2002 AMC 10P Problems/Problem 22
Problem
In how many zeroes does the number 
end?
Solution 1
We can solve this problem with an application of Legendre's formula.
We know that there will be an abundance of factors of 
compared to factors of 
so finding the amount of factors of 
is equivalent to finding how many factors of 
there are. Therefore, we plug in p=5 and n=2002, then plug in p=5 and n=1001 in:
![\[e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}\]](http://latex.artofproblemsolving.com/e/3/2/e321e2207deff3c5288006b566cc40a14eb2a944.png)
\begin{align*} e_5(2002!)=&\left\lfloor\frac{2002}{5}\right\rfloor+\left\lfloor\frac{2002}{5^2}\right\rfloor+\left\lfloor\frac {2002}{5^3}\right\rfloor+\left\lfloor\frac{2002}{5^4}\right\rfloor\\ =&400+80+16+3 \\ =&499 \end{align*}
or alternatively,
![$e_5(2002!)=\frac{2002-S_5(2002)}{5-1}=\frac{2002-S_5(31002_5)]}{4}=\frac{2002-6}{4}=499.$](http://latex.artofproblemsolving.com/d/c/1/dc121ebe3e5181feb2fbb033dc786a163424d83d.png)
Similarly,
\begin{align*} e_5(2002!)=&\left\lfloor\frac{1001}{5}\right\rfloor+\left\lfloor\frac{1001}{5^2}\right\rfloor+\left\lfloor\frac {1001}{5^3}\right\rfloor+\left\lfloor\frac{1001}{5^4}\right\rfloor\\ =&200+40+8+1 \\ =&299 \end{align*}
or alternatively,
![$e_5(1001!)=\frac{1001-S_5(1001)}{5-1}=\frac{1001-S_5(13001_5)]}{4}=\frac{1001-5}{4}=249.$](http://latex.artofproblemsolving.com/6/e/e/6ee0ad839c0857e2ca464310bf109cfd651e816d.png)
In any case, our answer is 
\boxed{\textbf{(B) } 1}.$
See also
| 2002 AMC 10P (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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