Finding solution for congruences

Question

Find the solution, if any exist, of:

$1)\quad 7x ≡ 3\pmod 9 $

$2)\quad 17x ≡ 4\pmod {36}$

1 and 2 are completely separate problems. Can anyone point me in the right direction? I don't get how to solve this.

Answer 1

mod is just another way of saying when you divide, what is the remainder you get. For example $5mod2=1$ because when you divide you get one as a remainder. So $$3mod9\equiv3\equiv7x$$ so then you just solve for x

Answer 2

Okay. So what we have is $7x \equiv 3$ $(\textrm{mod}\ 9)$, then by definition of mod, we can re-write this as $7x-9y=3$ and this can be solved. Similarly, the other problem follows the same. Can you take it from here?

Answer 3

For the first equation:

Find the inverse of $7$ modulo $9$ using Extended Euclidean algoritm, then multiply the whole equation by the inverse to get the result.

For example the inverse of $7$ modulo $9$ is $4$ (I found it by trying each number instead of the Euclidean algorithm since $9$ is small).

Then multiply $7x\equiv3(\text{mod } 9)$ by $4$ and get:

$x\equiv 3(\text{mod } 9)$

The second equation is solved the same. The application of Extended Euclidean algorithm gives $17^{-1}\equiv 17(\text{mod } 36)$ so by multiplying both terms by $17$ one gets:

$x\equiv 32(\text{mod } 36)$

This method works only if the coefficient of x is relatively prime to the modulo, which is your case $\gcd(7,9)=\gcd(17,36)=1$. If they are not prime, then more care is required when solving.