Create the page "AIME Problems/Problem 7" on this wiki! See also the search results found.

Page title matches

• 2006 AIME I Problems/Problem 7
  == Problem == defaultpen(linewidth(0.7)+fontsize(10));
  4 KB (709 words) - 01:31, 5 January 2025
• 2006 AIME A Problems/Problem 7
  #REDIRECT [[2006 AIME I Problems/Problem 7]]
  44 bytes (5 words) - 13:37, 2 July 2009
• 2005 AIME II Problems/Problem 7
  == Problem == ...y^6+y^4+y^2+1)(y+1) = (y^{15}+y^{14}+y^{13}+y^{12}+y^{11}+y^{10}+y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2+y+1)=\frac{y^{16}-1}{y-1}</cmath>
  2 KB (279 words) - 12:33, 27 October 2019
• 2005 AIME I Problems/Problem 7
  == Problem == label("$C$",(16.87,7),E);
  4 KB (567 words) - 20:20, 3 March 2020
• 2004 AIME I Problems/Problem 7
  == Problem == ...^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8</math></center>
  7 KB (1,099 words) - 13:41, 30 December 2024
• 2004 AIME II Problems/Problem 7
  == Problem == defaultpen(linewidth(0.7)+fontsize(10));
  9 KB (1,500 words) - 20:06, 8 October 2024
• 1983 AIME Problems/Problem 7
  == Problem == ...ights. (For example, if you pick <math>(5,6)</math>, you may not pick 4 or 7.) Thus, there are <math>25-4=21</math> ways to pick the third knight, for a
  9 KB (1,458 words) - 22:34, 2 February 2025
• 1984 AIME Problems/Problem 7
  == Problem == Open up a coding IDE and use recursion to do this problem. The idea is to define a function (I called it <math>f</math>, you can call
  4 KB (633 words) - 22:31, 6 October 2024
• 1985 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1985|num-b=6|num-a=8}}
  1 KB (222 words) - 11:04, 4 November 2022
• 1986 AIME Problems/Problem 7
  == Problem == ...term is <math>3^6</math> = <math>729</math> and the 128th term is <math>3^7</math> = <math>2187</math>. Writing out more terms of the sequence until th
  5 KB (866 words) - 00:00, 22 December 2022
• 1987 AIME Problems/Problem 7
  == Problem == ...p = 4</math> and at least one of <math>m, j</math> equal to 3. This gives 7 possible triples <math>(j, m, p)</math>: <math>(0, 3, 4), (1, 3, 4), (2, 3,
  3 KB (547 words) - 10:00, 18 June 2024
• 1988 AIME Problems/Problem 7
  == Problem == In [[triangle]] <math>ABC</math>, <math>\tan \angle CAB = 22/7</math>, and the [[altitude]] from <math>A</math> divides <math>BC</math> in
  1 KB (178 words) - 23:25, 20 November 2023
• 1989 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1989|num-b=6|num-a=8}}
  2 KB (320 words) - 15:50, 12 September 2024
• 1990 AIME Problems/Problem 7
  == Problem == ...{}^{}=(-8,5)</math>, <math>Q_{}^{}=(-15,-19)</math>, and <math>R_{}^{}=(1,-7)</math>. The [[equation]] of the [[angle bisector|bisector]] of <math>\angl
  8 KB (1,319 words) - 11:34, 22 November 2023
• 1991 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1991|num-b=6|num-a=8}}
  2 KB (285 words) - 12:20, 17 February 2025
• 1992 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1992|num-b=6|num-a=8}}
  1 KB (164 words) - 04:29, 3 February 2025
• 1993 AIME Problems/Problem 7
  == Problem == Note that the <math>1000</math> in the problem, is not used, and is cleverly bypassed in the solution, because we can call
  5 KB (772 words) - 09:04, 7 January 2022
• 1994 AIME Problems/Problem 7
  == Problem == draw((1,7)--(-1,-7),red); draw((7,1)--(5,-5), green);
  3 KB (442 words) - 19:51, 8 January 2024
• 1995 AIME Problems/Problem 7
  == Problem == {{AIME box|year=1995|num-b=6|num-a=8}}
  3 KB (427 words) - 09:23, 13 December 2023
• 1996 AIME Problems/Problem 7
  == Problem == Two [[square]]s of a <math>7\times 7</math> checkerboard are painted yellow, and the rest are painted green. Two
  5 KB (712 words) - 18:25, 31 August 2024

