Thread: Numbers of the form 41s+r View Single Post
 2019-01-18, 12:11 #2 ATH Einyen     Dec 2003 Denmark 2×1,657 Posts 41*21 = 1 (mod 43) so 41^(-1) = 21 (mod 43): 41s+1: --------- 41s+1 = 0 (mod 43) => 41s = -1 (mod 43) => s=(-1)*41^(-1) = (-1)*21 = -21 = 22 (mod 43). so s=22+43n and the numbers are then: 41*(43n+22)+1 = 1763n+903 41s+10: --------- 41s+10 = 0 (mod 43) => 41s = 33 (mod 43) => s=33*21 = 5 (mod 43) so s=5+43n and the numbers are then: 41*(43n+5)+10 = 1763n+215 41s+16: --------- 41s+16 = 0 (mod 43) => 41s = 27 (mod 43) => s=27*21 = 8 (mod 43) so s=8+43n and the numbers are then: 41*(43n+8)+16 = 1763n+344 41s+18: --------- 41s+18 = 0 (mod 43) => 41s = 25 (mod 43) => s=25*21 = 9 (mod 43) so s=9+43n and the numbers are then: 41*(43n+9)+18 = 1763n+387 41s+37: --------- 41s+37 = 0 (mod 43) => 41s = 6 (mod 43) => s=6*21 = 40 (mod 43) so s=40+43n and the numbers are then: 41*(43n+40)+37 = 1763n+1677 So 1763n + m, where m is 215,344,387,903,1677 Last fiddled with by ATH on 2019-01-18 at 12:16