems/Problem 24
  ==Problem== W = (18,0); X = (30,0); Y = (38,6); Z = (26,6);
  1 KB (185 words) - 13:33, 21 April 2024
• 2003 AMC 8 Problems/Problem 18
  ==Problem== <asy>/* AMC8 2003 #18 Problem */
  2 KB (250 words) - 22:17, 5 January 2024
• 1984 AHSME Problems/Problem 26
  ==Problem== ...9 \qquad \mathrm{(B) \ }12 \qquad \mathrm{(C) \ } 15 \qquad \mathrm{(D) \ }18 \qquad \mathrm{(E) \ } \text{Not uniquely determined} </math>
  2 KB (303 words) - 20:28, 2 October 2023
• 1984 AHSME Problems/Problem 30
  ==Problem== ...circ \qquad \text{(C) } \frac{1}{9}\cos40^\circ \qquad \text{(D) }\frac{1}{18}\cos20^\circ \qquad \text{(E) } \text{None of these} </math>
  3 KB (447 words) - 21:21, 17 July 2020
• 1989 AHSME Problems/Problem 17
  ==Problem== {{AHSME box|year=1989|num-b=16|num-a=18}}
  1 KB (171 words) - 13:33, 24 October 2014
• 1950 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1950 AHSME Problems/Problem 1|Problem 1]]
  3 KB (254 words) - 14:26, 20 February 2020
• 1950 AHSME Problems
  == Problem 1 == [[1950 AHSME Problems/Problem 1|Solution]]
  22 KB (3,306 words) - 19:50, 3 May 2023
• 1968 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1968 AHSME Problems/Problem 1|Problem 1]]
  2 KB (196 words) - 14:13, 20 February 2020
• 2006 AMC 8 Problems/Problem 1
  == Problem == ...bf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18 </math>
  752 bytes (103 words) - 13:14, 18 March 2020
• 1994 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1994 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 03:45, 29 September 2014
• 2006 AMC 8 Problems/Problem 11
  == Problem == ...bf{(A)}\ 13\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19 </math>
  971 bytes (130 words) - 13:44, 25 April 2021
• 1985 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1985 AHSME Problems/Problem 1|Problem 1]]
  2 KB (175 words) - 00:33, 2 October 2014
• 1985 AHSME Problems
  ==Problem 1== ...qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19 </math>
  17 KB (2,488 words) - 03:26, 20 March 2024
• 2003 AMC 8 Problems/Problem 9
  == Problem == <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 75\qquad\tex
  2 KB (304 words) - 19:14, 28 August 2016
• 2003 AMC 8 Problems/Problem 10
  == Problem == ...bf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 24</math>
  2 KB (245 words) - 00:47, 5 July 2013
• 1985 AHSME Problems/Problem 1
  ==Problem== ...qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19 </math>
  729 bytes (91 words) - 18:21, 19 March 2024
• 1950 AHSME Problems/Problem 6
  == Problem== (y+3)^2 -6y-18+10y+2&=0\\
  976 bytes (151 words) - 11:57, 5 July 2013
• 1985 AHSME Problems/Problem 17
  ==Problem== {{AHSME box|year=1985|num-b=16|num-a=18}}
  2 KB (340 words) - 22:39, 19 March 2024
• 1985 AHSME Problems/Problem 18
  ==Problem== Six bags of marbles contain <math> 18, 19, 21, 23, 25 </math> and <math> 34 </math> marbles, respectively. One ba
  2 KB (241 words) - 22:50, 19 March 2024
• 1985 AHSME Problems/Problem 19
  ==Problem== {{AHSME box|year=1985|num-b=18|num-a=20}}
  3 KB (519 words) - 23:10, 19 March 2024
• 1985 AHSME Problems/Problem 28
  ==Problem== ...t{11}}{9} \cdot \frac{25-11}{36} \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{7}{18} \\ &= \frac{35\sqrt{11}}{162}.\end{align*}</cmath>
  4 KB (657 words) - 02:54, 20 March 2024
• 1950 AHSME Problems/Problem 17
  == Problem == {{AHSME 50p box|year=1950|num-b=16|num-a=18}}
  843 bytes (134 words) - 00:51, 12 October 2020
• 2006 AMC 8 Problems/Problem 19
  == Problem == {{AMC8 box|year=2006|n=II|num-b=18|num-a=20}}
  1 KB (195 words) - 21:47, 2 January 2023
• 2010 AMC 8 Problems
  ==Problem 1== [[2010 AMC 8 Problems/Problem 1 | Solution]]
  18 KB (2,768 words) - 21:05, 9 January 2024
• 1950 AHSME Problems/Problem 24
  == Problem == <math>x^2 - 9x + 18 = 0</math> Get all terms on one side
  2 KB (343 words) - 00:57, 12 October 2020
• 2011 AMC 8
  ** [[2011 AMC 8 Problems/Problem 1]] ** [[2011 AMC 8 Problems/Problem 2]]
  1 KB (102 words) - 18:46, 21 November 2018
• 2003 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2003|num-b=16|num-a=18}}
  2 KB (265 words) - 23:23, 22 March 2022
• 2011 AMC 8 Problems
  ==Problem 1== [[2011 AMC 8 Problems/Problem 1|Solution]]
  16 KB (2,371 words) - 17:34, 9 January 2024
• 2003 AMC 8 Problems/Problem 5
  ==Problem== <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 30 </math>
  638 bytes (94 words) - 00:46, 5 July 2013
• 2011 AMC 8 Problems/Problem 14
  ==Problem== <math> \textbf{(A) } \dfrac7{18} \qquad\textbf{(B) } \dfrac7{15} \qquad\textbf{(C) } \dfrac{22}{45} \qquad\
  2 KB (250 words) - 15:10, 17 December 2023
• 2011 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2011|num-b=16|num-a=18}}
  924 bytes (137 words) - 15:08, 17 December 2023
• 2011 AMC 8 Problems/Problem 19
  ==Problem== J=(18,-20);
  1 KB (158 words) - 15:06, 17 December 2023
• 2003 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2003|num-b=18|num-a=20}}
  737 bytes (88 words) - 00:47, 5 July 2013
• 2006 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2006|num-b=16|num-a=18}}
  2 KB (330 words) - 05:08, 22 July 2018
• 1950 AHSME Problems/Problem 26
  == Problem== More simply, we can just simulate the problem, if we have <math>m = 10</math>, that means the right side must be 1, so th
  2 KB (251 words) - 19:38, 29 October 2020
• Mock Geometry AIME 2011 Problems/Problem 4
  ==Problem== ...qrt{3}}{\sqrt{19}}</math>. Solving for <math>s</math> yields <math>s=\frac{18}{\sqrt{19}}</math>. Then the perimeter of the triangle is <math>3s=\frac{54
  2 KB (370 words) - 03:38, 25 December 2022
• Mathtime Version 1 Issue 2
  ===Problem 1=== ===Problem 2===
  15 KB (2,444 words) - 21:46, 1 January 2012
• Mock Geometry AIME 2011 Problems/Problem 5
  ==Problem== ...<math>BD+CD=BC=40</math>. Solving this system of equations yields <math>BD=18,CD=22</math>. Thus, <math>DH=CD-CH=22-20=2</math>.
