# Thread: GCD and least common multiple problem -02

1. ## GCD and least common multiple problem -02

i. a = b(mod m)=> (a,m) = (b,m)
ii. a = b(mod m)=> a^n = b^n (mod m)
iii. ak=bk(mod m)=> b{mod[m/(k,m)]}

2. Originally Posted by dhammikai
i. a = b(mod m)=> (a,m) = (b,m)
ii. a = b(mod m)=> a^n = b^n (mod m)
iii. ak=bk(mod m)=> b{mod[m/(k,m)]}
Can we see where are you stuck here exactly plz?

3. I am trying to solve this form following please see, but i am stuck with how I implement it

4. Originally Posted by dhammikai
I am trying to solve this form following please see, but i am stuck with how I implement it
Plz avoid images as much as possible. At times you just don't feel like clicking and opening them.

Why don't you start. So. let's see some working for (1).
I will give you a hint if x|y and y|x => x=y.

5. (i) is talking about the Euclidean Algorithem.
(ii) and (iii) are talking about the property of congruence concernning the power and division operation.
So just follow the definition and the hint, you will get closer to the conclusion.

6. 1.) a = b(mod m)=> (a,m) = (b,m)

"Why don't you start. So. let's see some working for (1).
I will give you a hint if x|y and y|x => x=y."

then,
If x = y (mod m) then y = x+sm for some s,

is true?

7. Originally Posted by dhammikai
1.) a = b(mod m)=> (a,m) = (b,m)

"Why don't you start. So. let's see some working for (1).
I will give you a hint if x|y and y|x => x=y."

then,
If x = y (mod m) then y = x+sm for some s,

is true?
yes, more exactly (x-y)=sm
but you have just stated the definition of the congruence relation