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

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== Problem ==
 
== Problem ==
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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>\displaystyle m!</math> ends with exactly <math>\displaystyle n</math> zeroes. How many positiive integers less than <math>\displaystyle 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 positiive integers less than <math>1992</math> are not factorial tails?
  
 
== Solution ==
 
== Solution ==
−
{{solution}}
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The number of zeros at the end of <math>m!</math> is <math>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</math>.
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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>.
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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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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.
  
 
== 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 20:56, 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 $f(m) = 1992$ is greater than $7968$. Testing values greater than this yields $f(7980)=1992$.

There are $\frac{7980}{5} = 1596$ distinct values of $f(m)$ less than or equal to $1992$. Thus, there are $1992-1596 = \boxed{396}$ 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
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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