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  • ...as Solution 1, we find that 1999 has position <math>1024</math> in a <math>2048</math> card stack, where the fake cards towards the front. ...we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were <math>928 - 1 = \boxed{927}</math> cards a
    15 KB (2,673 words) - 18:16, 6 January 2024
  • ...}</math>, so the sum of these elements is <math>\sum_{i=0}^{5} {2i \choose i} = 1 + 2 +6 + 20 + 70 + 252 = 351</math>. ...ing out the <math> 44</math> numbers between <math> 2003</math> and <math> 2048</math> gives <math> 1155</math>. Thus the answer is <math> 155</math>.
    4 KB (651 words) - 18:42, 7 October 2023
  • ...we have <math>z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}</math>. ...could let <math>z = a + bi</math> and solve to get <math>z=\cos 3^\circ + i\sin 3^\circ</math>.
    4 KB (671 words) - 19:24, 2 November 2024
  • <math>W_S = \sum_{i=1}^{|D|} \tbinom{6}{x_i}\tbinom{6}{y_i}</math> if and only if there exists ...>S</math> is divisible by 3. Therefore, by the fact that <math>W_S = \sum_{i=1}^{|D|} \tbinom{6}{x_i}\tbinom{6}{y_i}</math>, we have that;
    26 KB (4,062 words) - 12:03, 19 January 2025
  • ===AIME styled=== (i) <math>1</math> ring changed position (i.e., that ring is transferred from one peg to another)
    64 KB (10,594 words) - 21:50, 9 January 2025