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It's quite straightforward to find out number of trailing zeros in n!.

But what if the reverse question is asked?

n! has 13 trailing zeros, what are the possible values of n ?

How should we approach the above question ?

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What can you say about when the number of trailing zeros increases - try the first few values of n! if you can't think of anything ... and make a hypothesis. – Mark Bennet Jun 26 '12 at 16:41
The number $13$ is very small. By $50$ we have a dozen $5$'s. – André Nicolas Jun 26 '12 at 16:43
@AndréNicolas: What is trailing zeros in problem? – Basil R Jun 26 '12 at 16:51
This is the number of $0$'s at the end of the decimal expansion of $n!$. So, counting from the right, it is the number of $0$'s up to the first non-zero digit. For example, $7!=5040$, so it has $1$ trailing $0$. You can see that $12!$ has $2$, $15!$ has $3$, $21!$ has $4$, and (first complication) $25!$ has $6$. – André Nicolas Jun 26 '12 at 17:12
up vote 9 down vote accepted

Write $n$ in base $5$ as $n = a_0 + 5a_1 + 25a_2 + 125a_3 + \cdots$ where $0 \leq a_k \leq 4$. $$\left\lfloor \dfrac{n}5\right\rfloor = a_1 + 5a_2 + 25a_3 + \cdots$$ $$\left\lfloor \dfrac{n}{25}\right\rfloor = a_2 + 5a_3 + \cdots$$ $$\left\lfloor \dfrac{n}{125}\right\rfloor = a_3 + \cdots$$ Hence, $$\left\lfloor \dfrac{n}5\right\rfloor + \left\lfloor \dfrac{n}{25}\right\rfloor + \left\lfloor \dfrac{n}{125}\right\rfloor + \cdots = a_1 + 6 a_2 + 31 a_3 + \cdots$$ Note that the coefficients of the terms from $a_3$ are greater than $13$. Since the sum must give us $13$, we get that $a_k = 0$, for all $k \geq 3$. Hence, we need to find the number of integer solutions to $a_1 + 6a_2 =13$ with $0 \leq a_1,a_2 \leq 4$. The constraint $0 \leq a_1,a_2 \leq 4$ further implies that $a_2$ cannot be $0$ and $1$. Hence, $a_2 = 2$. This gives us $a_1 = 1$.

Hence, $n = a_0 + 5a_1 +25a_2 = a_0 + 5 \times 1 + 25 \times 2 = 55+a_0$ where $a_0 \in \{0,1,2,3,4\}$. Hence, the desired values of $n$ are $$\{55,56,57,58,59\}$$ The same idea in principle will work when the trailing number of zeros is any number not just $13$.

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Why is this condition there in the first line 0≤ak≤4. – vinoth Jun 26 '12 at 17:33
@sun123 When you write a number in base $b$, its digits go from $0$ to $b-1$. Just like in decimal (base $10$), where the digits go from $0$ to $9$. – user17762 Jun 26 '12 at 17:58

To get a very good estimate, note that the number of trailing $0$'s is $$\left\lfloor \frac{n}{5}\right\rfloor+ \left\lfloor \frac{n}{5^2}\right\rfloor+ \left\lfloor \frac{n}{5^3}\right\rfloor+\cdots.$$ This is less than the infinite sum $$\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+\cdots.$$ The infinite geometric series has sum $\frac{n}{4}$. So if we want $z$ trailing zeros, then $n \gt 4z$.

A computation using $4z$ will tell us how much we need to go forward from $4z$. The difference between $\frac{n}{5^k}$ and $\left\lfloor \frac{n}{5^k}\right\rfloor$ is always less than $1$. So the amount we may need to go forward to find the appropriate $n$ (if there are any) is logarithmic in $z$.

Note that the first $n$ that works (if there is one) is a multiple of $5$. And then $n+1$, $n+2$, $n+3$, and $n+4$ are the others.

By that criterion, when $z=13$, our first candidate is $55$, the first multiple of $5$ after $(4)(13)$, and it works. Thereofore Of so do $56$, $57$, $58$, and $59$, and that's all.

Computations are equally straightforward for $z$ up to a few hundred. Calculate the number of trailing $0$'s for the first multiple of $5$ greater than $4z$, and make any minor adjustments necessary.

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Why the first $n$ must be multiple of 5 ? – MathMan Sep 6 '13 at 8:49
@MathMan: Definitely $n$ need not be a multiple of $5$. We are using the "floor" ir greatest integer function. So $\lfloor \frac{88}{5}\rfloor=17$. – André Nicolas Sep 6 '13 at 13:35

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