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• 2006 AIME I Problems/Problem 4
  ...oduct <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is divided by <math>1000</math>. ...<math>5</math> dividing it, for <math>76</math> extra; every n! for <math>n\geq 50</math> has one more in addition to that, for a total of <math>51</ma
  2 KB (278 words) - 08:33, 4 November 2022
• 2006 AIME I Problems/Problem 3
  ...math> so we take <math>N_0 = 25</math> and <math>n = 2.</math> Then <cmath>N = 7 \cdot 10^2 + 25 = \boxed{725},</cmath> and indeed, <math>725 = 29 \cdot ...ow that <math>N<1000</math> (because this is an AIME problem). Thus, <math>N</math> has <math>1,</math> <math>2</math> or <math>3</math> digits. Checkin
  4 KB (622 words) - 03:53, 10 December 2022
• 2022 AIME I Problems/Problem 1
  Q(20) &= &\hspace{1mm}-800 + 20c + d &= 53, &&(4) ...]</math> from <math>5\cdot[(1)+(2)]</math> to get <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath>
  4 KB (670 words) - 13:03, 13 November 2023
• 2021 JMC 10
  ...bf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 7</math> ==Problem 4==
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  ...ne <math>x\spadesuit y = (x + y)(x - y)</math>. What is <math>3\spadesuit(4\spadesuit 5)</math>? == Problem 4 ==
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2006 AMC 12A Problems
  == Problem 4 == [[2006 AMC 12A Problems/Problem 4|Solution]]
  15 KB (2,223 words) - 13:43, 28 December 2020
• 2005 AMC 12A Problems
  ...8 \qquad (\mathrm {B}) \ -4 \qquad (\mathrm {C})\ 2 \qquad (\mathrm {D}) \ 4 \qquad (\mathrm {E})\ 8 == Problem 4 ==
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2004 AMC 12A Problems
  == Problem 4 == [[2004 AMC 12A Problems/Problem 4|Solution]]
  13 KB (1,953 words) - 00:31, 26 January 2023
• 2003 AMC 12A Problems
  Members of the Rockham Soccer League buy socks and T-shirts. Socks cost &#36;4 per pair and each T-shirt costs &#36;5 more than a pair of socks. Each memb <math> \mathrm{(A) \ } 4.5\qquad \mathrm{(B) \ } 9\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 18
  13 KB (1,955 words) - 21:06, 19 August 2023
• 2002 AMC 12A Problems
  <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> <math> \mathrm{(A) \ } \frac{7}{2}\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 13 </
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2000 AMC 12 Problems
  ...00^{4000} \qquad \textbf{(D)}\ 4,000,000^{2000} \qquad \textbf{(E)}\ 2000^{4,000,000}</math> == Problem 4 ==
  13 KB (1,948 words) - 12:26, 1 April 2022
• 2001 AMC 12 Problems
  Let <math>P(n)</math> and <math>S(n)</math> denote the product and the sum, respectively, of the digits ...example, <math>P(23) = 6</math> and <math>S(23) = 5</math>. Suppose <math>N</math> is a
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  \qquad\mathrm{(C)}\ 4 when <math>x=4</math>?
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2003 AMC 12B Problems
  <cmath>\frac{2-4+6-8+10-12+14}{3-6+9-12+15-18+21}?</cmath> == Problem 4 ==
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  == Problem 4 == [[2004 AMC 12B Problems/Problem 4|Solution]]
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  ...x</math> has the property that <math>x\%</math> of <math>x</math> is <math>4</math>. What is <math>x</math>? \mathrm{(B)}\ 4 \qquad
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 14
  ...h>, <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? <math>253=N(4B+5J)</math>
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 18
  ...e object only makes <math>1</math> move, it is obvious that there are only 4 possible points that the object can move to. At this point we can guess that for n moves, there are <math>(n + 1)^2</math> different endpoints. Thus, for 10 moves, there are <math>11^2
  2 KB (354 words) - 16:57, 28 December 2020
• 2006 AMC 12B Problems/Problem 20
  ...th hold at the same time if and only if <math>10^k \leq x < \frac{10^{k+1}}4</math>. ...math>k</math> the length of the interval <math>\left[ 10^k, \frac{10^{k+1}}4 \right)</math> is <math>\frac 32\cdot 10^k</math>.
  3 KB (485 words) - 14:09, 21 May 2021
• 2006 AMC 12B Problems/Problem 22
  ...divisible by <math>10</math>. What is the smallest possible value of <math>n</math>? n{5}\right\rfloor +
  5 KB (881 words) - 15:52, 23 June 2021

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