Can someone show why the order of 10 is 11 while considering the group: Integers mod 55
thanks
presumably you mean the integers mod 55 under addition.
in this group, a "power" of 10, 10^k, is 10 added k times:
10^2 = 10 + 10
10^3 = 10 + 10 + 10
(etc.)
so we will write 10^k as k*10.
the most direct route is to show that k = 11, is the smallest positive k that will work.
1*10 = 10
2*10 = 20
3*10 = 30
4*10 = 40
5*10 = 50
6*10 = 5 (remember, we are working mod 55)
7*10 = 15
8*10 = 25
9*10 = 35
10*10 = 45
11*10 = 0 (110 = 0 mod 55, since 110 = (55)(2) + 0).