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• 2006 AIME I Problems/Problem 6
  ...e digits, 0 through 9, is 45. So the sum of all the numbers is <math>\frac{45\times72\times111}{999}= \boxed{360} </math>. {{AIME box|year=2006|n=I|num-b=5|num-a=7}}
  2 KB (237 words) - 19:14, 20 November 2023
• 2006 AIME I Problems/Problem 2
  ...e possible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>. ...math>, so the number of possible values of T, and therefore S, is <math>955-55+1=\boxed{901}</math>.
  1 KB (189 words) - 20:05, 4 July 2013
• 2021 JMC 10
  <math>\textbf{(A) }\pi-e \qquad\textbf{(B) }2\pi-2e\qquad\textbf{(C) }2e\qquad\textbf{(D) }2\pi \qquad\textbf{(E) }2\pi +e</m ...h>74</math> and <math>83</math> are pretentious. How many pretentious three-digit numbers are odd?
  12 KB (1,784 words) - 16:49, 1 April 2021
• 2006 AMC 12B Problems
  ...1.99</math>, and <math>\textdollar0.99</math>. Mary will pay with a twenty-dollar bill. Which of the following is closest to the percentage of the <ma How many even three-digit integers have the property that their digits, read left to right, are
  13 KB (2,058 words) - 12:36, 4 July 2023
• 2005 AMC 12A Problems
  ...they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met? ...on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math
  13 KB (1,971 words) - 13:03, 19 February 2020
• 2002 AMC 12A Problems
  <math>(2x+3)(x-4)+(2x+3)(x-6)=0 </math> ..., and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.
  12 KB (1,792 words) - 13:06, 19 February 2020
• 2001 AMC 12 Problems
  two-digit number such that <math>N = P(N)+S(N)</math>. What is the units digit o A telephone number has the form <math>\text{ABC-DEF-GHIJ}</math>, where each letter represents
  13 KB (1,957 words) - 12:53, 24 January 2024
• 2002 AMC 12B Problems
  The positive integers <math>A, B, A-B, </math> and <math>A+B</math> are all prime numbers. The sum of these four For how many integers <math>n</math> is <math>\dfrac n{20-n}</math> the square of an integer?
  10 KB (1,547 words) - 04:20, 9 October 2022
• 2004 AMC 12B Problems
  In the expression <math>c\cdot a^b-d</math>, the values of <math>a</math>, <math>b</math>, <math>c</math>, and Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2006 AMC 12B Problems/Problem 10
  \text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47 {{AMC12 box|year=2006|ab=B|num-b=9|num-a=11}}
  977 bytes (156 words) - 13:57, 19 January 2021
• 2006 AMC 12B Problems/Problem 23
  ...arrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. ...rrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}.
  7 KB (1,169 words) - 14:04, 10 June 2022
• 2006 AMC 12A Problems/Problem 17
  But it can also be seen that <math>\angle BDA = 45^\circ</math>. Therefore, since <math>D</math> lies on <math>\overline{BE}</ ...\circ.</math> Also, <math>ED = EG,</math> which implies <math>\angle EGD = 45^\circ</math>, so <math>\triangle EDG</math> is an isosceles right triangle.
  6 KB (958 words) - 23:29, 28 September 2023
• 2006 AMC 12A Problems/Problem 22
  Project any two non-adjacent and non-opposite sides of the [[hexagon]] to the [[circle]]; the [[arc]] between the [[Image:2006_12A_AMC-22.png]]
  2 KB (343 words) - 15:39, 14 June 2023
• 2005 AMC 12B Problems/Problem 15
  The sum of four two-digit numbers is <math>221</math>. None of the eight digits is <math>0</math <math>221</math> can be written as the sum of four two-digit numbers, let's say <math>\overline{ae}</math>, <math>\overline{bf}</ma
  2 KB (411 words) - 21:02, 21 December 2020
• 2005 AMC 12B Problems/Problem 18
  ..., which is <math>\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}</math>, which is approximately <math>51</math>. The answer {{AMC12 box|year=2005|ab=B|num-b=17|num-a=19}}
  2 KB (262 words) - 21:20, 21 December 2020
• 2006 AMC 10A Problems
  Define <math>x\otimes y=x^3-y</math>. What is <math>h\otimes (h\otimes h)</math>? Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half
  13 KB (2,028 words) - 16:32, 22 March 2022
• 2006 AMC 10A Problems/Problem 15
  <math>\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad</math> ...et exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are <math>\frac{30}{\frac{2\pi}{5}} \approx 23.8</ma
  3 KB (532 words) - 17:49, 13 August 2023
• 2006 AMC 10A Problems/Problem 17
  ...>BZ = \frac 12AH = 1</math>, so <math>\triangle BWZ</math> is a <math>45-45-90 \triangle</math>. Hence <math>WZ = \frac{1}{\sqrt{2}}</math>, and <math>[ ...ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]].
  6 KB (1,066 words) - 00:21, 2 February 2023
• Newman's Tauberian Theorem
  (which is well-defined by this formula for <math>\Re s>0</math>) admits an to some [[open set | open]] domain <math>E</math> containing the closed half-plane
  6 KB (1,034 words) - 07:55, 12 August 2019
• 2005 AIME II Problems/Problem 5
  ...e case <math> b=a^2 </math>, note that <math> 44^2=1936 </math> and <math> 45^2=2025 </math>. Thus, all values of <math>a</math> from <math>2</math> to < ...re <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 4
  3 KB (547 words) - 19:15, 4 April 2024

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