1. ## find all x

Find all x such that
a) 2x is congruent to x(mod5)

(When i did this one I there is no such x, but i am not sure)

b) -25 < x < 25 and x is congrent to 3(mod 5)

(I get confused on the negative side, I know on the positive side you have 3, 8, 13, 18, 23 but how would it go for the negative side, would it be -3, -8, -13, -18, -23 or am i doing this wrong?)

2. Originally Posted by mandy123
Find all x such that
a) 2x is congruent to x(mod5)

(When i did this one I there is no such x, but i am not sure)
Subtract x from both sides.

b) -25 < x < 25 and x is congrent to 3(mod 5)
If $x\equiv 3 ~ (5)$ then $x = 5k+3$ for some $k$. You want $-25 < 5k +3 < 25$ so $-28 < 5k < 21$. Thus, $-5\leq k \leq 4$. Put all those number into $5k+3$ and you get all the $x$.

3. ok so when I subtract x from both sides should it look like this

2x=5k+x for some k
2x-x=5k+x-x
x=5k
so then it would be x congruent to 0(mod5)?
and then x would be every multiple of 5?

4. Simply subtracting x from the congruence suffices:
$\begin{array}{rcl} 2x & \equiv & x \ (\text{mod } 5) \\ x & \equiv & 0 \ (\text{mod } 5) \qquad \text{(Subtracted x from both sides)} \end{array}$

So every x that is a multiple of 5 will satisfy your congruence. For example, take x = 20:
$2(20) = 40 \equiv 20 \ (\text{mod } 5)$