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• 2014 AMC 10B Problems
  What is the greatest power of <math>2</math> that is a factor of <math>10^{1002} - 4^{501}</math>? A list of <math>11</math> positive integers has a mean of <math>10</math>, a median of <math>9</math>, and a unique mode of <math>
  13 KB (2,011 words) - 21:54, 8 November 2022
• 2014 AMC 10B Problems/Problem 17
  What is the greatest power of <math>2</math> that is a factor of <math>10^{1002} - 4^{501}</math>? ...+ 1)</math> would end in ...<math>126</math> and thus would contribute one power of two to the answer, but not more.
  7 KB (998 words) - 10:51, 1 June 2024
• 2014 AMC 12B Problems/Problem 15
  ...k}</math>, the number <math>e^p</math> is an integer. What is the largest power of <math>2</math> that is a factor of <math>e^p</math>? Raising <math>e</math> to the power of this quantity eliminates the natural logarithm, which leaves us with
  1 KB (203 words) - 23:18, 3 January 2023
• Descartes Rule of Signs
  ...ts is the number of sign changes after multiplying the coefficients of odd-power terms by −1, or fewer than that by a positive even number.
  942 bytes (147 words) - 10:44, 20 September 2016
• 2014 AIME I Problems/Problem 15
  Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <m And then by Power of a Point, we have
  10 KB (1,643 words) - 22:30, 28 January 2024
• 1993 AHSME Problems/Problem 10
  Raising both sides to the <math>\frac{1}{b}</math> power we get that <math>27a^3=ax</math> or <math>x=27a^2</math>
  770 bytes (136 words) - 22:00, 27 May 2021
• 1992 AHSME Problems/Problem 29
  ...{3} \right)^{50}</math> and then divide by <math>2</math> because the even power terms that were not canceled were expressed twice. Thus, we have <cmath>P=\
  3 KB (472 words) - 19:29, 20 June 2021
• 1992 AHSME Problems/Problem 27
  Applying Power of a Point on <math>P</math>, we find that <math>PC=9</math> and thus <math
  2 KB (318 words) - 17:12, 7 June 2019
• 1992 AHSME Problems
  ...th>N=192021\cdots9192</math>. Suppose that <math>3^k</math> is the highest power of 3 that is a factor of <math>N</math>. What is <math>k</math>?
  16 KB (2,548 words) - 13:40, 19 February 2020
• 1990 AHSME Problems/Problem 18
  ...>1,3,9,7</math> in opposite directions, and as <math>4\mid 100</math> each power's units digit is equally probable. There are <math>16</math> ordered pairs
  987 bytes (146 words) - 11:52, 4 February 2016
• 1970 AHSME Problems/Problem 1
  The fourth power of <math>\sqrt{1+\sqrt{1+\sqrt{1}}}</math> is
  589 bytes (84 words) - 21:45, 16 February 2016
• 1952 AHSME Problems/Problem 29
  ...the center of the circle, <math>ON = a</math> and <math>CN = x</math>. By power of a point on <math>N</math>, we have
  1 KB (191 words) - 13:20, 15 January 2018
• 1970 AHSME Problems
  The fourth power of <math>\sqrt{1+\sqrt{1+\sqrt{1}}}</math> is:
  15 KB (2,366 words) - 07:52, 26 December 2023
• 1958 AHSME Problems/Problem 42
  ...th>. <math>CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}</math>. Using Power of a Point on <math>E</math>, <math>(BE)(EC) = (AE)(ED)</math> (there isn't
  1 KB (195 words) - 11:01, 29 June 2024
• 2007 iTest Problems/Problem 11
  Consider the "tower of power" <math>2^{2^{2^{\cdot^{\cdot^{\cdot^{2}}}}}}</math>, where there are 2007 t
  931 bytes (130 words) - 04:31, 10 June 2018
• 1969 IMO Problems/Problem 4
  ...<math>AB</math>. Similarly, in an inversion with the center B and positive power <math>r_B^2 = BC^2 = a^2</math> (<math>r_B</math> being the inversion circl
  5 KB (904 words) - 13:42, 29 January 2021
• 2007 IMO Problems/Problem 4
  == Alternative END Solution (Power of a Point) == ...t <math>P</math> into lengths of <math>r+x</math> and <math>r-x</math>. By power of point, <math>RP*PC=(r+x)(r-x)</math>. Similarly, <math>RQ*QC=(r+x)(r-x)
  8 KB (1,480 words) - 14:52, 5 August 2022
• 2014 IMO Problems/Problem 4
  ...tangent to the circumcircle of the <math>APC.</math> Therefore, we can use power of a point to solve for side ratios. We have <cmath>A=(1,0,0), B=(0,1,0), C
  13 KB (1,811 words) - 14:36, 1 June 2024
• 2013 UNCO Math Contest II Problems/Problem 2
  ...ath>2 \times 12</math>, which is <math>24</math>. Because we only have one power of three, we need two more, so we multiply <math>24 \times 3 \times 3</math To find a perfect 6th power, we multiply <math>36</math> by <math>216</math> to get <math>7776</math>.
  1 KB (217 words) - 19:47, 11 December 2023
• 2011 UNCO Math Contest II Problems/Problem 7
  Since every odd term is 1 less than a power of two, we know each group will be of the form <math>(2^{n}-2)+(2^{n}-1)</m
  992 bytes (150 words) - 13:10, 17 June 2021

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