Difference between revisions of "1992 AIME Problems/Problem 15"

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== Problem ==
 
== Problem ==
Define a positive integer <math>n^{}_{}</math> to be a factorial tail if there is some positive integer <math>m^{}_{}</math> such that the decimal representation of <math>m!</math> ends with exactly <math>n</math> zeroes. How many positiive integers less than <math>1992</math> are not factorial tails?
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Define a positive integer <math>n^{}_{}</math> to be a factorial tail if there is some positive integer <math>m^{}_{}</math> such that the decimal representation of <math>m!</math> ends with exactly <math>n</math> zeroes. How many positive integers less than <math>1992</math> are not factorial tails?
  
 
== Solution ==
 
== Solution ==

Revision as of 16:41, 4 March 2012

Problem

Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?

Solution

The number of zeros at the end of $m!$ is $f(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \left\lfloor \frac{m}{125} \right\rfloor + \left\lfloor \frac{m}{625} \right\rfloor + \left\lfloor \frac{m}{3125} \right\rfloor + \cdots$.

Note that if $m$ is a multiple of $5$, $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$.

Since $f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots  = \frac{m}{4}$, a value of $m$ such that $f(m) = 1991$ is greater than $7964$. Testing values greater than this yields $f(7975)=1991$.

There are $\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$, less than $1992$. Thus, there are $1991-1595 = \boxed{396}$ positive integers less than $1992$ than are not factorial tails.

See also

1992 AIME (ProblemsAnswer KeyResources)
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