2008 Mock ARML 2 Problems/Problem 7
Problem
Let 
equal the number of zeroes to the right of the rightmost non-zero digit in the decimal form of 
, and let 
. Given that 
can be written as 
, where 
and 
are relatively prime positive integers, is less than 
, and is less than 
, find 
.
Solution
Note that 
is an integer. From Legendre's Formula, we see that
![\[f(n)=\sum_{i=1}^{2007} \lfloor \frac{n}{5^i}\rfloor\]](http://latex.artofproblemsolving.com/f/0/9/f09ebe80aa4618ff8ca0d5fd013c9fd88380896e.png)
Now note that the largest multiple of 15 that is less than 
is 
. Therefore
![\[\lfloor \frac{n}{5}\rfloor=\frac{5^{2008}-10}{15}=\frac{5^{2007}-2}{3}\]](http://latex.artofproblemsolving.com/a/6/2/a62133d51ca431673f3799c25c4f5da0b48099ec.png)
We do the same process again: The largest multiple of 15 less than 
is 
, so
![\[\lfloor \frac{n}{25}\rfloor =\lfloor \frac{\lfloor \frac{n}{5}\rfloor}{5}\rfloor = \lfloor\frac{5^{2007}-2}{15}\rfloor\]](http://latex.artofproblemsolving.com/9/6/9/969fd629b12b937f1c794f1af078b1b2393f2c28.png)
![\[=\frac{5^{2007}-5}{15}=\frac{5^{2006}-1}{3}\]](http://latex.artofproblemsolving.com/2/3/a/23a95cee61f4acf769b375c1f219c7fa2c156364.png)
Similarly, 
. We then see a pattern;
![\[f(n)=\frac{5^{2007}-2}{3}+\frac{5^{2006}-1}{3}+\frac{5^{2005}-2}{3}+\cdots +\frac{5-2}{3}\]](http://latex.artofproblemsolving.com/8/b/e/8bef4526aa017b6ac2ac962cd6c55007338e72fc.png)
![\[=\frac{5^{2007}+5^{2006}+5^{2005}+\cdots +5}{3}-\frac{2\cdot 1004 + 1003}{3}\]](http://latex.artofproblemsolving.com/3/5/1/351c1806386b298969a3040b89bb8ec7d244e157.png)
Now note that 
, so
![\[f(n)=\frac{5^{2008}-5}{12}-\frac{3011}{3}=\frac{5^{2008}-12049}{12}\]](http://latex.artofproblemsolving.com/7/c/a/7ca4eefa01eedff7b879a9298b59452ad748fa58.png)
This shows that 
, so 
.
See also
| 2008 Mock ARML 2 (Problems, Source) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | ||
