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- ~Williamgolly4 KB (623 words) - 19:32, 15 February 2021
- Solution by Williamgolly5 KB (766 words) - 19:26, 15 February 2021
- ~Solution by Williamgolly3 KB (515 words) - 13:46, 14 February 2021
- ~solution by williamgolly2 KB (307 words) - 20:06, 15 July 2020
- Note: for each combination of 7 numbers, exactly 1 is in increasing order -Williamgolly1 KB (195 words) - 16:29, 29 November 2019
- ~Williamgolly3 KB (547 words) - 01:16, 3 March 2021
- Explanation of the bijection by WIlliamgolly:4 KB (695 words) - 21:34, 28 February 2021
- ...we can solve for DC and see it is congruent to a 30 60 90 triangle by SSS ~Williamgolly2 KB (371 words) - 11:20, 28 December 2020
- ...ferences. (This is due to the fact that we are calculating |d| at the end ~Williamgolly) Hence we can [[WLOG]] assume that <math>b=2</math>.3 KB (455 words) - 00:41, 19 January 2021
- Motivation(by williamgolly, solution by MAA):15 KB (2,229 words) - 18:09, 27 December 2020
- ...circle's interior (this is because we want no two chords to be parallel ~Williamgolly). Therefore, the answer is <math>{{8}\choose{6}}</math>, which is equivalen1 KB (171 words) - 14:15, 10 August 2020
- Sidenote by Williamgolly:1 KB (189 words) - 09:44, 11 July 2020
- Note from williamgolly:6 KB (968 words) - 23:02, 2 February 2021
- Note by Williamgolly:5 KB (845 words) - 22:34, 4 December 2020
- Williamgolly6 KB (990 words) - 18:19, 11 January 2021
- Note by Williamgolly:934 bytes (151 words) - 16:43, 13 July 2020
- Note from Williamgolly: To see why symmetry occurs here, we group the factors of 21!into 2 groups,2 KB (285 words) - 18:11, 17 January 2021
- -Williamgolly ~Williamgolly5 KB (866 words) - 19:55, 2 February 2021
- Note from Williamgolly: We can WLOG assume <math>a_1,a_2... a_{2017} \equiv 0 \pmod 6</math> and h5 KB (820 words) - 21:09, 30 January 2021
- Revised solution by Williamgolly (includes bijections):5 KB (836 words) - 23:32, 2 February 2021