20190118, 09:53  #1 
Mar 2018
17·31 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 2k 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 20190118 at 09:56 
20190118, 12:11  #2 
Einyen
Dec 2003
Denmark
2×3×5×101 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 20190118 at 12:16 
20190213, 09:40  #3 
Mar 2018
20F_{16} 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 ? 
20190213, 14:04  #4  
Feb 2017
Nowhere
10332_{8} Posts 
Quote:
The PariGP function chinese() [named in honor of the Chinese Remainder Theorem, or CRT] is made for such things. The conditions 43k, k == 1 (mod 41), 2k, and k == 6 (mod 13) give ? chinese([Mod(0,43),Mod(1,41),Mod(0,2),Mod(6,13)]) %1 = Mod(23822, 45838) while 43k, 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. 

20190213, 21:55  #5  
Einyen
Dec 2003
Denmark
2×3×5×101 Posts 
Quote:
41s+10: 22919n+215, 22919n+14319 41s+16: 22919n+344, 22919n+9159 41s+18: 22919n+10965, 22919n+19780 41s+37: 22919n+8729, 22919n+17544 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Factors of numbers of special form  CRGreathouse  Abstract Algebra & Algebraic Number Theory  7  20190607 06:48 
Numbers of the form 1!+2!+3!+...  ricky  Factoring  41  20181001 11:54 
Most Abundant form of Prime Numbers  a1call  Information & Answers  17  20170226 22:01 
Does n have the form (a^p+b^p)/(a+b) for p > 2?  carpetpool  carpetpool  1  20170130 13:36 
least common multiple of numbers of the form a^x1  juergen  Math  2  20040417 12:19 