Find $a,b \in \mathbb{N}$ with $\mathrm{lcm}(a,b)=12\gcd(a,b)\,$ and $\,a\bmod b = 5$ 
Find $a,b \in\mathbb{N}$ such that:
Remainder of $a$ divided by $b$ equals $5$ 
$\mathrm{lcm}(a,b)=12\gcd(a,b)$

I don't know how or where to start. The most similar problem I found was Find $a, b \in \mathbb{N}$ that verify $a + b = 1271, lcm(a, b) = 330.gcd(a, b).$
Still I can't figoure out how to solve my problem, any help will be appreciated, thanks.

Remember $lcm(a,b) = \frac {ab}{\gcd(a,b)}$.
So $\frac {ab}{\gcd(a,b)}= 12\gcd(a,b)$ and 
$\frac {a}{\gcd(a,b)}\frac {b}{\gcd(a,b)}= 12$.
Let $a' = \frac a{\gcd(a,b)}; b' =\frac b{\gcd(a,b)} = 12$ so
$(a',b') = \{(1,12), (2,6), (3,4),(4,3), (6,2), (12,1)\}$.
But $\gcd(a',b') = 1$ so $(a',b') \ne (2,6)$ or $(6,2)$.
Now $a= mb + 5$ for some integer $m$
So we have $a'\gcd(a,b) = mb'\gcd(a,b) + 5$.  Let $d = \gcd(a,b)$ and we have:
$a'd =md + 5$
$(a'-mb')d = 5$
$d = \frac {5}{a'-mb'}$
$5$ is prime so $a'-mb' = 1,5$.
We have 4 cases $(a',b') = (1,12), (3,4), (4,3), (12,1)$.
Case 1: $a'=1;b'=12$ and $1-12m = \pm 1, \pm 5$.  That can only be $m=0$ and $d=5$ and $a=1*5=5; b=12*5= 60$.  Then $5 = 0*60 + 5$ and $lcm(5,60)= 60=12*5=12\gcd(5,60)$.
Case 2: $a'=3;b'=4$ and $3-4m =1,5$. That's impossible.
Case 3: $a'=4;b'=3$ and $4-3m=1,5$.  That can only be $m=1$ and $d=5$ and $a=4*5=20$ and $b=3*5 =15$. and $20 = 1*15+5$ and $lcm(20,15) =60=12*5=12\gcd(20,15)$
Case 4: $a'=12; b'=1$ and $12-3m = 1,5$.  That's impossible.
