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

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== 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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Let the number of zeros at the end of <math>m!</math> be <math>f(m)</math>. We have <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>.  
  
 
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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== Solution 2 ==
 
== Solution 2 ==
  
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After testing various values of <math>m</math> in <math>f(m)</math> to determine <math>m</math> for which <math>f(m) = 1992</math>, we find that <math>m \in \{7980, 7981, 7982, 7983, 7984\}</math>. WLOG, we select <math>7980</math>. Furthermore, note that every time <math>k</math> reaches a multiple of <math>25</math>, <math>k!</math> will gain two or more additional factors of <math>5</math> and will thus skip one or more numbers.
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After testing various values of <math>m</math> in <math>f(m)</math> of solution 1 to determine <math>m</math> for which <math>f(m) = 1992</math>, we find that <math>m \in \{7980, 7981, 7982, 7983, 7984\}</math>. WLOG, we select <math>7980</math>. Furthermore, note that every time <math>k</math> reaches a multiple of <math>25</math>, <math>k!</math> will gain two or more additional factors of <math>5</math> and will thus skip one or more numbers.
  
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With this logic, we realize that the desired quantity is simply <math>\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots</math>, where the first term accounts for every time <math>1</math> number is skipped, the second term accounts for each time <math>2</math> numbers are skipped, and so on. Evaluating this gives us <math>319 + 63 + 12 + 2 = \boxed{396}</math>. - Spacesam
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With this logic, we realize that the desired quantity is simply <math>\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots</math>, where the first term accounts for every time <math>1</math> number is skipped, the second term accounts for each time <math>2</math> numbers are skipped, and so on. Evaluating this gives us <math>319 + 63 + 12 + 2 = \boxed{396}</math>. - Spacesam(edited by srisainandan6)
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=14|after=Last Question}}
 
{{AIME box|year=1992|num-b=14|after=Last Question}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:54, 2 October 2020

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

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

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 $1992$ that are not factorial tails.

Solution 2

After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$, we find that $m \in \{7980, 7981, 7982, 7983, 7984\}$. WLOG, we select $7980$. Furthermore, note that every time $k$ reaches a multiple of $25$, $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers.

With this logic, we realize that the desired quantity is simply $\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots$, where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = \boxed{396}$. - Spacesam(edited by srisainandan6)

See also

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
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