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  • * Prove that <math>n^2 + 3n + 5</math> is never divisible by 121 for any positive integer <math>{n}</math>.
    3 KB (532 words) - 21:00, 13 January 2024
  • 121 take algebra and history. <math>=243+323+143+241+300-213-264-144-121-111-90-80-60-70-60</math>
    9 KB (1,703 words) - 00:20, 7 December 2024
  • <math>\text(AC)^2 = (AD)^2 - 121</math>
    7 KB (1,198 words) - 08:36, 8 December 2024
  • ...g the former). Lop off the ending 0's and repeat. 1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==> 1 NOPE. Works in general with numbers that are relative
    10 KB (1,572 words) - 21:11, 22 September 2024
  • ...e are no solutions when <math>x \geq 3</math>. Thus, there are <math>68+53=121</math> positive integer solutions to the equation. <math>\square</math>
    5 KB (709 words) - 16:40, 24 September 2024
  • ...5 96 98 99 100 102 104 105 106 108 110 111 112 114 115 116 117 118 119 120 121 122 123 124 125 126 128 129 130 132 133 134 135 136 138 140 141 142 143 144
    6 KB (350 words) - 11:58, 26 September 2023
  • ...enominator are relatively prime), are divisible by five? <cmath>\frac{4^b +121^b}{2^a -3^b}</cmath>
    12 KB (1,784 words) - 15:49, 1 April 2021
  • \mathrm{(B)}\ 121 \mathrm{(D)}\ 121
    13 KB (2,058 words) - 11:36, 4 July 2023
  • \mathrm{(B)}\ 121 ...x|</math> valid choices for <math>y</math>, giving a total of <math>\boxed{121}</math> possible positions.
    2 KB (354 words) - 15:57, 28 December 2020
  • \mathrm{(D)}\ 121 PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}
    7 KB (1,182 words) - 13:31, 1 September 2024
  • ...= 2^{10}\cdot 5^{10}</math> so <math>10^{10}</math> has <math>11\cdot11 = 121</math> [[divisor]]s. Now, we use the [[Principle of Inclusion-Exclusion]]. We have <math>121 + 64 + 276</math> total potential divisors so far, but we've overcounted th
    3 KB (377 words) - 17:36, 1 January 2024
  • ...ac{10}{8}</math> days. It has been <math>1+\frac{10}{9}+\frac{10}{8}=\frac{121}{36}</math> days since the job started, and we still have <math>\frac{23}{3
    4 KB (592 words) - 18:02, 26 September 2020
  • A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allo
    7 KB (1,045 words) - 00:18, 5 January 2025
  • <math>16 \cdot (x - 49)^2 = 121 \cdot (x - 9)^2</math>.
    5 KB (982 words) - 23:57, 5 December 2024
  • ...^2}}\Longrightarrow\frac{11}{8}=\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{121}{64}=4-\frac{9}{r^2}</cmath> <cmath>\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}</cmath>
    5 KB (784 words) - 20:05, 8 December 2024
  • A sample of 121 [[integer]]s is given, each between 1 and 1000 inclusive, with repetitions ...= \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right|
    5 KB (851 words) - 17:01, 28 December 2022
  • ...We find the former is true for <math>(x,y) = (5,11)</math>. <math>x^2+y^2=121+25=146</math>.
    4 KB (628 words) - 21:05, 7 June 2021
  • <math>(a_5,b_5)=\left(\frac{122}{243}, \frac{121}{243}\right)</math>
    7 KB (1,058 words) - 19:57, 22 December 2020
  • <math>a^2 + 121 = b^2 + 1369 \longrightarrow a^2 = b^2 + 1248</math>. Expanding the second
    5 KB (789 words) - 20:30, 1 January 2025
  • 37&79&121&&& \
    3 KB (436 words) - 11:34, 14 June 2024

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