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Difference between revisions of "1992 AIME Problems/Problem 15"
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Note that if <math>m</math> is a multiple of <math>5</math>, <math>f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)</math>.  
 
Note that if <math>m</math> is a multiple of <math>5</math>, <math>f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)</math>.  
  
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Since <math>f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots  = \frac{m}{4}</math>, a value of <math>m</math> such that <math>f(m) = 1992</math> is greater than <math>7968</math>. Testing values greater than this yields <math>f(7980)=1992</math>.
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Since <math>f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots  = \frac{m}{4}</math>, a value of <math>m</math> such that <math>f(m) = 1991</math> is greater than <math>7964</math>. Testing values greater than this yields <math>f(7975)=1991</math>.
  
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There are <math>\frac{7980}{5} = 1596</math> distinct values of <math>f(m)</math> less than or equal to <math>1992</math>. Thus, there are <math>1992-1596 = \boxed{396}</math> positive integers less than <math>1992</math> than are not factorial tails.  
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There are <math>\frac{7975}{5} = 1595</math> distinct positive integers, <math>f(m)</math>, less than <math>1992</math>. Thus, there are <math>1991-1595 = \boxed{396}</math> positive integers less than <math>1992</math> than are not factorial tails.  
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=14|after=Last Question}}
 
{{AIME box|year=1992|num-b=14|after=Last Question}}

Revision as of 21:02, 12 June 2008

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 positiive 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 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
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All AIME Problems and Solutions
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