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• 2015 AIME II Problems/Problem 3
  ...pmod{9}</math>. In this problem, we know <math>17\mid m</math>, or <math>m=17k</math> for a positive integer <math>k</math>. Also, we know that <math>m\equiv 17\equiv -1\pmod{9}</math>, or <math>17k\equiv -k\equiv -1\pmod{9}</math>.
  5 KB (741 words) - 14:03, 14 February 2023
• 2017 Indonesia MO Problems
  .../math> not greater than 2017 such that <math>n</math> divides <math>20^n + 17k</math> for some positive integer <math>k</math>.
  3 KB (531 words) - 00:07, 13 August 2018
• 2017 Indonesia MO Problems/Problem 6
  .../math> not greater than 2017 such that <math>n</math> divides <math>20^n + 17k</math> for some positive integer <math>k</math>. ...can also show that if <math>k \equiv 0 \pmod{5^b},</math> the value <math>17k</math> is congruent to <math>0</math> modulo <math>5^b.</math>
  2 KB (286 words) - 01:33, 13 August 2018
• 2021 AIME II Problems/Problem 14
  ...linear. Let <math>\angle ABC=13k,\angle BCA=2k</math> and <math>\angle XOY=17k.</math> ...le OAM \sim \triangle OXY</math> by AA, so <math>\angle AOM = \angle XOY = 17k.</math>
  10 KB (1,526 words) - 16:50, 25 December 2022
• 2024 AIME I Problems/Problem 13
  So \(m\equiv\pm2\), \(\pm8\pmod{17}\). If \(m\equiv2\pmod{17}\), let \(m=17k+2\), by the binomial theorem, 0&\equiv(17k+2)^4+1\equiv\mathrm {4\choose 1}(17k)(2)^3+2^4+1=17(1+32k)\pmod{17^2}\\[3pt]
  4 KB (616 words) - 19:30, 23 May 2024