elementary number theory - Not understanding Simple Modulus Congruency

Hi this is my first time posting on here... so please bear with me :P

I was just wondering how I can solve something like this:

$$25x ≡ 3 \pmod{109}.$$

If someone can give a break down on how to do it would be appreciated (I'm a slow learner...)!

Here is proof that I've attempted:

1. Using definition of modulus we can rewrite $$25x ≡ 3 \pmod{109}$$ as $25x = 3 + 109y$ (for some integer $y$). We can rearrange that to $25x - 109y = 3$.




2. We use Extended Euclidean Algorithm (not sure about this part, I keep messing things up), so this is where I'm stuck at.

Thanks!

Answer

The extended euclidean algorithm is used to find x and y such that ax + by = gcd of a and b.

In our case $a = 109$ and $b = 25$.

So we start as follows.

Find remainder and quotient when we divide $109$ by $25$ and write the remainder on the left hand side.

So we get

9 = 109 - 25*4.

Now we get two new numbers $25$ and $9$. Write the remainder on the left hand side again.

7 = 25 - 9*2.

So we have two new numbers, 9 and 7.

In the extended algorithm, we use the formula for 9 in the first step

7 = 25 - (109 - 25*4)*2 = 25*9 - 109*2.

Now

2 = 9 - 7*1

= (109-25*4) - (25*9 - 109*2) = 109*3 - 25*13

Now write

1 = 7 - 3*2

i.e.

1 = (25*9 - 109*2) - 3*(109*3 - 25*13)

i.e.
1 = 25*48 - 109*11

Thus $25x - 109y = 1$ for $x = 48$ and $y = 11$.

So $25x - 109y = 3$ for x = 48*3 = 144 and y = 11*3 = 33.

Therefore 144*25 = 3 (mod 109).

If you need a number $ \le 109,$

$144 = 109 + 35$.

So we have (109+35)*25 = 3 (mod 109).

Which implies 35*25 = 3 (mod 109).

Thus $x = 35$ is a solution to your equation, which we found using the extended euclidean algorithm.

Hope that helps.