# Find all values of $n$ with $0 ≤ n ≤ 35$ such that the congruence $24x$ $≡$ $n$ $(mod$ $36)$ has a solution.

I was checking the following number theory excercise:

Find all values of $$n$$ with $$0 ≤ n ≤ 35$$ such that the congruence $$24x$$ $$≡$$ $$n$$ $$(mod$$ $$36)$$ has a solution.

I made the $$gcd$$ between $$12$$ and $$36$$ and my conclusion was that the numbers are $$12$$ and $$24$$ but I think is not correct.

Any help will be really appreciated.

Take a look at the expression:

$$c \equiv a \pmod b$$

If $$\gcd(a,b)=d \rightarrow \gcd(c,d) = d$$

Since $$24 = 2^3\cdot 3$$ and $$36 = 2^2\cdot 3^2$$

Then $$\gcd(24, 36) = 2^2\cdot 3 = 12$$.

So the remainder has the form $$12y$$. How many $$y's$$ are there until $$35$$?

Right: $$12\cdot 0, 12\cdot1$$ and $$12\cdot2=24$$

$$n$$ has to be a multiple of $$\operatorname{gcd}(24,36)=12$$. So you left out $$0$$.