Exercise (6.1).1

Find the order of 2

(a) modulo 7

(b) modulo 11

(c) modulo 17

(d) modulo 23

We do this exercise by brute force, as per the textbook's examples at the beginning of Chapter 6.

(a) The following calculations show $2^3 \equiv 8 \equiv 1 \pmod 7$.

n	2^n	2^n mod 7
1	2	2
2	4	4
3	8	1

Since $\gcd(2,7)=1$, the order of 2 modulo 7 is 3.

(b) The following calculations show $2^{10} \equiv 1024 \equiv 1 \pmod {11}$.

n	2^n	2^n mod 11
1	2	2
2	4	4
3	8	8
4	16	5
5	32	10
6	64	9
7	128	7
8	256	3
9	512	6
10	1024	1

Since $\gcd(2,11)=1$, the order of 2 modulo 11 is 10.

(c) The following calculations show $2^8 \equiv 256 \equiv 1 \pmod {17}$.

n	2^n	2^n mod 17
1	2	2
2	4	4
3	8	8
4	16	16
5	32	15
6	64	13
7	128	9
8	256	1

Since $\gcd(2,17)=1$, the order of 2 modulo 17 is 8.

(d) The following calculation show $2^11 \equiv 2048 \equiv 1 \pmod {23}$.

n	2^n	2^n mod 23
1	2	2
2	4	4
3	8	8
4	16	16
5	32	9
6	64	18
7	128	13
8	256	3
9	512	6
10	1024	12
11	2048	1

Since $\gcd(2,23)=1$, the order of 2 modulo 23 is 11.