  4 KB (665 words) - 03:40, 25 December 2022
• 2011 AMC 8 Problems/Problem 13
  ==Problem== <math> \textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 25 </math>
  1 KB (185 words) - 10:00, 13 January 2024
• 1967 AHSME Problems/Problem 1
  ==Problem== ...+ b + 9</math> is a multiple of <math>9</math>. So, <cmath>5 + b + 9 = 9, 18, 27, 36...</cmath>
  977 bytes (141 words) - 01:45, 16 August 2023
• 1973 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[1973 AHSME Problems/Problem 1|Problem 1]]
  2 KB (203 words) - 13:52, 20 February 2020
• 1992 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1992 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 00:31, 2 October 2014
• 1991 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1991 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 00:31, 2 October 2014
• 1988 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1988 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 00:32, 2 October 2014
• 1989 AHSME Problems
  == Problem 1 == [[1989 AHSME Problems/Problem 1|Solution]]
  15 KB (2,247 words) - 13:44, 19 February 2020
• 2012 AMC 10A
  **[[2012 AMC 10A Problems/Problem 1|Problem 1]] **[[2012 AMC 10A Problems/Problem 2|Problem 2]]
  2 KB (173 words) - 06:29, 7 November 2022
• 2012 AMC 10A Problems/Problem 23
  == Problem == ...can have at most 4 friends since they cannot be all friends (stated in the problem).
  5 KB (815 words) - 17:53, 12 October 2023
• 2012 AMC 10A Problems/Problem 17
  ==Problem== {{AMC10 box|year=2012|ab=A|num-b=16|num-a=18}}
  6 KB (1,024 words) - 01:35, 1 October 2023
• 2012 AMC 10A Problems/Problem 19
  ==Problem 19== {{AMC10 box|year=2012|ab=A|num-b=18|num-a=20}}
  3 KB (422 words) - 17:11, 21 August 2021
• 2012 AMC 10A Problems
  == Problem 1 == [[2012 AMC 10A Problems/Problem 1|Solution]]
  13 KB (1,994 words) - 01:31, 22 December 2023
• 2012 AMC 12A
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2012 AMC 12A Problems/Problem 1|Problem 1]]
  2 KB (201 words) - 21:46, 6 October 2014
• 2012 AMC 10A Problems/Problem 18
  ...12A Problems|2012 AMC 12A #14]] and [[2012 AMC 10A Problems|2012 AMC 10A #18]]}} == Problem 14 ==
  5 KB (775 words) - 22:33, 22 October 2023
• 2012 AMC 12A Problems
  == Problem 1 == [[2012 AMC 12A Problems/Problem 1|Solution]]
  14 KB (2,197 words) - 13:34, 12 August 2020
• 2012 AMC 12A Problems/Problem 14
  #REDIRECT [[2012 AMC 10A Problems/Problem 18]]
  46 bytes (5 words) - 14:29, 12 February 2012
• 2012 AMC 12A Problems/Problem 17
  == Problem == ...bf{(B)}\ 13\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 </math>
  3 KB (519 words) - 19:01, 30 March 2024
• 1991 AHSME Problems/Problem 17
  == Problem == {{AHSME box|year=1991|num-b=16|num-a=18}}
  1 KB (195 words) - 12:38, 13 December 2016
• 1992 AHSME Problems/Problem 17
  == Problem == {{AHSME box|year=1992|num-b=16|num-a=18}}
  2 KB (342 words) - 13:19, 4 January 2021
• 1989 AHSME Problems/Problem 19
  == Problem == <math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(
  4 KB (649 words) - 10:04, 20 May 2021
• 1989 AHSME Problems/Problem 6
  == Problem == ...b}=\frac{1}{2}\cdot\frac{6}{a}\cdot\frac{6}{b}</math> so that <math> \frac{18}{ab} = 6</math>, simplifying we would get <math>ab=3</math>. Hence the answ
  911 bytes (141 words) - 21:12, 8 September 2023
• 1989 AHSME Problems/Problem 10
  == Problem == ...} \qquad \mathrm{(C) \ 16 } \qquad \mathrm{(D) \ 17 } \qquad \mathrm{(E) \ 18 } </math>
  825 bytes (125 words) - 13:48, 3 February 2016
• 2012 AMC 10B
  **[[2012 AMC 10B Problems/Problem 1|Problem 1]] **[[2012 AMC 10B Problems/Problem 2|Problem 2]]
  2 KB (175 words) - 21:13, 18 July 2021
• 2012 AMC 12B
  ...contains the full set of test problems. The rest contain each individual problem and its solution. **[[2012 AMC 12B Problems/Problem 1|Problem 1]]
  2 KB (197 words) - 21:19, 6 October 2014
• 2012 AMC 12B Problems/Problem 1
  == Problem == Each third-grade classroom at Pearl Creek Elementary has <math>18</math> students and <math>2</math> pet rabbits. How many more students than
  1 KB (142 words) - 21:05, 8 February 2014
• 2012 AMC 12B Problems/Problem 14
  == Problem== ...ded 50 to any number before this to obtain a number in the range <math>[16,18]</math>, hence the minimum <math>N</math> is 16 with the sum of digits bein
  3 KB (463 words) - 09:46, 24 April 2024
• 2012 AMC 10B Problems
  == Problem 1 == Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 rabbits. How many more students than rabbits are there in a
  18 KB (2,350 words) - 18:48, 9 July 2023
• 2012 AMC 10B Problems/Problem 1
  == Problem == Each third-grade classroom at Pearl Creek Elementary has <math>18</math> students and <math>2</math> pet rabbits. How many more students than
  633 bytes (90 words) - 13:32, 16 February 2016
• 2012 AMC 10B Problems/Problem 5
  == Problem 5 == <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\tex
  836 bytes (132 words) - 15:02, 1 May 2021
• Mathematical Red Rocket Competition Problems
  Problem 1 Problem 2
  390 bytes (37 words) - 17:34, 27 September 2012
• 2012 AMC 12B Problems
  == Problem 1 == Each third-grade classroom at Pearl Creek Elementary has 18 students and 2 pet rabbits. How many more students than rabbits are there i
  20 KB (2,681 words) - 09:47, 29 June 2023
• 2012 AMC 12B Problems/Problem 17
  ==Problem== {{AMC12 box|year=2012|ab=B|num-b=16|num-a=18}}
  12 KB (2,183 words) - 21:05, 23 December 2023
• 2012 AMC 12B Problems/Problem 7
  == Problem == <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 18.5\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 20.5\qquad\textbf{(E)}\ 22.5 </m
  1 KB (180 words) - 07:37, 29 June 2023
• 2012 AMC 12B Problems/Problem 19
  ==Problem== {{AMC12 box|year=2012|ab=B|num-b=18|num-a=20}}
  5 KB (815 words) - 21:59, 19 September 2023
• 2012 AMC 12B Problems/Problem 12
  ==Problem== ...is, there are 20 strings with 1 zero, 19 strings with 2 consecutive zeros, 18 strings with 3 consecutive zeros, ..., 1 string with 20 consecutive zeros.
