Finding the sum of two numbers given the difference of LCM and GCD. 
l-h=57
Let, the two numbers be a and b.
Then,
     ab=lh

How to proceed further? Please give me just hints? They have asked the minimum value of the sum so, I think that I have to take different possibilities.
 A: You want to minimize $lh$ subject to the condition $l-h=57$. So $l=57+h$, and the formula you want to minimize is $lh=(57+h)h$. Since we are looking at natural numbers, this will be the smallest possible when $h=1$, and then $lh=(57+1)\times1=58$. 
A: Start from $h=1, l=58$ - this gives us two possibilities $(1, 58)$ and $(2, 29)$ for $(a, b)$. The minimum of these is $2+29=31$.
Now consider $h=2, l=59$. $h$ must divide $l$, so this is not valid.
Next we have $h=3, l=60$. This is possibility since $h$ divides $l$. We need to assign $a = 3 \times ?$ and $b = 3 \times ?$, where the product of the question marks is $20$. The minimum such assignment gives $a = 3 \times 4 = 12$ and $b = 3 \times 5 = 15$, and a sum of 27.
For $h = 4, 5, \ldots, 13$, we have that $h$ does not divide $h + 57$. Thus these are not possible. For $h \geq 14$, $a, b$ are both multiples of $h$ and must be at least $14$, meaning that their sum will exceed 27.
Thus the answer is 27.
A: How about a=2 and b=29?
So lcm(a,b)-gcd(a,b)=57 and I'm just letting gcd(a,b)=1 so lcm(a,b)=58; since 58=2*29, therefore ...
