# Order

• Oct 11th 2011, 08:49 PM
jzellt
Order
Can someone show why the order of 10 is 11 while considering the group: Integers mod 55

thanks
• Oct 11th 2011, 08:56 PM
Deveno
Re: Order
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).
• Oct 11th 2011, 09:13 PM
jzellt
Re: Order
Thank! One more thing. What would the order of 0 and 1 and 5 be while looking at the integers mod 6?
Order(2) = 3
Order(3) = 2
Order(4) = 3
But I'm not sure about 0,1 and 5
• Oct 11th 2011, 09:43 PM
Deveno
Re: Order
why don't you try and calculate?

for 0, what is:

0
0+0
0+0+0
0+0+0+0
0+0+0+0+0
0+0+0+0+0+0

for 1, what is:
1
1+1
1+1+1
1+1+1+1
1+1+1+1+1
1+1+1+1+1+1

for 5, what is
5
5+5
5+5+5
5+5+5+5
5+5+5+5+5
5+5+5+5+5+5?
• Oct 11th 2011, 09:59 PM
jzellt
Re: Order
So,
Order(1) = 6
Order(5) = 6
Order(0) = 0?
• Oct 11th 2011, 10:25 PM
Deveno
Re: Order
very close....the order of 0 is 1 (the order is always a positive number). you got the other two right.