Page text matches

• Pythagorean Theorem
  ...e many tools in a good geometer's arsenal. A very large number of geometry problems can be solved by building right triangles and applying the Pythagorean Theo E = (7,10);
  6 KB (978 words) - 12:02, 6 March 2025
• United States of America Mathematical Olympiad
  ...dministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the MAA. ...ficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}
  6 KB (836 words) - 04:57, 3 February 2025
• Completing the square
  ...[[perfect square]], <math>(x-3)^2=x^2-6x+9</math>. To do this, add <math>7</math> to each side of the equation to get ...+9=7\Rightarrow(x-3)^2=7\Rightarrow x-3=\pm{\sqrt7}\Rightarrow x=3\pm\sqrt{7}</math>
  2 KB (422 words) - 16:20, 5 March 2023
• Factorial
  * <math>7! = 5040</math> ...reciprocal of an [[exponential function]]. For example, the power of <math>7</math> in <math>100!</math> is just
  10 KB (809 words) - 16:40, 17 March 2024
• 2006 AIME II
  ...contains the full set of test problems. The rest contain each individual problem and its solution. * [[2006 AIME II Problems]]
  1 KB (133 words) - 12:32, 22 March 2011
• Fermat's Little Theorem
  In contest problems, Fermat's Little Theorem is often used in conjunction with the [[Chinese Re === Proof 7 (Field Theory) ===
  15 KB (2,618 words) - 12:03, 19 February 2025
• Parabola
  draw(graph(f,(7+2*sqrt(13))/3,(7-2*sqrt(13))/3),red+linewidth(1)); ==Problems==
  3 KB (551 words) - 16:22, 13 September 2023
• Euler's Totient Theorem
  ==Problems== ...ast two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]])
  4 KB (569 words) - 22:34, 30 December 2024
• Brahmagupta's Formula
  == Problems == ...rs. Determine <math>p + q</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 7|Source]])
  3 KB (543 words) - 19:35, 29 October 2024
• Ptolemy's theorem
  ...uality]]. Ptolemy's theorem frequently shows up as an intermediate step in problems involving inscribed figures. == Problems ==
  6 KB (922 words) - 17:34, 13 January 2025
• Constructive counting
  This is a problem where constructive counting is not the simplest way to proceed. This next e '''Solution''': We can model this situation as a row of 7 boxes, like this: <cmath>\square \square \square \square \square \square \s
  13 KB (2,018 words) - 15:31, 10 January 2025
• Casework
  ..., then adding together the totals of each part. Casework is a very general problem-solving approach, and as such has wide applicability. ...emonstrate casework in action. Unlike the selections in this article, most problems cannot be completely solved through casework. However, it is crucial as an
  5 KB (709 words) - 17:40, 24 September 2024
• Function
  ==Problems== ([[2006 AMC 10A Problems/Problem 2|Source]])
  10 KB (1,761 words) - 03:16, 12 May 2023
• Complementary counting
  ...ng may lead to a quick solution is the phrase "not" or "at least" within a problem statement. ''[[2006 AMC 10A Problems/Problem 21 | 2006 AMC 10A Problem 21]]: How many four-digit positive integers have at least one digit that is
  8 KB (1,192 words) - 17:20, 16 June 2023
• Prime factorization
  ...hen doing long division again, <math>151</math> is not a multiple of <math>7</math>, so we move on. ...the prime factorization of <math>38052</math> is <math>2^2 \cdot 3^2 \cdot 7 \cdot 151</math>.
  3 KB (496 words) - 22:14, 5 January 2024
• Ellipse
  MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E); MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);
  5 KB (892 words) - 21:52, 1 May 2021
• Base numbers
  ...e number 2746. This number can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> == Example Problems ==
  4 KB (547 words) - 17:23, 30 December 2020
• Fibonacci sequence
  ==Problems== ...iv style="text-align:right">([[2000 AMC 12 Problems/Problem 4|2000 AMC 12, Problem 4]])</div>
  7 KB (1,111 words) - 14:57, 24 June 2024
• Arithmetic sequence
  ...etic sequence with common difference <math>-8</math>; however, <math>7, 0, 7, 14</math> and <math>4, 12, 36, 108, \ldots</math> are not arithmetic seque == Problems ==
  4 KB (736 words) - 02:00, 7 March 2024
• Diophantine equation
  ...s for any given Diophantine equations. This is known as [[Hilbert's tenth problem]]. The answer, however, is no. ...en, the theorem was finally proven by [[Andrew Wiles]] after he spent over 7 years working on the 200-page proof, and another year fixing an error in th
  9 KB (1,434 words) - 01:15, 4 July 2024

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