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• 2020 AIME II Problems/Problem 5
  ...h this step, note that <math>2^{10} = 1024 \equiv 24 \pmod{1000}</math>) ~ TopNotchMath
  3 KB (414 words) - 22:01, 23 March 2023
• 2020 AIME II Problems/Problem 11
  ...= R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath
  6 KB (1,015 words) - 21:16, 28 December 2023