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2008 Mock ARML 2 Problems/Problem 7

Problem

Let $f(x)$ equal the number of zeroes to the right of the rightmost non-zero digit in the decimal form of $x!$ , and let $n = \frac {5^{2008} + 2}{3}$ . Given that $f(n)$ can be written as $\frac {5^{a} - b}{c}$ , where $b$ and $c$ are relatively prime positive integers, $b$ is less than $10^5$ , and $c$ is less than $10^2$ , find $a + b + c$ .

Solution

Note that $n$ is an integer. From Legendre's Formula, we see that

$f(n)=\sum_{i=1}^{2007} \lfloor \frac{n}{5^i}\rfloor$

Now note that the largest multiple of 15 that is less than $5^{2008}+2$ is $5^{2008}+2-12=5^{2008}-10$ . Therefore

$\lfloor \frac{n}{5}\rfloor=\frac{5^{2008}-10}{15}=\frac{5^{2007}-2}{3}$

We do the same process again: The largest multiple of 15 less than $5^{2007}-2$ is $5^{2007}-2-3=5^{2007}-5$ , so

$\lfloor \frac{n}{25}\rfloor =\lfloor \frac{\lfloor \frac{n}{5}\rfloor}{5}\rfloor = \lfloor\frac{5^{2007}-2}{15}\rfloor$

$=\frac{5^{2007}-5}{15}=\frac{5^{2006}-1}{3}$

Similarly, $\lfloor\frac{n}{125}\rfloor=\frac{5^{2005}-2}{3}$ . We then see a pattern;

$f(n)=\frac{5^{2007}-2}{3}+\frac{5^{2006}-1}{3}+\frac{5^{2005}-2}{3}+\cdots +\frac{5-2}{3}$

$=\frac{5^{2007}+5^{2006}+5^{2005}+\cdots +5}{3}-\frac{2\cdot 1004 + 1003}{3}$

Now note that $5^{2007}+5^{2006}+\cdots +5 = \frac{5^{2008}-5}{3}$ , so

$f(n)=\frac{5^{2008}-5}{12}-\frac{3011}{3}=\frac{5^{2008}-12049}{12}$

This shows that $(a,b,c)=(2008, 12049, 12)$ , so $a+b+c=\boxed{14069}$ .

See also

2008 Mock ARML 2 (Problems, Source)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8

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