high school math, senior A division contest Larry selects a 20-digit number while David selects a 14-digit number. When larry divides his number by David's number, the quotient is an integer with n digits. Compute all possible value of n.
This is a question on the senior contest, but I'm a sophomore, I really have no idea how to crack this question, do I need to know trig and calculus in order to get it?
Can u help me with it? I need easy and detailed explanation in order to fully understand this type of question.
 A: The smallest integer with $20$ digits is $10^{19}$ and the largest integer with $14$ digits is $10^{14}-1$. Thus, the result of the division is at least $$\frac{10^{19}}{10^{14}-1}>\frac{10^{19}}{10^{14}}=10^5$$ 
The largest integer with $20$ digits is $10^{20}-1$ and the smallest integer with $14$ digits is $10^{13}$. Thus, the result of the division is at most $$\frac{10^{20}-1}{10^{13}}<\frac{10^{20}}{10^{13}}=10^7$$ 
Consequently $10^5< q < 10^7$ or equivalently $$10^5+1\le q\le 10^7-1$$ where $q$ is the quotient with $n$ digits. Since $10^5+1$ has $6$ digits and $10^7-1$ hat $7$ digits and $n$ is an integer, this gives that $n \in \{6,7\}$.
A: Largest solution would be 
$$\left\lfloor \frac{999\ldots 99}{10\ldots 0} \right\rfloor = \left\lfloor \frac{10^{20}-1}{10^{13}} \right\rfloor = \frac{10^{20}-10^{13}}{10^{13}} = 10^7-1.$$
Smallest solution would be
$$\left\lceil \frac{100\ldots 00}{99\ldots 9} \right\rceil = \left\lceil \frac{10^{19}}{10^{14}-1} \right\rceil = \left\lceil \frac{10^5(10^{14}-1) + 10^5}{10^{14}-1} \right\rceil = \left\lceil 10^5 + \frac{10^5}{10^{14}-1} \right\rceil = 10^5 + 1.$$
One can easily argue that all quotients inbetween occur.
So the set of all possible integer quotients is $\{q \in \mathbb{N} \ | \ 10^5 < q < 10^7 \}$. Therefore the quotient is at least a $6$-digit but at most a $7$-digit number, so $n \in \{6,7\}$ holds.