  5 KB (822 words) - 20:15, 5 May 2024
• 2010 AMC 8 Problems/Problem 25
  ==Problem== <math> \textbf{(A)}\ 13 \qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24 </math>
  4 KB (595 words) - 13:38, 4 April 2024
• 2010 AMC 8 Problems/Problem 19
  == Problem == {{AMC8 box|year=2010|num-b=18|num-a=20}}
  2 KB (279 words) - 09:04, 10 March 2023
• 2004 AMC 8 Problems/Problem 3
  == Problem == ...ered one meal. The portions were so large, there was enough food for <math>18</math> people. If they shared, how many meals should they have ordered to h
  700 bytes (106 words) - 00:54, 5 July 2013
• 2012 AMC 10B Problems/Problem 19
  ==Problem== We know these points from the problem statement:
  3 KB (509 words) - 14:23, 23 August 2022
• 2012 AIME I Problems/Problem 3
  ==Problem== The problem we must solve is to distribute meals <math>\text{BBCCCFFF}</math> to orders
  3 KB (572 words) - 18:56, 13 June 2023
• 2010 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2010|num-b=16|num-a=18}}
  3 KB (459 words) - 21:04, 23 December 2023
• 2012 AIME I Problems
  == Problem 1 == [[2012 AIME I Problems/Problem 1|Solution]]
  10 KB (1,617 words) - 14:49, 2 June 2023
• 2012 AIME I Problems/Problem 7
  ==Problem== pair G = 3.6*dir(18), H = 3.6*dir(90), I = 3.6*dir(162), J = 3.6*dir(234), K = 3.6*dir(306);
  6 KB (1,058 words) - 01:49, 25 November 2023
• 2012 AIME II Problems/Problem 15
  == Problem 15 == ...is. Most of the people who've done EGMO Chapter 8 should recognize this as problem 8.32 (2009 Russian Olympiad) with the computational finish afterwards.
  13 KB (2,298 words) - 12:56, 10 September 2023
• Mock AIME II 2012 Problems/Problem 2
  ==Problem== ...ct square will be <math>k=18</math> and will take more than <math>60+(19^2-18^2)=97</math> as the value of <math>m</math>, and hence all other perfect sq
  4 KB (641 words) - 00:06, 6 October 2017
• 2017 AMC 8 Problems
  ==Problem 1== [[2017 AMC 8 Problems/Problem 1|Solution
  12 KB (1,771 words) - 21:13, 20 January 2024
• Mock AIME II 2012 Problems/Problem 9
  ==Problem== draw((0,0)--(18,10));
  7 KB (1,069 words) - 14:04, 27 December 2012
• Mock AIME II 2012 Problems/Problem 15
  ==Problem:== ...us <math>f(3)-f(2)=-6</math>, <math>f(4)-f(3)=-12</math>, <math>f(5)-f(4)=-18, f(6)-f(5)=-24</math>. Let <math>f(x)=b_n*x^n+b_{n-1}*x^{n-1}+b_{n-2}*x^{n
  4 KB (660 words) - 15:55, 8 March 2015
• Mock AIME I 2012 Problems/Problem 15
  ==Problem== The problem is equivalent to finding the number of complex numbers <math>z_i</math> suc
  3 KB (488 words) - 20:05, 10 March 2015
• 1951 AHSME Problems/Problem 7
  == Problem == <math> \textbf{(A)}\ \text{greater by }.18 \qquad\textbf{(B)}\ \text{the same} \qquad\textbf{(C)}\ \text{less}</math>
  905 bytes (141 words) - 12:20, 5 July 2013
• 1951 AHSME Problems/Problem 17
  == Problem == {{AHSME 50p box|year=1951|num-b=16|num-a=18}}
  1 KB (170 words) - 20:18, 30 April 2015
• 1951 AHSME Problems/Problem 19
  == Problem == {{AHSME 50p box|year=1951|num-b=18|num-a=20}}
  819 bytes (112 words) - 08:13, 31 December 2023
• 1966 AHSME Problems/Problem 18
  == Problem == {{AHSME box|year=1966|num-b=17|num-a=18}}
  492 bytes (69 words) - 03:33, 15 February 2019
• 1974 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[1974 AHSME Problems/Problem 1|Problem 1]]
  2 KB (178 words) - 03:51, 29 September 2014
• 1974 AHSME Problems
  ==Problem 1== [[1974 AHSME Problems/Problem 1|Solution]]
  15 KB (2,151 words) - 14:04, 19 February 2020
• 1974 AHSME Problems/Problem 17
  ==Problem== {{AHSME box|year=1974|num-b=16|num-a=18}}
  549 bytes (75 words) - 12:43, 5 July 2013
• 1974 AHSME Problems/Problem 19
  ==Problem== {{AHSME box|year=1974|num-b=18|num-a=20}}
  2 KB (250 words) - 12:43, 5 July 2013
• 2006 SMT/General
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[2006 SMT/General Problems/Problem 1|Problem 1]]
  2 KB (167 words) - 14:19, 27 May 2012
• 2006 SMT/General Problems
  ==Problem 1== [[2006 SMT/General Problems/Problem 1|Solution]]
  10 KB (1,477 words) - 16:02, 27 May 2012
• 1952 AHSME Problems
  == Problem 1 == [[1952 AHSME Problems/Problem 1|Solution]]
  23 KB (3,556 words) - 15:35, 30 December 2023
• 1953 AHSME Problems
  ==Problem 1== [[1953 AHSME Problems/Problem 1|Solution]]
  21 KB (3,123 words) - 14:24, 20 February 2020
• 1954 AHSME Problems
  == Problem 1== [[1954 AHSME Problems/Problem 1|Solution]]
  23 KB (3,535 words) - 16:29, 24 April 2020
• 1955 AHSME Problems
  == Problem 1== [[1955 AHSME Problems/Problem 1|Solution]]
  22 KB (3,509 words) - 21:29, 31 December 2023
• 1980 AHSME Problems
  == Problem 1 == [[1980 AHSME Problems/Problem 1|Solution]]
  15 KB (2,302 words) - 10:47, 30 April 2021
• 1989 AHSME Problems/Problem 30
  ==Problem== We approach this problem using Linearity of Expectation. Consider a pair of two people standing next
  2 KB (293 words) - 17:13, 24 August 2020
• 2020 Mock Combo AMC 10
  ==Problem 1== [[2020 Mock Combo AMC 10 II Problems/Problem 1|Solution]]
  15 KB (2,452 words) - 03:03, 4 July 2020
• 2006 AMC 8 Problems/Problem 23
  ==Problem== divided by <math>5</math> are <math>3,8,13,18,23,28,33, \cdots</math> So <math>28</math> is the smallest possible number
  2 KB (383 words) - 20:15, 26 November 2023
• 2003 AMC 12B Problems/Problem 9
  == Problem == \text {(A) } 12 \qquad \text {(B) } 18 \qquad \text {(C) } 24 \qquad \text {(D) } 30 \qquad \text {(E) } 36
  818 bytes (128 words) - 19:05, 7 July 2020
• 2003 AMC 12B Problems/Problem 8
  ==Problem== ...y=\clubsuit (x)</math>. Since <math>x \leq 99</math>, we have <math>y \leq 18</math>. Thus if <math>\clubsuit (y)=3</math>, then <math>y=3</math> or <mat
  1 KB (159 words) - 15:38, 30 March 2020
• 2008 AMC 8 Problems/Problem 23
  ==Problem== ...)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20} </math>
  3 KB (418 words) - 21:28, 26 December 2023
• 2008 AMC 8 Problems
  ==Problem 1== [[2008 AMC 8 Problems/Problem 1|Solution]]
  14 KB (2,035 words) - 15:23, 26 January 2024
• 2012 AMC 8
  ** [[2012 AMC 8 Problems/Problem 1]] ** [[2012 AMC 8 Problems/Problem 2]]
  1 KB (102 words) - 18:47, 21 November 2018
• 2012 AMC 8 Problems
  ==Problem 1== [[2012 AMC 8 Problems/Problem 1|Solution]]
  13 KB (1,835 words) - 08:51, 8 March 2024
• 2012 AMC 8 Problems/Problem 17
  ==Problem== ...ares with area <math> 4 </math>, which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the or
  1 KB (163 words) - 00:42, 18 February 2024
• 2012 AMC 8 Problems/Problem 19
  ==Problem== ...2(r+g+b)=6+8+4=18</math>. It gives us all of the marbles are <math>r+g+b = 18/2 = 9</math>. So the answer is <math>\boxed{\textbf{(C)}\ 9}</math>.
