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• 1995 IMO Problems/Problem 2
  By the [[Power Mean Inequality]],
  6 KB (1,121 words) - 05:32, 1 June 2024
• 1985 AIME Problems/Problem 13
  ...ath>n</math>, we can multiply the left integer, <math>100+n^2</math>, by a power of two without affecting the greatest common divisor. Since the <math>n^2</
  4 KB (671 words) - 20:04, 6 March 2024
• 1985 AIME Problems/Problem 7
  ...[perfect power | perfect fourth power]], <math>b</math> is a perfect fifth power, <math>c</math> is a [[perfect square]] and <math>d</math> is a [[perfect c
  1 KB (222 words) - 11:04, 4 November 2022
• 1986 AIME Problems/Problem 11
  We want the coefficient of the <math>y^2</math> term of each power of each binomial, which by the binomial theorem is <math>{2\choose 2} + {3\ ...c{d^2f}{dx^2} = 2\cdot 1 - 3\cdot 2x+\cdots-17\cdot 16x^{15}</math> by the power rule.
  6 KB (872 words) - 16:51, 9 June 2023
• 1986 AIME Problems/Problem 8
  ...question asks for proper divisors, we exclude <math>2^65^6</math>, so each power is actually <math>141</math> times. The answer is thus <math>S = \log 2^{14
  3 KB (487 words) - 20:52, 16 September 2020
• 1986 AIME Problems/Problem 7
  ...math>k</math>th term after the <math>n</math>th power of 3 is equal to the power plus the <math>k</math>th term in the entire sequence. Thus, the <math>100< ...= <math>2187</math>. Writing out more terms of the sequence until the next power of 3 again (81) we can see that the (<math>2^n</math>+<math>2^{n+1}</math>)
  5 KB (866 words) - 00:00, 22 December 2022
• 1987 AIME Problems/Problem 11
  ...of the two factors will be a power of three, and the other will be twice a power of three. <math>(2n + m + 1)</math> will represent the greater factor while
  3 KB (418 words) - 18:30, 20 January 2024
• 1989 AIME Problems/Problem 9
  ...>'s units digit is <math>0, 2, 4, 6,</math> or <math>8.</math> When to the power of <math>5,</math> they each give <math>0, 2, 4, 6,</math> and <math>8</mat
  6 KB (874 words) - 15:50, 20 January 2024
• 1990 AIME Problems/Problem 13
  '''Lemma''': For all positive integers n, there's exactly one n-digit power of 9 that does not have a left-most digit 9 ...ove by contradiction that there must be at least either one or two n-digit power of 9 for all n.
  5 KB (762 words) - 01:18, 10 February 2023
• 1990 AIME Problems/Problem 5
  ...<math>5</math>, the greatest of all the factors, to be raised to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^
  1 KB (175 words) - 03:45, 21 January 2023
• 1990 AIME Problems/Problem 2
  The <math>3/2</math> power is quite irritating to work with so we look for a way to eliminate that. No
  5 KB (765 words) - 23:00, 26 August 2023
• 2006 AMC 10B Problems/Problem 25
  ...visibility by any powers lower than these means indivisibility by a higher power of the prime (for example, indivisibility by <math>2^2=4</math> means indiv
  5 KB (878 words) - 14:39, 3 December 2023
• 1996 AIME Problems/Problem 3
  ...will contain <math>(n+1)^2</math> terms, as each term will have an unique power of <math>x</math> or <math>y</math> and so none of the terms will need to b
  3 KB (515 words) - 04:29, 27 November 2023
• 1998 AIME Problems/Problem 13
  ...re <math>i^2 = - 1.</math> Let <math>S_n</math> be the sum of the complex power sums of all nonempty [[subset]]s of <math>\{1,2,\ldots,n\}.</math> Given t ...(for now, including the empty subset, which we will just define to have a power sum of zero) with <math>9</math> in it is equal to the number of subsets wi
  2 KB (384 words) - 14:47, 14 June 2024
• 1998 AIME Problems/Problem 1
  ...rs an be found by writing out their factorizations and taking the greatest power for each factor. <math>[6^6,8^8] = 2^{24}3^6</math>. Therefore <math>12^{12 ...</math> wouldn't be <math>12</math>) and <math>0\le a\le 24</math> (or the power of <math>2</math> in the <math>\operatorname{lcm}</math> would be <math>a</
  2 KB (289 words) - 22:50, 23 April 2024
• 1999 AIME Problems/Problem 4
  ..., etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length <math>1</math> in the
  3 KB (398 words) - 13:27, 12 December 2020
• 2000 AIME I Problems/Problem 15
  To simplify matters, we want a power of <math>2</math>. Hence, we will add <math>48</math> 'fake' cards which we ...<math>31-15=16</math> cards remaining. Since <math>16</math> is a perfect power of <math>2</math>, we will not need to worry about this scenario again in t
  15 KB (2,673 words) - 19:16, 6 January 2024
• 2000 AIME I Problems/Problem 11
  ...hen divide by 1000. To do this, write the corresponding divisor under each power. e.g. 2 - 500, 4 - 250, 5 - 200, etc. Call this the "partner" of any diviso ...minator, then every power of five will be multiplied by the partner of the power of 2. Essentially, all we have to do is a large scale distributive property
  4 KB (667 words) - 13:58, 31 July 2020
• 2000 AIME I Problems/Problem 1
  ...>2</math>s and the <math>5</math>s separated, so we need to find the first power of 2 or 5 that contains a 0.
  1 KB (163 words) - 17:44, 16 December 2020
• 2001 AIME I Problems/Problem 5
  ...is a diameter of the unit circle. Then <math>XC=2-2n\sqrt{3}.</math> Using power of a point on X,
  6 KB (1,043 words) - 10:09, 15 January 2024

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