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  • <math>\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 144</math>
    1 KB (216 words) - 23:23, 5 September 2024
  • ...s+(N-1)+N=\frac{N(N+1)}{2}=2016.</cmath> Noticing that <math>\frac{63\cdot 64}{2}=2016,</math> we have <math>N=63,</math> so our answer is <math>\boxed{\ ...learn that when we subtract <math>70, 69, 68, 67, 66, 65,</math>and <math>64, N = 63.</math> Adding those two digits, we get the answer <math>\boxed{\te
    2 KB (395 words) - 22:29, 3 December 2024
  • ...32 33 34 35 36 38 39 40 42 44 45 46 48 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 So, this concludes that all even positive integers except 2,4 and 64 are sum of 2 composite numbers
    6 KB (350 words) - 11:58, 26 September 2023
  • <math>\log_{225}x+\log_{64}y=4</math> <math>\log_{x}225-\log_{y}64=1</math>
    4 KB (680 words) - 11:54, 16 October 2023
  • ...000</math> implies that <math>\frac{n+1}{64}\le 15</math>, so <math>n+1=64,64\cdot 3^2</math>.
    10 KB (1,702 words) - 21:23, 25 July 2024
  • ...} 47 \qquad \textbf{(B) } 51 \qquad \textbf{(C) } 63 \qquad \textbf{(D) } 64 \qquad \textbf{(E) } 75</math>
    12 KB (1,784 words) - 15:49, 1 April 2021
  • ...\qquad\textrm{(C)}\ 48\qquad\textrm{(D)}\ 32+16\sqrt{3}\qquad\textrm{(E)}\ 64 </math>
    13 KB (1,955 words) - 20:06, 19 August 2023
  • ...{(A) } 0 \qquad \text {(B) } 13 \qquad \text {(C) } 37 \qquad \text {(D) } 64 \qquad \text {(E) } 83
    13 KB (1,987 words) - 17:53, 10 December 2022
  • \qquad\mathrm{(B)}\ -64 \qquad\mathrm{(D)}\ 64
    13 KB (2,049 words) - 12:03, 19 February 2020
  • ...are 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}
    1 KB (167 words) - 00:27, 26 September 2024
  • ...ore time: <math>8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0</math>. The only positive value for <math
    2 KB (299 words) - 14:29, 5 July 2022
  • ...ually works, consider the quadratics <math>x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0</math>.
    2 KB (322 words) - 01:57, 11 November 2024
  • ...{(A)}\ \frac{35}{2}\qquad\mathrm{(B)}\ 15\sqrt{2}\qquad\mathrm{(C)}\ \frac{64}{3}\qquad\mathrm{(D)}\ 16\sqrt{2}\qquad\mathrm{(E)}\ 24</math>
    13 KB (2,028 words) - 15:32, 22 March 2022
  • ...{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad</math> Simplifying, we have that <math>64 = 8a^2 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8} = 2 \sqrt{2}.</math>
    5 KB (811 words) - 10:44, 30 November 2024
  • ...> ways for the first digit to be unoccupied. There are <math>2(3 \cdot (56+64))</math> = <math>720</math> arrangements. ...for the non-<math>2</math> or <math>3</math> digits. We have <math>6 \cdot 64</math> = <math>384</math> arrangements. If the <math>2</math> or the <math>
    3 KB (525 words) - 19:25, 30 April 2024
  • <math>\text{(A)}\ \frac{1}{1024} \qquad \text{(B)}\ \frac{15}{64} \qquad \text{(C)}\ \frac{243}{1024} \qquad \text{(D)}\ \frac{1}{4} \qquad
    17 KB (2,246 words) - 12:37, 19 February 2020
  • <math>15^7 = 3^7\cdot5^7</math> so <math>15^7</math> has <math>8\cdot8 = 64</math> divisors. Now, we use the [[Principle of Inclusion-Exclusion]]. We have <math>121 + 64 + 276</math> total potential divisors so far, but we've overcounted those f
    3 KB (377 words) - 17:36, 1 January 2024
  • ...stinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)
    3 KB (547 words) - 18:15, 4 April 2024
  • ...corners+ the area of an isoceles triangle. Adding these all up gives <math>64=\frac{d^2}{4}+\frac{d^2}{4}+(8-\frac{d\sqrt{2}}{2}+\frac{d}{2})(8-\frac{d}{
    4 KB (707 words) - 10:11, 16 September 2021
  • 8&0&0&0&1&1&2&4&8&16&32&64\ 9&0&0&1&1&2&4&8&16&32&64&128\
    9 KB (1,491 words) - 00:23, 26 December 2022

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