Can someone show why the order of 10 is 11 while considering the group: Integers mod 55

thanks

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- October 11th 2011, 09:49 PMjzelltOrder
Can someone show why the order of 10 is 11 while considering the group: Integers mod 55

thanks - October 11th 2011, 09:56 PMDevenoRe: 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). - October 11th 2011, 10:13 PMjzelltRe: 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 - October 11th 2011, 10:43 PMDevenoRe: 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? - October 11th 2011, 10:59 PMjzelltRe: Order
So,

Order(1) = 6

Order(5) = 6

Order(0) = 0? - October 11th 2011, 11:25 PMDevenoRe: Order
very close....the order of 0 is 1 (the order is always a positive number). you got the other two right.