简明信息安全数学基础第三章

一，判断题

1.×
2.√
5.√
6.√

二，单选题

2.D
3.C
4.D

三，多选题

1.BD
5.AC
12.ABE

四，综合题

(1)3x≡2 (mod 7);

(3,7)=1
(3,7)|2  ∴有解
3s+7t=1
s=5,t=-2 ∴x0=5
x1=x0*(b/(a,m)) (mod m/(a,m))=5*(2/1) (mod 7/1)=10 (mod7)=3 (mod7)
x=x1+t*(m/(a,m)) (mod m)=3+(7/1)*t (mod 7)=3+7t (mod 7)=3+t (mod 7)   t=0; //t=0,1,...,(a,m)-1 

(2)6x≡3 (mod 9);

(6,9)=3
(6,9)|3 ∴有解
2x≡1 (mod 3)
(2,3)=1 2s+3t=1
s=2,t=-1 x0=2
x1=2*(3/3) (mod 9/3)=2 (mod 3)
x=2+(9/3)*t (mod 9)=2+3t (mod 9) t=0.1.2

m=5*6*7*11=2310
M1=6*7*11=462
M1`M1≡1 (mod m1)  462M1`≡1 (mod 5)   2M1`≡1 (mod 5)  
∴M1`=3

M2=5*7*11=385
M2`M2≡1 (mod m2)  385M2`≡1 (mod 6)   1M2`≡1 (mod 6)  
∴M2`=1

M3=5*6*11=330
M3`M3≡1 (mod m3)  330M3`≡1 (mod 7)   1M3`≡1 (mod 7)  
∴M3`=1

M4`M4≡1 (mod m4)  210M4`≡1 (mod 11)   1M4`≡1 (mod 11)  
∴M4`=1
X=M1M1`b1+M2M2`b2+M3M3`b3+M4M4`b4 (mod m)
X=462*3*b1+385*1*b2+330*1*b3+210*1*b4 (mod 2310)

(1) 23x≡1 (mod 140)

23x≡1 (mod 4)   x≡3 (mod 4)  
23x≡1 (mod 5)   x≡2 (mod 5)  
23x≡1 (mod 7)   x≡4 (mod 7)
m=4*5*7=140
M1=5*7=35
M1`M1≡1 (mod m1)  35M1`≡1 (mod 4)   3M1`≡1 (mod 4)  
∴M1`=3

M2=4*7=28
M2`M2≡1 (mod m2)  28M2`≡1 (mod 5)   3M2`≡1 (mod 5)  
∴M2`=2

M3=4*5=20
M3`M3≡1 (mod m3)  20M3`≡1 (mod 7)   6M3`≡1 (mod 7)  
∴M3`=6
X=M1M1`b1+M2M2`b2+M3M3`b3+M4M4`b4 (mod m)
X=35*3*3+28*2*2+20*4*6 (mod 140)
 =67 (mod 140)

编程题

19.编写程序实现2^200为的RSA公钥密码系统