Let $a$ and $b$ be positive integers such that $a+b=57$ and LCM $[a,b]=680$. Find $a$ and $b$. Let $a$ and $b$ be positive integers such that $a+b=57$ and LCM $[a,b]=680$. Find $a$ and $b$.
Step by step explanation please! 
I know $lcm*gcd=ab$ but I'm not sure how to incorporate this or if i need to incorporate this to solve the problem. 
Thanks! 
 A: Note that $a$ and $b$ are relatively prime. For any common divisor of $a$ and $b$ must divide $a+b$ and the lcm of $a$ and $b$. But $57$ and $680$ are relatively prime. 
It follows that $ab=680$. Now we can substitute $57-a$ for $b$ in $ab=680$ and get a quadratic equation. More simply,
$$(a-b)^2=(a+b)^2-4ab= 57^2-4(680)=529.$$
That gives $a-b=\pm 23$, and now solving for $a$ and $b$ is straightforward. 
A: Let $a=pd$ and $b=qd$, where $d$ is the H.C.F of $a,b$ and so $p$ is prime to $q$.
Also $680.d = pd.qd$ or $680=pqd$.
Given $d(p+q)=57$.
Divide to get $\frac{p+q}{pq}=\frac{57}{680}$ or $(p+q)680=pq(57)$
From here $57$ is prime $680$, therefore $57$ divides $(p+q)$ also $(p+q)$ is prime to $pq$ , therefore $(p+q)$ divides $57$, hence we get $p+q=57$ and $pq=680$ solving we get $p=17, q=40$ or $p=40, q=17$ hence $d=1$.
Hence $(a,b)=(40,17),(17,40)$
A: Hint $\,\gcd(a,b)=1$ so we seek a factorization of $\,17\cdot 8\cdot 5\,$ into coprime factors $\,a,b\,$ with sum  $57.$ It can only be $\ (a,b) = (17,\,8\cdot5)\,$ or its swap, since $\, 57 < 17\cdot 8, 17\cdot 5.$