  4 KB (620 words) - 00:30, 18 February 2024
• 2012 AMC 12B Problems/Problem 18
  == Problem 18 == This problem is worded awkwardly. More simply, it asks: “How many ways can you order n
  4 KB (715 words) - 00:50, 27 December 2022
• 2012 AMC 12B Problems/Problem 20
  == Problem 20 == <math>S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27</math>
  4 KB (670 words) - 07:14, 27 December 2022
• 2012 AMC 12B Problems/Problem 24
  == Problem == ...bf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 17\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 19 </math>
  6 KB (969 words) - 10:06, 5 November 2021
• 2004 AMC 8 Problems/Problem 20
  ==Problem== <math>\textbf{(A)}\ 12\qquad \textbf{(B)}\ 18\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36</math>
  827 bytes (131 words) - 22:43, 25 March 2022
• 2004 AMC 8 Problems/Problem 14
  ==Problem== <math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 18\frac12 \qquad \textbf{(C)}\ 22\frac12 \qquad \textbf{(D)}\ 27 \qquad \textb
  2 KB (238 words) - 10:13, 19 January 2024
• 2004 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2004|num-b=16|num-a=18}}
  3 KB (498 words) - 00:35, 30 December 2023
• 2004 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2004|num-b=18|num-a=20}}
  904 bytes (132 words) - 01:00, 5 July 2013
• 2004 AMC 8 Problems/Problem 15
  ==Problem== ...(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
  1 KB (215 words) - 20:20, 12 October 2020
• 2005 AMC 8 Problems/Problem 12
  ==Problem== ...(n+12)+(n+18)+(n+24)</math> is equal to <math>100</math>, as stated in the problem. We can write a very simple equation, that is: <math>5n+60=100</math>.
  2 KB (293 words) - 11:48, 4 May 2022
• 2005 AMC 8 Problems/Problem 17
  == Problem == {{AMC8 box|year=2005|num-b=16|num-a=18}}
  1 KB (160 words) - 13:34, 19 October 2020
• 2005 AMC 8 Problems/Problem 19
  ==Problem== pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0);
  1 KB (180 words) - 21:49, 2 January 2023
• 2005 AMC 8 Problems/Problem 21
  ==Problem== <math> \textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24 </math>
  1 KB (187 words) - 18:19, 8 January 2024
• 1993 AHSME Problems/Problem 21
  == Problem == \text{(B) } 18\quad
  1,013 bytes (161 words) - 17:49, 26 August 2017
• 2012 AMC 10B Problems/Problem 17
  ==Problem== {{AMC10 box|year=2012|ab=B|num-b=16|num-a=18}}
  2 KB (364 words) - 00:51, 17 January 2021
• 2013 AMC 12A Problems/Problem 24
  == Problem== <cmath> = \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286} </cmath
  3 KB (395 words) - 15:54, 8 November 2022
• 2013 AMC 10A
  **[[2013 AMC 10A Problems/Problem 1|Problem 1]] **[[2013 AMC 10A Problems/Problem 2|Problem 2]]
  2 KB (172 words) - 13:29, 26 January 2020
• 2013 AMC 12A
  **[[2013 AMC 12A Problems/Problem 1|Problem 1]] **[[2013 AMC 12A Problems/Problem 2|Problem 2]]
  2 KB (170 words) - 21:44, 6 October 2014
• 2013 AMC 12A Problems
  == Problem 1 == [[2013 AMC 12A Problems/Problem 1|Solution]]
  14 KB (2,206 words) - 19:31, 15 May 2024
• 2013 AMC 12A Problems/Problem 6
  == Problem== <math> \textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36 </math>
  3 KB (419 words) - 11:39, 10 March 2024
• 2013 AMC 10A Problems/Problem 11
  ==Problem== ...bf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25 </math>
  2 KB (336 words) - 13:29, 27 July 2021
• 2013 AMC 12A Problems/Problem 17
  == Problem 17 == {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}
  3 KB (502 words) - 01:36, 11 October 2020
• 2013 AMC 12A Problems/Problem 15
  == Problem == We tackle the problem by sorting it by how many stores are involved in the transaction.
  3 KB (441 words) - 10:10, 4 August 2020
• 2013 AMC 12A Problems/Problem 19
  == Problem== {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}
  3 KB (546 words) - 15:24, 19 September 2021
• 2013 AMC 10A Problems/Problem 9
  ==Problem== <math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </mat
  2 KB (257 words) - 13:07, 1 July 2023
• 2013 AMC 10A Problems/Problem 13
  ==Problem== We use a casework approach to solve the problem. These three digit numbers are of the form <math>\overline{xyx}</math>.(<ma
  2 KB (248 words) - 11:15, 21 January 2021
• 2013 AMC 10A Problems/Problem 15
  ==Problem== ...f{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math>
  2 KB (407 words) - 01:12, 22 September 2022
• 2013 AMC 10A Problems/Problem 17
  ==Problem== We get <math>30+18+24=72</math>.
  3 KB (458 words) - 16:10, 30 August 2023
• 2013 AMC 10A Problems/Problem 19
  ==Problem== ...tbf{(B)}\ 9\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 18 </math>
  1 KB (185 words) - 13:22, 2 April 2023
• 2013 AMC 10A Problems/Problem 21
  ==Problem== {{AMC12 box|year=2013|ab=A|num-b=16|num-a=18}}
  4 KB (698 words) - 16:28, 25 November 2020
• 2013 AMC 10B Problems/Problem 22
  ==Problem== When <math>J=9, S=18, A+E=B+F=C+G=D+H=9</math>
  4 KB (746 words) - 17:29, 30 September 2023
• 2013 AMC 10A Problems/Problem 23
  ==Problem== {{AMC12 box|year=2013|ab=A|num-b=18|num-a=20}}
  5 KB (846 words) - 23:02, 21 August 2023
• 2013 AMC 10A Problems
  ==Problem 1== [[2013 AMC 10A Problems/Problem 1|Solution]]
  12 KB (1,894 words) - 15:59, 3 January 2024
• 2013 AMC 12A Problems/Problem 20
  == Problem 20 == <cmath>\sum_{k=11}^{18} 3\cdot 19 \cdot (19-k) = 3\cdot 19 \cdot (1+\cdots+8) = 3\cdot 19\cdot 36<
  5 KB (885 words) - 10:14, 29 October 2023
• 2013 AMC 10B
  **[[2013 AMC 10B Problems/Problem 1|Problem 1]] **[[2013 AMC 10B Problems/Problem 2|Problem 2]]
  2 KB (175 words) - 18:17, 6 October 2014
• 2013 AMC 10B Problems/Problem 25
  ==Problem== Using the same logic for <math>b</math>, if <math>b < 18</math>, <math>2N \equiv 2b \pmod {36}</math>, and in the other case <math>2
  10 KB (1,623 words) - 15:44, 31 August 2022
• 2013 AMC 10B Problems/Problem 19
  ==Problem== {{AMC10 box|year=2013|ab=B|num-b=18|num-a=20}}
  5 KB (969 words) - 19:14, 15 August 2023
• 2013 AMC 10B Problems/Problem 16
  ==Problem== ...en <math>[ABC]=3\cdot 6=18</math>. Finally, <math>[AEDC]=\frac{3}{4}\times 18=\boxed{13.5}</math>.
