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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.

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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 $n \in \{6,7\}$.

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    $\begingroup$ That's not right! You have shown that $10^5 < q < 10^7$, where $q$ is the quotient. That doesn't mean that $q$ necessarily has $6$ digits. $\endgroup$ – TonyK Nov 8 '14 at 20:05
  • $\begingroup$ @TonyK Ok, thanks, I corrected it. $\endgroup$ – Jimmy R. Nov 8 '14 at 22:22
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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.

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  • $\begingroup$ $10^5$ has $6$ digits, not $5$. $\endgroup$ – TonyK Nov 8 '14 at 20:07
  • $\begingroup$ You are right, damn off-by-one error. Corrected. $\endgroup$ – GDumphart Nov 8 '14 at 20:09
  • $\begingroup$ Also, those ceilings should be floors, shouldn't they? So $10^5+1$ should be $10^5$. $\endgroup$ – TonyK Nov 8 '14 at 20:10
  • $\begingroup$ @TonyK I really don't think so. That would yield a quotient of $10^5$ which is not achieveable with a 6-digit-difference, the denominator would need 15 digits for that. $\endgroup$ – GDumphart Nov 8 '14 at 20:14
  • $\begingroup$ Yes, OK. The wording of the question suggests that the division is exact, in which case you are correct. I was looking at the word "quotient" and assuming that there might also be a remainder, but on re-reading the question, I think that is ruled out. $\endgroup$ – TonyK Nov 8 '14 at 20:22

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