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• 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
• 2002 AIME I Problems/Problem 13
  By the [[Power of a Point Theorem]] on <math>E</math>, we get <math>EF = \frac{12^2}{27} = Using Power of a Point, we have
  6 KB (974 words) - 18:10, 22 June 2024
• 2024 AMC 8 Problems/Problem 8
  ==Video Solution 1 (easy to digest) by Power Solve==
  2 KB (384 words) - 22:57, 17 February 2024
• 2003 AIME II Problems/Problem 13
  ====4.1.4: Using the JCF to calculate the transition matrix to the power of any n, large or small==== ...then we must find the sum of the coefficients that share a variable with a power divisible by <math>3</math>.
  15 KB (2,406 words) - 23:56, 23 November 2023
• 2002 AIME II Problems/Problem 14
  ...\triangle MPA</math> are similar. Also note that <math>AM = BM</math> by [[power of a point]]. Using the fact that the ratio of corresponding sides in simil
  4 KB (658 words) - 19:15, 19 December 2021
• 2001 AIME II Problems/Problem 8
  ...proach is to consider the graph of <math>f(x)</math>, which iterates every power of <math>3</math>, and resembles the section from <math>1 \le x \le 3</math
  3 KB (545 words) - 23:41, 14 June 2023
• 2000 AIME II Problems/Problem 13
  ...(or in this case, nearly symmetric) polynomials is to divide through by a power of <math>x</math> with half of the polynomial's degree (in this case, divid ...can see that the number of zeros in a term more or less correlates to the power of <math>x^2</math>. Thus, we let <math>y=10x^2</math>. The equation then b
  6 KB (1,060 words) - 17:36, 26 April 2024
• 2000 AIME II Problems/Problem 4
  ...her is (<math>6 = 2 \cdot 3</math>) a prime raised to the <math>5</math>th power, or two primes, one of which is squared. The smallest example of the former ...e now divide all of the odd factors from <math>n</math>; then we require a power of <math>2</math> with <math>\frac{18}{6} = 3</math> factors, namely <math>
  2 KB (397 words) - 15:55, 11 May 2022
• 2006 USAMO Problems/Problem 5
  ...> or to <math>n+2^{m_n+1}</math> where <math>2^{m_n}</math> is the largest power of 2 that is a factor of <math>n</math>. Show that if <math>k\ge 2</math> i ...made from a number that is divisible by <math>2^M</math> (and by no higher power of 2). Thus we must have <math>M < i_0</math>, since otherwise a number div
  7 KB (1,280 words) - 17:23, 26 March 2016
• Holomorphic function
  <math>z</math> if <math>f</math> has a convergent [[power series]] expansion on some its power series diverges when <math>\lvert x \rvert > 1</math>. But in the
  9 KB (1,537 words) - 21:04, 26 July 2017
• Minnesota ARML
  ...f Wayzata HS (1991-2006), founder and long-time problem writer of the ARML Power Contest, and Mike Reiners of Christ's Household of Faith School (2007-14),
  4 KB (680 words) - 16:45, 10 June 2015

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