  5 KB (761 words) - 19:33, 11 January 2024
• 2013 AMC 12B Problems
  ==Problem 1== [[2013 AMC 12B Problems/Problem 1|Solution]]
  16 KB (2,459 words) - 02:46, 30 January 2021
• 2013 AMC 12B
  **[[2013 AMC 12B Problems/Problem 1|Problem 1]] **[[2013 AMC 12B Problems/Problem 2|Problem 2]]
  2 KB (178 words) - 22:07, 30 September 2020
• 2013 AMC 12B Problems/Problem 10
  ==Problem== We can approach this problem by assuming he goes to the red booth first. You start with <math>75 \text{R
  4 KB (611 words) - 20:26, 7 August 2023
• 2013 AMC 12B Problems/Problem 12
  == Problem == ...)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18</math>
  3 KB (533 words) - 22:11, 3 October 2022
• 2013 AMC 12B Problems/Problem 17
  ==Problem== {{AMC12 box|year=2013|ab=B|num-b=16|num-a=18}}
  7 KB (1,225 words) - 14:59, 8 August 2021
• 2013 AMC 12B Problems/Problem 19
  ==Problem== <math> \textbf{(A)}\ 18\qquad\textbf{(B)}\ 21\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\tex
  6 KB (1,004 words) - 22:38, 18 June 2023
• 2013 AMC 12B Problems/Problem 24
  ==Problem== ...let's deduct some convenient conditions that seem sufficient to solve the problem.
  11 KB (1,876 words) - 00:08, 12 October 2023
• 2013 AMC 10B Problems
  ==Problem 1== [[2013 AMC 10B Problems/Problem 1|Solution]]
  12 KB (1,926 words) - 21:54, 6 October 2022
• 1981 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[1981 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 03:49, 29 September 2014
• 2013 AIME I Problems/Problem 3
  == Problem == ...ares (<math>a^2+b^2</math>) is <math>\frac{9}{10}</math> (as stated in the problem) the area of the whole square, it is clear that the
  5 KB (942 words) - 03:51, 23 January 2023
• 2013 AIME I Problems/Problem 7
  ==Problem== You get <math>\sqrt{25a^2+576} = 30</math>, giving <math>a=\frac{18}{5}</math>, so our answer is <math>\boxed{041}</math>
  6 KB (934 words) - 20:06, 24 January 2021
• 1991 AHSME Problems
  == Problem 1 == [[1991 AHSME Problems/Problem 1|Solution]]
  16 KB (2,451 words) - 04:27, 6 September 2021
• 2003 AMC 12B Problems/Problem 12
  ...12B Problems|2003 AMC 12B #12]] and [[2003 AMC 10B Problems|2003 AMC 10B #18]]}} ==Problem==
  2 KB (357 words) - 16:43, 29 June 2021
• 2013 USAMO Problems/Problem 2
  == Problem == fill(shift((18,0))*unitrect,mediumgray);
  14 KB (2,076 words) - 20:29, 10 July 2023
• 1989 AHSME Problems/Problem 27
  == Problem == ...rm{(C)}\ 16 \text{ or } 17\ \mathrm{(D)}\ 17 \text{ or } 18\ \mathrm{(E)}\ 18 \text{ or } 19</math>
  2 KB (285 words) - 12:39, 5 July 2021
• Mock AIME 6 2006-2007 Problems
  ==Problem 1== [[Mock AIME 6 2006-2007 Problems/Problem 1|Solution]]
  7 KB (1,173 words) - 21:04, 7 December 2018
• 2014 AIME II Problems/Problem 3
  ==Problem== B = (0,18);
  3 KB (454 words) - 22:00, 24 January 2024
• 2014 AIME II Problems/Problem 7
  ==Problem== ...en <math>n</math>, it telescopes to <math>\frac{n+2}{2}</math> where <math>18</math> is the only possible <math>n</math> value. Thus the answer is <math>
  3 KB (436 words) - 02:10, 12 December 2023
• 1967 AHSME Problems/Problem 16
  == Problem == ...nal Root Theorem, the only possible integer roots are <math>1, 2, 3, 6, 9, 18, 27, 54</math>. Bases <math>1, 2, 3, 6</math> do not have a <math>6</math>
  1 KB (176 words) - 01:39, 16 August 2023
• 1967 AHSME Problems/Problem 19
  == Problem == {{AHSME 40p box|year=1967|num-b=18|num-a=20}}
  1 KB (211 words) - 01:40, 16 August 2023
• 2013 USAJMO Problems/Problem 6
  ==Problem 6== Again, we experiment. If x = 2, y = 3, and z = 3, then <math>18 > 7 + 4\sqrt{6}</math>.
  3 KB (517 words) - 20:02, 30 April 2014
• Mock AIME I 2015 Problems
  == Problem 1 == ...and Richard states the product of the numbers in boxes C and D to be <math>18</math>. Finally, Palmer announces the product of the numbers in boxes D an
  9 KB (1,463 words) - 14:48, 12 February 2017
• 1966 AHSME Problems/Problem 19
  == Problem == {{AHSME box|year=1966|num-b=18|num-a=20}}
  659 bytes (111 words) - 02:30, 15 September 2014
• 1967 AHSME Problems/Problem 15
  == Problem == The difference in the areas of two similar triangles is <math>18</math> square feet, and the ratio of the larger area to the smaller is the
  1 KB (231 words) - 01:39, 16 August 2023
• 1967 AHSME Problems/Problem 17
  == Problem == {{AHSME 40p box|year=1967|num-b=16|num-a=18}}
  861 bytes (132 words) - 01:40, 16 August 2023
• 2016 AMC 12A
  **[[2016 AMC 12A Problems/Problem 1|Problem 1]] **[[2016 AMC 12A Problems/Problem 2|Problem 2]]
  2 KB (178 words) - 20:56, 12 December 2022
• 2016 AMC 10A
  **[[2016 AMC 10A Problems/Problem 1|Problem 1]] **[[2016 AMC 10A Problems/Problem 2|Problem 2]]
  2 KB (178 words) - 17:55, 3 February 2016
• 2016 AMC 10A Problems
  ==Problem 1== [[2016 AMC 10A Problems/Problem 1|Solution]]
  14 KB (2,104 words) - 22:26, 16 September 2022
• 1962 AHSME Problems/Problem 6
  ==Problem== ...f <math>18</math>. It follows that each side of the square is <math>\dfrac{18}{4}=4.5</math>. If we draw the diagonal, we create a 45-45-90 triangle, who
  1 KB (174 words) - 15:55, 18 April 2018
• 1962 AHSME Problems/Problem 17
  ==Problem== {{AHSME 40p box|year=1962|num-b=16|num-a=18}}
  772 bytes (114 words) - 20:48, 31 May 2018
• 1962 AHSME Problems/Problem 19
  ==Problem== {{AHSME 40p box|year=1962|before=Problem 18|num-a=20}}
  992 bytes (161 words) - 22:17, 3 October 2014
• 1962 AHSME Problems/Problem 38
  ==Problem== ...ath> <math>=</math> <math>36</math> or <math>b</math> <math>=</math> <math>18</math>, which means <math>a</math> <math>=</math> <math>15</math>.
