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  • ...th>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>.
    4 KB (569 words) - 21:34, 30 December 2024
  • ...rithmetic sequence with common difference <math>1</math> and <math>99, 91, 83, 75</math> is an arithmetic sequence with common difference <math>-8</math> * [[2012 AIME I Problems/Problem 2]]
    4 KB (736 words) - 01:00, 7 March 2024
  • \text{\bf Fold 1.} & 942 \text{ is to the right.} & 942 \mapsto 83 & \text{There is } 2^1 - 1 - 0=\boxed{1} \text{ square below it.} \ \textbf{Fold 2.} & 83 \text{ is to the left.} & 83 \mapsto 83 & \text{There is } \boxed{1} \text{ square below it.} \
    6 KB (899 words) - 19:58, 12 May 2022
  • {{AIME Problems|year=1983}} [[1983 AIME Problems/Problem 1|Solution]]
    7 KB (1,104 words) - 02:13, 27 May 2024
  • ...th>a_n=6^{n}+8^{n}</math>. Determine the remainder upon dividing <math>a_ {83}</math> by <math>49</math>. ...umbers in terms of <math>7</math>, we get <math>a_{83} = (7-1)^{83}+(7+1)^{83}</math>.
    3 KB (361 words) - 19:20, 14 January 2023
  • 2. <math>166 = 83 \cdot 2 => 81 + 85</math> - refuted [https://youtu.be/8_C0lmbpRWA?si=1d1wRlity1w_nHHf 1984 AIME #14]
    8 KB (1,365 words) - 14:38, 10 December 2024
  • 41&83&125&&& \ {{AIME box|year=1995|num-b=9|num-a=11}}
    3 KB (436 words) - 11:34, 14 June 2024
  • ...math>\frac{332}{1996}=\frac{83}{499}</math>. The answer is then <math>499+83=\boxed{582}</math>. ...work. Thus, <math>m/n = 332/1996 = 83/499.</math> So our answer is <math>83 + 499 = \boxed{582}.</math>
    6 KB (932 words) - 14:55, 16 November 2024
  • ...Therefore, <math>CB = MB</math>, so <math>\angle CMB = \angle MCB = \boxed{83^\circ}</math>. ...]] [[measure]]s: <math>m\angle AMC = 150^\circ</math>, <math>m\angle MCB = 83^\circ</math>. If we define <math>m\angle CMB = \theta</math> then we also
    7 KB (1,082 words) - 01:08, 30 September 2024
  • We set up a trivial coordinate bash. Let A = 0,0, C = 48,0, B = 83/2, 13sqrt3/2. We find the coordinates of the circumcenter to be 24, -11sqrt *[[Mock AIME 4 2006-2007 Problems/Problem 14| Previous Problem]]
    5 KB (734 words) - 13:46, 27 December 2024
  • ...iple of <math>12</math> to <math>1000</math> is <math>996</math> (<math>12*83</math>), so we know that this is the last term in the sequence. Therefore, {{AIME box|year=2007|n=I|before=First Question|num-a=2}}
    1 KB (204 words) - 12:56, 7 February 2023
  • ...can partition <math>S</math> into <math>S \cap \{3, 4, 5, 6, 7, 8, 81, 82, 83, 84 ... 242\}</math> and <math>S \cap \{9, 10, 11 ... 80\}</math>, and in n [https://youtu.be/bjHBaOeFt6g?si=ZcIOVMm6PpVdv0Ii 2010 AIME I #12]
    3 KB (532 words) - 15:29, 26 December 2024
  • &=2^{22}(\sin(1)\sin(89)\sin(3)\sin(87)\sin(5)\sin(85)\sin(7)\sin(83)\cdots\sin(41)\sin(49)\sin(43)\sin(47))\ ...(1)\sin(3)\sin(5)\sin(7)\cdots\sin(41)\sin(43))(\sin(47)\sin(49)\cdots\sin(83)\sin(85)\sin(87)\sin(89))\end{align*}</cmath>
    11 KB (1,634 words) - 18:51, 18 February 2025
  • 83. Gmaas ate cat-food today. EDIT: Gmaas also ate it yesterday. EDIT: Only be 152. The reason why AIME cutoffs aren't out yet is that Gmaas refused to grade them due to too much
    69 KB (11,805 words) - 19:49, 18 December 2019
  • A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997); A = (18,83); B = (18,39); C = (78,99); D = (56, 104.908997);
    13 KB (2,303 words) - 21:41, 2 January 2025
  • 83. Almighty Gmaas ate cat-food today. EDIT: Almighty Gmaas also ate it yester 152. The reason why AIME cutoffs aren't out yet is that Almighty Gmaas refused to grade them due to
    99 KB (14,096 words) - 22:49, 19 February 2025
  • So, <math>m\in\{83,166,332\}</math>. Testing the cases, only <math>332</math> fails. This leav ...use <math>n+5 > 17</math>, the only choices for <math>n+5</math> are <math>83, 166,</math> and <math>332.</math> Checking all three cases verifies that <
    6 KB (918 words) - 09:27, 21 December 2024
  • Given all conditions above, the possible <math>q</math> are 74, 83, 88, 92, 97, 101, 106, 109, 116, 118, 127. [[File:AIME-I-2022-14a.png|400px|right]]
    16 KB (2,730 words) - 01:56, 4 January 2023
  • ...eed <math>2^n \equiv n \equiv 2^3 \pmod{25}</math>; clearly <math>n \equiv 83 \pmod{100}</math>. In the case that <math>n \equiv 17 \pmod{20}</math>, by In the final step, we need to calculate <math>2^{97}</math> and <math>2^{83}</math> modulo <math>125</math>:
    19 KB (2,691 words) - 04:58, 17 February 2025
  • ...ns, etc. Thus, we have: <cmath>f(3) = f(2)+7+(12+10+8+...+2)=34+7+6\cdot 7=83.</cmath> [[File:AIME 2022 II 9-min.png|350px|right]]
    12 KB (2,025 words) - 13:56, 25 January 2024

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