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- 1 KB (186 words) - 18:28, 21 November 2018
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- *AIME floor: 81 (top ~15%) *Average score: 57.8117 KB (1,900 words) - 19:38, 3 January 2025
- *(BorealBear) Find the last two digits of <math> 7^{81}-3^{81} </math>. ([[Euler's Totient Theorem Problem 1 Solution|Solution]])4 KB (569 words) - 21:34, 30 December 2024
- A hexagon is inscribed in a circle. Five of the sides have length <math>81</math> and the sixth, denoted by <math>\overline{AB}</math>, has length <ma7 KB (1,198 words) - 08:36, 8 December 2024
- ...ded and the first digit's number removed. Then there are <math>9 + 3 \cdot 81 = 252</math> of these numbers. ...ath>9</math> options, as does the second. Then there are <math>9 + 7 \cdot 81 = 576</math> of these numbers.5 KB (709 words) - 16:40, 24 September 2024
- ...49 50 51 52 54 55 56 57 58 60 62 63 64 65 66 68 69 70 72 74 75 76 77 78 80 81 82 84 85 86 87 88 90 91 92 93 94 95 96 98 99 100 102 104 105 106 108 110 116 KB (350 words) - 11:58, 26 September 2023
- ...c 2 3</math>. Thus, the integral is not much less than <math>\left(\frac 4{81}\right)^n</math>.8 KB (1,469 words) - 20:11, 16 September 2022
- ...ath>z_0</math> is even. QED [Oops, this doesn't work. 21 (or <math>3^4 = 81</math>) are equal to <math>1\mod 5</math> and not even...]9 KB (1,434 words) - 00:15, 4 July 2024
- ...1}{729}\qquad \mathrm{(C) \ } \frac{2}{243}\qquad \mathrm{(D) \ } \frac{1}{81}\qquad \mathrm{(E) \ } \frac{5}{243}</math>15 KB (2,223 words) - 12:43, 28 December 2020
- ...{(B) } 4 \qquad \text {(C) } 36 \qquad \text {(D) } 49 \qquad \text {(E) }81</math>13 KB (1,953 words) - 23:31, 25 January 2023
- ...les defined by <math>(x-10)^2 + y^2 = 36</math> and <math>(x+15)^2 + y^2 = 81</math>12 KB (1,792 words) - 12:06, 19 February 2020
- \mathrm{(B)}\ \frac{81}{5}13 KB (2,049 words) - 12:03, 19 February 2020
- ...less than or equal to <math>120</math> are <math>\{0,1,4,9,16,25,36,49,64,81,100\}</math>, so there are <math>11</math> values for <math>120-\sqrt{x}</m1 KB (167 words) - 00:27, 26 September 2024
- ...rac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) } \frac{5}{243}</math> Therefore, there is only <math>1</math> way out of <math>81</math> to get to <math>D</math> last.6 KB (1,083 words) - 13:05, 25 November 2024
- ...\frac{1}{729}\qquad\mathrm{(C)}\ \frac{2}{243}\qquad\mathrm{(D)}\ \frac{1}{81}\qquad\mathrm{(E)}\ \frac{5}{243}</math>13 KB (2,028 words) - 15:32, 22 March 2022
- ...\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad \text{(E)}\ \frac{81}{256}</math>17 KB (2,246 words) - 12:37, 19 February 2020
- ...is to the right.} & 14 \mapsto 3 & \text{There are } 2^7 - 1 - 46 = \boxed{81} \text{ squares below it.}\ ...d 8.} & 3 \text{ is to the left.} & 3 \mapsto 3 & \text{There are } \boxed{81} \text{ squares below it.} \6 KB (899 words) - 19:58, 12 May 2022
- A hexagon is inscribed in a circle. Five of the sides have length 81 and the sixth, denoted by <math>\overline{AB}</math>, has length 31. Find t7 KB (1,106 words) - 21:05, 7 June 2021
- ...ur answer is then <cmath>\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}</cmath>3 KB (532 words) - 12:05, 29 November 2024
- ...lf of our terms cancel out and we are left with <math>2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)</math>. We realize that all of these terms are divisibl <math>6^{83} + 8^{83} = (6+8)(6^{82}-6^{81}8+\ldots-8^{81}6+8^{82})</math>3 KB (361 words) - 19:20, 14 January 2023
- 2. <math>166 = 83 \cdot 2 => 81 + 85</math> - refuted 3. <math>158 = 79 \cdot 2 => 77 + 81</math> - refuted8 KB (1,365 words) - 14:38, 10 December 2024