*I have the answer to (a) \"The element [5] has multiplicative order 4, and [7]
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*I have the answer to (a) "The element [5] has multiplicative order 4, and [7] has multiplicative order 2." However, I am not sure how exactly to go about solving it.
*For b, I have a way to do it, but I am not sure if it is correct. [2]^2 = [4]; [2]^3 = [8]; [2]^4 = [16]; [2]^5 = [15]; [2]^6 = [13]; [2]^7 = [9];[2]^8 = [1]. Hence, the multiplicative order of [2] in Z17 is 8. [5]^2 = [8]; [5]^3 = [6]; [5]^4 = [13]; [5]^5 = [14]; [5]^6 = [2]; [5]^7 = [10]; [5]^8 = [16];[5]^9 = [12]; [5]^10 = [9]; [5]^11 = [11]; [5]^12 = [4]; [5]^13 = [3]; [5]^14 = [15]; [5]^15 = [7];[5]^16 = [1]. Hence, the multiplicative order of [5] in Z17 is 16.
I am not sure how these were figured out though. But I figured this might help with (a).
Let (a,n)=1. The smallest positive integer k such that Zx17 . *I have the answer to (a) The element [5] has multiplicative order 4, and [7] has multiplicative order 2. However, I am not sure how exactly to go about solving it. *For b, I have a way to do it, but I am not sure if it is correct. [2]^2 = [4]; [2]^3 = [8]; [2]^4 = [16]; [2]^5 = [15]; [2]^6 = [13]; [2]^7 = [9];[2]^8 = [1]. Hence, the multiplicative order of [2] in Z17 is 8. [5]^2 = [8]; [5]^3 = [6]; [5]^4 = [13]; [5]^5 = [14]; [5]^6 = [2]; [5]^7 = [10]; [5]^8 = [16];[5]^9 = [12]; [5]^10 = [9]; [5]^11 = [11]; [5]^12 = [4]; [5]^13 = [3]; [5]^14 = [15]; [5]^15 = [7];[5]^16 = [1]. Hence, the multiplicative order of [5] in Z17 is 16. I am not sure how these were figured out though. But I figured this might help with (a). zx16 . (b) Find the multiplicative orders of [2] and [5] in Zxn . (a) Find the multiplicative orders of [5] and [7] in Ak = 1(modn) is called the multiplicative order of [a] inExplanation / Answer
a) 5^k cong 1(mod 16)
i.e 5^k = 1 + 16p for some integer p
p = (5^k - 1) / 16, Hence, smallest k = 4 for which p can be an integer.
Alternative way of writing
5^1 cong (5 mod 16)
5^2 cong (25 mod 16) => 5^2 cong (9 mod 16) [ 25 - 16 = 9]
5^3 cong (5*9 mod 16) => 5^3 cong( 13 mod 16)
5^4 cong (5*13 mod 16) => 5^4 cong(1 mod 16)
Hence k = 4
similarly for [7], p = (7^k - 1)/16, k = 2 => p is integer and it is the smallest
Therefore k should be = 2 (You can use the alternative way also as shown above).
b)Same way, p = (2^k - 1)/17 we see that the smallest k for which p is integer is 8. Hence k = 8
p = (5^k - 1)/17 => k = 16( Your method of calculation of the last digit of k^16 is absolutely viable)
Alternative way of writing it is
5 cong 5 mod 17
5^2 cong 8 mod 17
5^3 cong 8*5 mod 17 = 5^3 cong 6 mod 17
and so on upto
5^16 cong 1 mod 17
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