  4 KB (570 words) - 16:52, 28 June 2023
• 1962 AHSME Problems
  == Problem 1 == [[1962 AHSME Problems/Problem 1|Solution]]
  17 KB (2,459 words) - 22:40, 10 April 2023
• 1988 AHSME Problems/Problem 1
  ==Problem== <math> \sqrt{8}+\sqrt{18}= </math>
  585 bytes (72 words) - 01:42, 23 October 2014
• 2013 AMC 8
  ** [[2013 AMC 8 Problems/Problem 1]] ** [[2013 AMC 8 Problems/Problem 2]]
  1 KB (102 words) - 17:53, 7 November 2020
• 2013 AMC 8 Problems
  ==Problem 1== [[2013 AMC 8 Problems/Problem 1|Solution]]
  15 KB (2,162 words) - 20:05, 8 May 2023
• 2013 AMC 8 Problems/Problem 10
  ==Problem== ...that are in both factorizations and multiply. <math>3^2 \times 2</math> = 18.
  2 KB (293 words) - 22:42, 5 January 2024
• 2013 AMC 8 Problems/Problem 17
  ==Problem== {{AMC8 box|year=2013|num-b=16|num-a=18}}
  2 KB (256 words) - 20:36, 24 December 2022
• 2013 AMC 8 Problems/Problem 19
  ==Problem== {{AMC8 box|year=2013|num-b=18|num-a=20}}
  2 KB (259 words) - 23:19, 26 December 2023
• 2013 AMC 8 Problems/Problem 21
  ==Problem== ...B)}\ 6 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
  2 KB (263 words) - 19:21, 19 December 2023
• 1994 AHSME Problems
  == Problem 1 == [[1994 AHSME Problems/Problem 1|Solution]]
  14 KB (2,124 words) - 13:39, 19 February 2020
• 1952 AHSME Problems/Problem 17
  == Problem== {{AHSME 50p box|year=1952|num-b=16|num-a=18}}
  1 KB (186 words) - 01:44, 3 January 2014
• 1952 AHSME Problems/Problem 19
  == Problem== {{AHSME 50p box|year=1952|num-b=18|num-a=20}}
  1 KB (245 words) - 21:15, 9 January 2014
• 2014 AMC 10B
  **[[2014 AMC 10B Problems/Problem 1|Problem 1]] **[[2014 AMC 10B Problems/Problem 2|Problem 2]]
  2 KB (175 words) - 18:17, 6 October 2014
• 1981 AHSME Problems
  ==Problem 1== [[1981 AHSME Problems/Problem 1|Solution]]
  17 KB (2,633 words) - 15:44, 16 September 2023
• 1967 AHSME Problems/Problem 30
  == Problem == <math>\textbf{(A)}\ 18\qquad
  2 KB (265 words) - 01:40, 16 August 2023
• 1968 AHSME Problems
  ==Problem 1== [[1968 AHSME Problems/Problem 1|Solution]]
  16 KB (2,571 words) - 14:13, 20 February 2020
• 2014 AMC 10A
  **[[2014 AMC 10A Problems/Problem 1|Problem 1]] **[[2014 AMC 10A Problems/Problem 2|Problem 2]]
  2 KB (175 words) - 18:15, 6 October 2014
• 2014 AMC 10A Problems
  ==Problem 1== [[2014 AMC 10A Problems/Problem 1|Solution]]
  15 KB (2,190 words) - 15:21, 22 December 2020
• 2014 AMC 12A
  **[[2014 AMC 10A Problems/Problem 1|Problem 1]] **[[2014 AMC 12A Problems/Problem 2|Problem 2]]
  2 KB (182 words) - 21:43, 6 October 2014
• 2014 AMC 10A Problems/Problem 8
  ==Problem== ...frac{16!17!}2\qquad\textbf{(D)}\ \dfrac{17!18!}2\qquad\textbf{(E)}\ \dfrac{18!19!}2 </math>
  1 KB (186 words) - 00:09, 27 June 2023
• 2014 AMC 10A Problems/Problem 11
  ==Problem== Coupon 3: <math>18\%</math> off the amount by which the listed price exceeds <math>\textdollar
  2 KB (321 words) - 11:44, 7 September 2021
• 2014 AMC 10A Problems/Problem 12
  ==Problem== ...t{3}-9\pi \qquad \text{(B)}\ 27\sqrt{3}-6\pi \qquad \text{(C)}\ 54\sqrt{3}-18\pi \qquad \text{(D)}\ 54\sqrt{3}-12\pi \qquad \text{(E)}\ 108\sqrt{3}-9\pi
  3 KB (482 words) - 11:50, 7 September 2021
• 2014 AMC 10A Problems/Problem 14
  ==Problem== ...hat <math>AP=\sqrt{6^2+2^2}=2\sqrt{10}</math>, and that <math>AQ=\sqrt{6^2+18^2}=6\sqrt{10}</math>. <math>PQ</math> is obviously <math>10-(-10)=20</math>
  6 KB (1,001 words) - 13:07, 25 July 2022
• 2014 AMC 10A Problems/Problem 16
  ==Problem== <math> \textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12
  9 KB (1,411 words) - 19:51, 25 July 2023
• 2014 AMC 10A Problems/Problem 17
  ==Problem== {{AMC10 box|year=2014|ab=A|num-b=16|num-a=18}}
  3 KB (496 words) - 22:43, 21 November 2022
• 2014 AMC 10A Problems/Problem 19
  ==Problem== {{AMC10 box|year=2014|ab=A|num-b=18|num-a=20}}
  3 KB (377 words) - 21:23, 28 October 2023
• 2014 AMC 10A Problems/Problem 21
  ==Problem== <math> \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}
  2 KB (314 words) - 16:35, 30 July 2022
• 2014 AMC 10A Problems/Problem 22
  ==Problem== ...tbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20 </math>
  12 KB (1,821 words) - 18:16, 29 October 2023
• 2014 AMC 10A Problems/Problem 24
  ==Problem== <math>13,14,15,16,17,18</math> etc.
