希望大侠指点一下mod运算!S=(1+2+...+2^(2X))*(1+3+...+3^X)(1+167+...+167^X) (mod 29) =(1+2...+2^(2X))*(1+3+...+3^X)(1+22+...+22^X) (mod 29) =(2^(2X+1)-1)*(3^(X+1)-1)*2^(-1) *(22^(X+1)-1)*22^(-1) (mod 29)2^(-1) = 15 (mod 29) 22^(-1)=4 (mod 29)
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![希望大侠指点一下mod运算!S=(1+2+...+2^(2X))*(1+3+...+3^X)(1+167+...+167^X) (mod 29) =(1+2...+2^(2X))*(1+3+...+3^X)(1+22+...+22^X) (mod 29) =(2^(2X+1)-1)*(3^(X+1)-1)*2^(-1) *(22^(X+1)-1)*22^(-1) (mod 29)2^(-1) = 15 (mod 29) 22^(-1)=4 (mod 29)](/uploads/image/z/4001480-8-0.jpg?t=%E5%B8%8C%E6%9C%9B%E5%A4%A7%E4%BE%A0%E6%8C%87%E7%82%B9%E4%B8%80%E4%B8%8Bmod%E8%BF%90%E7%AE%97%21S%3D%281%2B2%2B...%2B2%5E%282X%29%29%2A%281%2B3%2B...%2B3%5EX%29%281%2B167%2B...%2B167%5EX%29+%28mod+29%29+%3D%281%2B2...%2B2%5E%282X%29%29%2A%281%2B3%2B...%2B3%5EX%29%281%2B22%2B...%2B22%5EX%29+%28mod+29%29+%3D%282%5E%282X%2B1%29-1%29%2A%283%5E%28X%2B1%29-1%29%2A2%5E%28-1%29+%2A%2822%5E%28X%2B1%29-1%29%2A22%5E%28-1%29+%28mod+29%292%5E%28-1%29+%3D+15+%28mod+29%29+22%5E%28-1%29%3D4+%28mod+29%29)
希望大侠指点一下mod运算!S=(1+2+...+2^(2X))*(1+3+...+3^X)(1+167+...+167^X) (mod 29) =(1+2...+2^(2X))*(1+3+...+3^X)(1+22+...+22^X) (mod 29) =(2^(2X+1)-1)*(3^(X+1)-1)*2^(-1) *(22^(X+1)-1)*22^(-1) (mod 29)2^(-1) = 15 (mod 29) 22^(-1)=4 (mod 29)
希望大侠指点一下mod运算!
S=(1+2+...+2^(2X))*(1+3+...+3^X)(1+167+...+167^X) (mod 29)
=(1+2...+2^(2X))*(1+3+...+3^X)(1+22+...+22^X) (mod 29)
=(2^(2X+1)-1)*(3^(X+1)-1)*2^(-1) *(22^(X+1)-1)*22^(-1) (mod 29)
2^(-1) = 15 (mod 29)
22^(-1)=4 (mod 29)
所以
S = (2^( (2X+1)(mod 28) )-1) *(3^ ((X+1)(mod 28)) -1)*15*(22^((X+1)(mod 28))-1)*4
下面这两个式子看不懂啊.
2^(-1) = 15 (mod 29)
22^(-1)=4 (mod 29)
希望大侠指点一下mod运算!S=(1+2+...+2^(2X))*(1+3+...+3^X)(1+167+...+167^X) (mod 29) =(1+2...+2^(2X))*(1+3+...+3^X)(1+22+...+22^X) (mod 29) =(2^(2X+1)-1)*(3^(X+1)-1)*2^(-1) *(22^(X+1)-1)*22^(-1) (mod 29)2^(-1) = 15 (mod 29) 22^(-1)=4 (mod 29)
是这么回事:
m的简约剩余系Z*中的余数a,总存在 b∈Z*,使得ab≡1(mod m),
即同余方程ax≡1(mod)总是有唯一解.a,b互称为对方的乘法逆元.
按通常的除法,方程的解写作x≡1/a(mod m),1/a按指数法就写为a^(-1).
1/a(mod m)在计算时是这么进行的:若c≡1(mod m),d≡a(mod),那么1/a≡c/d(mod m)
例如(mod7)1/5 ≡(14+1)/5≡3,即在模7中,5的逆元等于3.
或者(mod7)1/5 ≡(-6)/(-2)≡3
再如(mod29)1/22≡(-28)/(-7)≡4
(mod29)3/11≡3/(-18)≡1/(-6))≡30/(-6)≡-5≡24
但我后面不明白最后怎么成了(mod28),是不是键入错误?
mod就是取余