Number theory question 
$K$ is a three digit number such that the ratio of the number to the sum of its digits is least. What is the difference between the hundreds and the tens digits of $K$
   a) 9
   b) 8
   c) 7
   d) None of these   

I was able to solve this question by using hit and trial but could not think of a proper way to solve it. 
 A: We want to minimise $\cfrac {100a+10b+c}{a+b+c}$
This is equal to $$1+\frac{99a+9b}{a+b+c} $$ and this is clearly least when c is greatest. So we have $c=9$.
We then rewrite the fraction as $$100-\frac{90b+99c}{a+b+c}$$ which is least when $a$ is least. So we have $a=1$.
Isolating $b$ as above gives $$10+\frac {90a-9c}{a+b+c}$$ Since $a$ is non-zero the numerator is positive and the minimum is obtained when $b=9$.
A: Convince yourself that the higher the units digit and tens digit, the smaller the ratio will be. Then convince yourself that a higher hundreds digit will make the ratio bigger. From here, it is clear that the number must be $199$, and the solution is $|1-9| = 8$ hence (b).

Why you want the units digit to be higher:
Let the number be $x$ and the digital sum be $d$.
Increase the units digit by $1$ (without changing the hundreds digit).
Then $x/d > (x+10)/(d+1)$ iff $xd + x > xd + 10d$ iff $x > 10d$.
If $x = 100a + 10b + c$ is a three digit number, then:
$x > 10d$ iff $100a + 10b + c > 10a + 10b + 10c$ iff $90a - 9c > 0$ iff $9(10a - c) > 0$.
Since $a\neq0$ and $c$ are digits, we know that $10a > c$; hence the last inequality holds, so that the new ratio is, indeed, smaller. A similar analysis can be carried out for the other digit places in the event that "convince yourself" is not convincing enough.