  6 KB (893 words) - 21:33, 4 May 2024
• 1994 AHSME Problems/Problem 9
  ==Problem== ...textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ} </math>
  959 bytes (145 words) - 17:33, 9 January 2021
• 1994 AHSME Problems/Problem 16
  ==Problem== ...plies b=18.</cmath> Therefore, the total number of marbles is <cmath>r+b=4+18=\boxed{\textbf{(B) }22.}</cmath>
  1 KB (237 words) - 02:54, 28 May 2021
• 1994 AHSME Problems/Problem 17
  ==Problem== {{AHSME box|year=1994|num-b=16|num-a=18}}
  2 KB (287 words) - 03:07, 28 May 2021
• 1994 AHSME Problems/Problem 18
  ==Problem== ...f{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18 </math>
  1 KB (170 words) - 03:08, 28 May 2021
• 1994 AHSME Problems/Problem 19
  ==Problem== We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pi
  1 KB (209 words) - 03:10, 28 May 2021
• 1960 AHSME Problems
  == Problem 1== [[1960 AHSME Problems/Problem 1|Solution]]
  21 KB (3,242 words) - 21:27, 30 December 2020
• 2014 AMC 12B
  **[[2014 AMC 12B Problems/Problem 1|Problem 1]] **[[2014 AMC 12B Problems/Problem 2|Problem 2]]
  2 KB (178 words) - 16:53, 24 January 2020
• 2014 AMC 12A Problems
  ==Problem 1== [[2014 AMC 10A Problems/Problem 1|Solution]]
  12 KB (1,863 words) - 19:04, 11 April 2024
• 2014 AMC 12A Problems/Problem 19
  ==Problem== {{AMC12 box|year=2014|ab=A|num-b=18|num-a=20}}
  2 KB (272 words) - 20:33, 4 April 2024
• 2014 AMC 12A Problems/Problem 15
  ==Problem== \textbf{(B) }18\qquad
  4 KB (612 words) - 22:42, 2 August 2021
• 2014 AMC 12A Problems/Problem 17
  ==Problem== [[File:2014 AMC 12A Problem 17.JPG|none|500px|caption]]
  4 KB (602 words) - 02:42, 13 June 2022
• 2014 AMC 12B Problems
  ==Problem 1== [[2014 AMC 12B Problems/Problem 1|Solution]]
  13 KB (2,066 words) - 14:08, 1 November 2022
• 2014 AMC 10B Problems
  ==Problem 1== [[2014 AMC 10B Problems/Problem 1|Solution]]
  13 KB (2,011 words) - 21:54, 8 November 2022
• 2014 AMC 10B Problems/Problem 5
  ==Problem== fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray);
  2 KB (257 words) - 15:11, 20 March 2024
• 2014 AMC 10B Problems/Problem 8
  ==Problem== ...We see that he does <math>\dfrac{180}{t}</math> cycles of <math>\dfrac{b}{18}</math> yards. Multiplying, we get <math>\dfrac{180b}{18t}</math>, or <math
  1 KB (248 words) - 12:02, 2 July 2023
• 2014 AMC 10B Problems/Problem 17
  ==Problem== {{AMC10 box|year=2014|ab=B|num-b=16|num-a=18}}
  7 KB (988 words) - 18:47, 11 August 2023
• 2014 AMC 10B Problems/Problem 19
  ==Problem== {{AMC10 box|year=2014|ab=B|num-b=18|num-a=20}}
  2 KB (383 words) - 19:44, 28 April 2021
• 2014 AMC 10B Problems/Problem 21
  ==Problem== ...qrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26 </math>
  4 KB (652 words) - 09:18, 23 September 2021
• 2014 AMC 10B Problems/Problem 24
  {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #18]] and [[2014 AMC 10B Problems|2014 AMC 10B #24]]}} ==Problem==
  7 KB (1,133 words) - 12:46, 12 March 2022
• 2015 AMC 10A
  **[[2015 AMC 10A Problems/Problem 1|Problem 1]] **[[2015 AMC 10A Problems/Problem 2|Problem 2]]
  2 KB (178 words) - 19:55, 4 February 2015
• 2015 AMC 10B
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. ** [[2015 AMC 10B Problems/Problem 1]]
  1 KB (163 words) - 11:27, 2 March 2015
• 2014 AMC 12B Problems/Problem 5
  == Problem == fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray);
  1 KB (211 words) - 00:38, 15 November 2022
• 2014 AMC 12B Problems/Problem 2
  ==Problem== ...sets of two balloons, so the total number of balloons he can buy is <math>18\times2 \implies \boxed{\textbf{(C)}\ 36 }</math>
  2 KB (253 words) - 18:41, 30 May 2023
• 2014 AMC 12B Problems/Problem 20
  ==Problem== \textbf{(B) }18\qquad
  2 KB (269 words) - 22:03, 14 September 2023
• 2014 AMC 12B Problems/Problem 10
  ==Problem== ...9a = 5k</math>. Thus, k is divisible by <math>9</math>. Because <math>55 * 18 = 990</math>, <math>k</math> must be <math>9</math>, and therefore <math>c
  2 KB (321 words) - 19:46, 17 January 2021
• 2014 AMC 12B Problems/Problem 15
  ==Problem== ...\qquad\textbf{(B)}\ 2^{14}\qquad\textbf{(C)}\ 2^{16}\qquad\textbf{(D)}\ 2^{18}\qquad\textbf{(E)}\ 2^{20} </math>
  1 KB (203 words) - 23:18, 3 January 2023
• 2014 AMC 12B Problems/Problem 17
  ==Problem== {{AMC12 box|year=2014|ab=B|num-b=16|num-a=18}}
  2 KB (384 words) - 21:28, 13 September 2023
• 2014 AMC 12B Problems/Problem 22
  == Problem == as said in the problem.
  7 KB (1,082 words) - 22:35, 3 April 2024
• 2014 AMC 12B Problems/Problem 19
  == Problem == {{AMC12 box|year=2014|ab=B|num-b=18|num-a=20}}
  4 KB (703 words) - 16:24, 9 September 2022
• 2014 AMC 12B Problems/Problem 18
  {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #18]] and [[2014 AMC 10B Problems|2014 AMC 10B #24]]}} ==Problem==
  2 KB (331 words) - 04:43, 12 January 2021
• Mock AIME 3 2006-2007 Problems/Problem 11
  == Problem == ...3))^2 - 4\cdot 8 \cdot (a + 6)^2 \ge 0</math> simplifies to <math>a^2 \ge 18</math>.
