1)
a) 2^(2^2006) = 4 ^ 2006 = 1 ^ 2006 (mod 3) = 1
b) By Fermat’s little theorem, p = 31 is prime
5^30 = 1 (mod 31)
6^30 = 1 (mod 31)
Now, 5^30000 = 5 ^ rem(30000,30) (mod 31) = 1
6^123456 = 6^6*6^123450 = 6^6 *6^rem(123450,30) (mod 31) = 6^6 (mod 31) = 1
Hence 5^30000 – 6^123456 = (1-1) (mod 31) = 0.
Hence the difference is a multiple of 31.
2)The necessary and sufficient condition on x and c is gcd(x,c) =1 .
3)
Modular inverse of a (mod m) exists iff gcd(a,m) = 1
a) 20 (mod 79)
Let us find the gcd(20,79) using Extended Euclids theorem.
Iteration #
1
2
3
4
Q
0
3
1
19
a
20
79
20
19
1
From the table, gcd(20,79) = 1 = 4(20) -1(79)
Multiplicative inverse = 4
b) 3 (mod 62)
GCD(3,62)
A
(1,0)
(0,1)
(1,0)
(-3,1)
(4,-1)
b
79
20
19
1
0
B
(0,1)
(1,0)
(-3,1)
(4,-1)
(-76,20)
Iteration #
1
2
3
4
Q
0
0
20
1
2
a
3
62
3
2
1
A
(1,0)
(0,1)
(1,0)
(-20,1)
(21,-1)
b
62
3
2
1
0
B
(0,1)
(1,0)
(-20,1)
(21,-1)
(-62,3)
From the table, GCD(3,62) = 1 = (21)(3) + (-1)(62)
Multiplicative inverse = 21
c) 21 (mod 91)
As we can see, GCD(21,91) = 7 the multiplicative inverse does not exist because 21 and 91 are not
relatively prime.
d) 5 (mod 23)
GCD(5,23) using Extended Euclid’s Theorem
Iteration #
1
2
3
4
5
Q
0
0
4
1
1
2
a
5
23
5
3
2
1
From the table, GCD(5,23) = 1 = (-9)(5) + (2)(23)
Multiplicative inverse = -9.
4) GCD(210,588)
210
= 2 * 105
= 2 * 3 * 35
= 2* 3 * 5 * 7
= (2^1) * (3^1) * (5^1) * (7^1)
588
= 2 * 294
= 2 * 2 * 147
A
(1,0)
(0,1)
(1,0)
(-4,1)
(5,-1)
(-9,2)
b
23
5
3
2
1
0
B
(0,1)
(1,0)
(-4,1)
(5,-1)
(-9,2)
(23,-5)
= 2 * 2 * 3 * 49
=2*2*3*7*7
= (2^2) * (3^1) * (7^2)
Hence, GCD(210,588) = (2^min(1,2)) * (3^min(1,1)) * (7^min(1,2))
= 2*3*7
= 42
5) GCD(210,588) using Extended Euclids theorem
Iteration #
1
2
3
4
Q
0
0
2
1
4
a
210
588
210
168
42
From the table, GCD(210,588) = 42, x = 3 and y = -1.
A
(1,0)
(0,1)
(1,0)
(-2,1)
(3,-1)
B
588
210
168
42
0
B
(0,1)
(1,0)
(-2,1)
(3,-1)
(-14,5)