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• 1995 USAMO Problems/Problem 1
  ...)d</math> have the same number of tailing zeros, let's say there are <math>x</math> tailing zeros. and let <math>\dfrac{{0,d,2d,\cdots,(p-1)d}}{p^x}=S</math>
  4 KB (625 words) - 18:23, 22 March 2024
• 2010 AMC 12A Problems/Problem 20
  <cmath>\begin{align*}&\left(a_n\right) \Rightarrow \{1, x+1, 2x+1, 3x+1, ...\}\\ where <math>x</math> and <math>y</math> (<math>x\leq y</math>) are the common differences of each, respectively.
  5 KB (797 words) - 15:27, 3 July 2023
• 2010 AMC 12A Problems/Problem 23
  ...f in doubt, look at the answer choices), we only need to consider <math>P \mod 25</math>. ...articular number. <math>1 \cdot 2 \cdot 3 \cdot 4 \equiv -1 \text{ }(\text{mod }25)</math>, and so is <math>6 \cdot 7 \cdot 8 \cdot 9</math>. How can we g
  10 KB (1,525 words) - 09:44, 24 April 2024
• Characteristic polynomial
  <center><cmath>P(x) = x^k - c_{1}x^{k-1} - c_{2}x^{k-2} - \cdots -c_{k-1}x - c_k.</cmath></center> Then, its characteristic polynomial is <math>x^2 - x - 1</math>. <br><br>
  19 KB (3,412 words) - 14:57, 21 September 2022
• Pell equation
  A '''Pell equation''' is a type of [[diophantine equation]] in the form <math>x^2-Dy^2 = \pm1</math> for a [[natural number]] <math>D</math>. Generally, <m ...d using [[difference of squares]]. We would have <math>x^2 - Dy^2 = (x+dy)(x-dy) = 1</math>, from which we can use casework to quickly determine the sol
  6 KB (1,076 words) - 10:20, 29 October 2023
• 2010 AIME I Problems/Problem 10
  ...2010\ (\textrm{mod}\ 10)</math> and <math>a_1 \equiv 2010 - a_0\ (\textrm{mod}\ 100)</math>. It's easy to see that exactly 10 values in <math>0 \leq a_0 ...th>. However, <math>2010-10a_1-a_0 = 1000a_3 + 100a_2 \equiv 0\ (\textrm{mod}\ 100)</math>. Thus, we have <math>2010-10a_1-a_0 \geq 1000</math> always.
  7 KB (1,200 words) - 12:38, 19 June 2024
• Mock AIME 2 2010 Answer Key
  ...ree rightmost nonzero digits of the product of the coefficients of <math>P(x)</math>. (Alex Zhu) ...>b_i \in \{0,1\}</math>, is obtained from multiplying together all <math>i x^i</math> with <math>b_i = 1</math>.) Therefore, we are trying to find the p
  36 KB (6,214 words) - 20:22, 13 July 2023
• 2010 USAMO Problems/Problem 1
  Angles are directed mod 180 {don't worry}(take so that the solution can be applied to any configura // points X, Y, Z
  13 KB (2,271 words) - 14:08, 23 May 2024
• 1984 USAMO Problems/Problem 2
  ...numbers cancel out, the geometric means divided, or <math>{\sqrt[q]{\frac{x}{y}}}</math>, must be rational. This is a contradiction, so no such infinit ...emainder when divided by <math>q</math>. (Otherwise, because <math>v_p(n) \mod q</math> has <math>q</math> distinct values, <math>S</math> could contain a
  4 KB (855 words) - 03:12, 23 October 2022
• 2011 AMC 10A Problems/Problem 23
  ...0,\cdots.</math> She excludes <math>5\mod9.</math> Let each mod be <math>a\mod b.</math> We notice that, each time, <math>b</math> triples, and <math>a=\d ...ace in the next person's set. Let George's number be the item number <math>x</math> in his set (obviously <math>1</math>). Then Fatima places sets of th
  6 KB (926 words) - 23:38, 8 April 2024
• 2011 AIME I Problems/Problem 11
  ...quiv 2^j \pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation: S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.
  12 KB (1,905 words) - 18:00, 19 June 2024
• 2011 AIME I Problems/Problem 15
  For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</m <math>(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc</math>
  8 KB (1,302 words) - 04:07, 24 July 2023
• 2011 USAJMO Problems/Problem 1
  Let <math>2^n + 12^n + 2011^n = x^2</math>. Then <math>(-1)^n + 1 \equiv x^2 \pmod {3}</math>.
  6 KB (990 words) - 21:49, 1 October 2021
• 2003 AMC 10B Problems/Problem 25
  <cmath> I. x+y = 1, \{10\} = 1 </cmath> <cmath> II. x+y = 4, \{13, 22, 31, 40\} = 4 </cmath>
  6 KB (882 words) - 11:45, 12 November 2023
• 1979 USAMO Problems/Problem 3
  ...problem is equivalent to proving that <math>4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3</math>. Try proving this inequality for all nonnegative real numbers x, y, z; not just positive integers.
  2 KB (307 words) - 14:14, 7 June 2021
• 1964 AHSME Problems/Problem 3
  When a positive integer <math>x</math> is divided by a positive integer <math>y</math>, the quotient is <ma What is the remainder when <math>x+2uy</math> is divided by <math>y</math>?
  2 KB (296 words) - 06:04, 31 December 2023
• 1995 AJHSME Problems/Problem 15
  ...igits, digit number x = digit number (x mod 3), and the 100th digit = (100 mod 3)th digit = 1st digit = <math>\boxed{\text{(B)}\ 1}</math>
  577 bytes (81 words) - 00:23, 5 July 2013
• 2012 AMC 10A Problems/Problem 16
  ...l lap the slower runner exactly once, or run 500 meters farther. Let <math>x</math> be the time these runners run in seconds. <cmath>4.8x-4.4x=500 \Rightarrow x=1250</cmath>
  5 KB (761 words) - 00:21, 14 June 2024
• 2012 AMC 12A Problems/Problem 17
  ...e by <math>5</math> if paired with another element that is <math>0\ (\text{mod}\ 5)</math>. The final answer is <math>\boxed{\textbf{(B)}\ 13}</math>. ...mod 5), since we can still have one of those. The <math>k</math> and <math>x</math> values are largest values, not quantities. So there are <math>6</mat
  3 KB (519 words) - 19:01, 30 March 2024
• 1992 AHSME Problems/Problem 23
  The fact that <math>x \equiv 0 \mod 7 \Rightarrow 7 \mid x</math> is assumed as common knowledge in this answer. ...th>7</math> possible numbers equivalent to each of <math>2</math>-<math>6 \mod 7</math>.
  2 KB (246 words) - 13:25, 16 July 2024

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