  4 KB (699 words) - 01:53, 30 April 2022
• 1964 AHSME Problems/Problem 33
  ==Problem== Thus, <math>34 = 16 + x^2</math>, or <math>x = \sqrt{18} = 3\sqrt{2}</math>, which is option <math>\boxed{\textbf{(B)}}</math>
  2 KB (337 words) - 00:46, 15 June 2022
• 1964 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. ** [[1964 AHSME Problems/Problem 1|Problem 1]]
  2 KB (217 words) - 14:16, 20 February 2020
• 2014 AIME II Problems
  ==Problem 1== [[2014 AIME II Problems/Problem 1|Solution]]
  8 KB (1,410 words) - 00:04, 29 December 2021
• 2014 AIME I Problems/Problem 13
  ==Problem 13== <cmath>\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}</cmath>
  9 KB (1,404 words) - 21:07, 13 October 2023
• 1969 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1969 AHSME Problems/Problem 1|Problem 1]]
  2 KB (196 words) - 14:12, 20 February 2020
• 1993 AHSME Problems/Problem 15
  == Problem == \text{(B) } 18\quad
  897 bytes (140 words) - 13:39, 6 November 2020
• 1993 AHSME Problems/Problem 17
  == Problem == {{AHSME box|year=1993|num-b=18|num-a=19}}
  2 KB (301 words) - 03:42, 12 December 2018
• 1993 AHSME Problems/Problem 18
  == Problem == Day 18: Al works; Barb rests
  2 KB (294 words) - 16:49, 9 September 2020
• 1993 AHSME Problems/Problem 19
  == Problem == ...to factor the equation as follows: <math>(m-4)(n-2) = 8</math>. Since the problem only wants integer pairs <math>(m,n)</math>, the pairs are given by the fac
  904 bytes (137 words) - 23:30, 22 December 2020
• 1993 AHSME Problems/Problem 22
  == Problem == draw((3.5,7)--(4.5,9)--(9.5,9)--(14.5,9)--(19.5,9)--(18.5,7)--(19.5,9)--(19.5,7), linewidth(1));
  7 KB (780 words) - 08:19, 27 June 2021
• 1993 AHSME Problems/Problem 23
  == Problem == <math>\text{(A) } \cos(6^\circ)\cos(12^\circ)\sec(18^\circ)\quad\\
  2 KB (284 words) - 17:00, 29 August 2020
• 1993 AHSME Problems/Problem 24
  == Problem == <math>\dfrac{6*3}{35}=\dfrac{18}{35}</math>
  3 KB (558 words) - 00:25, 23 December 2020
• 1987 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1987 AHSME Problems/Problem 1|Problem 1]]
  2 KB (169 words) - 08:30, 23 October 2014
• 1986 AHSME
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[1986 AHSME Problems/Problem 1|Problem 1]]
  2 KB (169 words) - 09:03, 23 October 2014
• 1982 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1982 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 22:33, 1 October 2014
• 1979 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1979 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 22:26, 1 October 2014
• 1978 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1978 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 22:24, 1 October 2014
• 1977 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1977 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 22:21, 1 October 2014
• 1976 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1976 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 22:19, 1 October 2014
• 1975 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. *[[1975 AHSME Problems/Problem 1|Problem 1]]
  2 KB (174 words) - 22:17, 1 October 2014
• 1972 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[1972 AHSME Problems/Problem 1|Problem 1]]
  2 KB (203 words) - 14:09, 20 February 2020
• 1971 AHSME
  ...k contains the full set of test problems. The rest contain each individual problem and its solution. **[[1971 AHSME Problems/Problem 1|Problem 1]]
  2 KB (201 words) - 14:10, 20 February 2020
• 1973 AHSME Problems
  ==Problem 1== [[1973 AHSME Problems/Problem 1|Solution]]
  18 KB (2,788 words) - 13:55, 20 February 2020
• 1992 AHSME Problems/Problem 20
  == Problem == \text{(B) } 18\quad
  2 KB (281 words) - 12:28, 1 May 2021
• 1992 AHSME Problems/Problem 19
  == Problem == {{AHSME box|year=1992|num-b=18|num-a=20}}
  1 KB (182 words) - 03:11, 20 February 2018
• 1992 AHSME Problems/Problem 11
  == Problem == draw(circle((0,0),18),black+linewidth(.75));
  2 KB (324 words) - 12:02, 24 November 2016
• 1992 AHSME Problems
  == Problem 1 == [[1992 AHSME Problems/Problem 1|Solution]]
  16 KB (2,548 words) - 13:40, 19 February 2020
• 1991 AHSME Problems/Problem 19
  == Problem == {{AHSME box|year=1991|num-b=18|num-a=20}}
  3 KB (506 words) - 22:41, 23 November 2020
• 1968 AHSME Problems/Problem 17
  == Problem == {{AHSME 35p box|year=1968|num-b=16|num-a=18}}
  541 bytes (86 words) - 01:52, 16 August 2023
• 1968 AHSME Problems/Problem 19
  == Problem == {{AHSME 35p box|year=1968|num-b=18|num-a=20}}
  462 bytes (69 words) - 01:52, 16 August 2023
• 1968 AHSME Problems/Problem 24
  == Problem == A painting <math>18</math>" X <math>24</math>" is to be placed into a wooden frame with the lon
  1 KB (191 words) - 01:53, 16 August 2023
• 1990 AHSME Problems/Problem 19
  == Problem == {{AHSME box|year=1990|num-b=18|num-a=20}}
  2 KB (272 words) - 01:24, 9 August 2022
• 1990 AHSME Problems/Problem 17
  == Problem == {{AHSME box|year=1990|num-b=16|num-a=18}}
  665 bytes (94 words) - 05:42, 4 February 2016
• 1990 AHSME Problems
  == Problem 1 == [[1990 AHSME Problems/Problem 1|Solution]]
  14 KB (2,099 words) - 01:15, 10 September 2021
• 1969 AHSME Problems/Problem 17
  == Problem == {{AHSME 35p box|year=1969|num-b=16|num-a=18}}
  969 bytes (149 words) - 16:30, 10 July 2015
• 1969 AHSME Problems/Problem 19
  == Problem == {{AHSME 35p box|year=1969|num-b=18|num-a=20}}
  971 bytes (157 words) - 16:14, 10 July 2015
• 1969 AHSME Problems/Problem 20
  == Problem == ...h>3.6*3.4>10</math>, so we can further simplify the product with <math>10^{18}*10^{14}*10=10^{33}</math> which means the product has <math>\fbox{34 (C)}<
  1 KB (170 words) - 08:51, 27 May 2020
• 1969 AHSME Problems/Problem 30
  == Problem == label("$x-a$",(14,18),E);
  3 KB (444 words) - 23:12, 21 June 2018
• 1969 AHSME Problems
  ==Problem 1== [[1969 AHSME Problems/Problem 1|Solution]]
  16 KB (2,662 words) - 14:12, 20 February 2020
• 1970 AHSME Problems/Problem 11
  == Problem == <math>3h = 18</math>
  915 bytes (150 words) - 21:26, 13 July 2019
• 1970 AHSME Problems/Problem 17
  == Problem == {{AHSME 35p box|year=1970|num-b=16|num-a=18}}
  2 KB (282 words) - 02:37, 17 December 2021
• 1970 AHSME Problems/Problem 19
  == Problem == So we can apply this to the conditions given by the problem.
  1 KB (186 words) - 21:14, 20 February 2019
• 1970 AHSME Problems/Problem 25
  == Problem == ...s, the cost is <math>12</math> cents. Option <math>D</math> remains <math>18</math> cents, while option <math>E</math> gives <math>12</math> cents, the
  1 KB (244 words) - 06:54, 15 July 2019
• 1970 AHSME Problems
  ==Problem 1== [[1970 AHSME Problems/Problem 1|Solution]]
  15 KB (2,366 words) - 07:52, 26 December 2023
• 1980 AHSME Problems/Problem 17
  == Problem == {{AHSME box|year=1980|num-b=16|num-a=18}}
  729 bytes (116 words) - 19:10, 18 June 2021
• 1980 AHSME Problems/Problem 19
  == Problem == {{AHSME box|year=1980|num-b=18|num-a=20}}
  2 KB (350 words) - 16:41, 12 September 2021
• 1988 AHSME Problems
  ==Problem 1== <math>\sqrt{8}+\sqrt{18}= </math>
  17 KB (2,535 words) - 13:45, 19 February 2020

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