Find the remainder of the division of 1112131 41516171819 by 99
. (b) Find the remainder 111213141516171819 by 101
For a) I think I just need to sum up two digit such as 135 mod 99 =36
b.) sum up three digit each such as
1971 mod 101 then can i sum 197+1 mod 101?
Is this right?
Answer
N=19+18×100+17×1002+16×1003+15×1004+14×1005+13×1006+12×1007+11×1008
a:100≡1 mod99 ⇒ 100k≡1mod99
⇒ N≡(11+12+13+14+15+16+17+18+19)mod99≡135 mod99≡36 mod99
b:100≡−1 mod101 ⇒ 1002k≡[(−1)2k=1] mod101; 1002k+1≡[(−1)2k+1=−1] mod101
⇒ N≡(11−12+13−14+15−16+17−18+19)mod101≡15 mod101
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