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 2019-01-18, 09:53 #1 enzocreti   Mar 2018 2×263 Posts Numbers of the form 41s+r Let be k an integer I search the values of k such that: 1- k is a multiple of 43 2-k is of the form 41s+r where r can be one of these numbers: 1, 10, 16, 18, 37. s in the integers obviously Somebody can give me a routine for SAGE for exampel to find them out? Last fiddled with by enzocreti on 2019-01-18 at 09:56
 2019-01-18, 12:11 #2 ATH Einyen     Dec 2003 Denmark 56308 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
 2019-02-13, 09:40 #3 enzocreti   Mar 2018 2×263 Posts congruent to 6 or 7 mod 13 And what if I search for values of k such that k is multiple of 43 k is of the form 41s+r with r=1,10,16,18,37 then if k is even, k have to be congruent to 6 mod 13 if k is odd, k have to be congruent to 7 mod 13 ?
2019-02-13, 14:04   #4
Dr Sardonicus

Feb 2017
Nowhere

DF616 Posts

Quote:
 Originally Posted by enzocreti And what if I search for values of k such that k is multiple of 43 k is of the form 41s+r with r=1,10,16,18,37 then if k is even, k have to be congruent to 6 mod 13 if k is odd, k have to be congruent to 7 mod 13 ?

The Pari-GP function chinese() [named in honor of the Chinese Remainder Theorem, or CRT] is made for such things. The conditions 43|k, k == 1 (mod 41), 2|k, and k == 6 (mod 13) give

? chinese([Mod(0,43),Mod(1,41),Mod(0,2),Mod(6,13)])
%1 = Mod(23822, 45838)

while 43|k, k == 1 (mod 41), k == 1 (mod 2), and k == 7 (mod 13) give

? chinese([Mod(0,43),Mod(1,41),Mod(1,2),Mod(7,13)])
%2 = Mod(32637, 45838)

and similarly for the others.

2019-02-13, 21:55   #5
ATH
Einyen

Dec 2003
Denmark

23·7·53 Posts

Quote:
 Originally Posted by enzocreti And what if I search for values of k such that k is multiple of 43 k is of the form 41s+r with r=1,10,16,18,37 then if k is even, k have to be congruent to 6 mod 13 if k is odd, k have to be congruent to 7 mod 13 ?
41s+1: 22919n+903, 22919n+9718

41s+10: 22919n+215, 22919n+14319

41s+16: 22919n+344, 22919n+9159

41s+18: 22919n+10965, 22919n+19780

41s+37: 22919n+8729, 22919n+17544

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