1992 AIME Problems/Problem 15: Difference between revisions
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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 positive integers less than <math>1992</math> are not factorial tails? | |||
== Solution == | == Solution == | ||
{{solution}} | 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>. | |||
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>. | |||
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> that are not factorial tails. | |||
== Solution 2 == | |||
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. | |||
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}} | |||
[[Category:Intermediate Number Theory Problems]] | |||
{{MAA Notice}} | |||
Latest revision as of 00:54, 2 October 2020
Problem
Define a positive integer 
to be a factorial tail if there is some positive integer 
such that the decimal representation of 
ends with exactly 
zeroes. How many positive integers less than 
are not factorial tails?
Solution
Let the number of zeros at the end of be 
. We have 
.
Note that if 
is a multiple of 
, 
.
Since 
, a value of such that 
is greater than 
. Testing values greater than this yields 
.
There are 
distinct positive integers, , less than . Thus, there are 
positive integers less than that are not factorial tails.
Solution 2
After testing various values of in of solution 1 to determine for which 
, we find that 
. WLOG, we select 
. Furthermore, note that every time 
reaches a multiple of 
, 
will gain two or more additional factors of and will thus skip one or more numbers.
With this logic, we realize that the desired quantity is simply 
, where the first term accounts for every time 
number is skipped, the second term accounts for each time 
numbers are skipped, and so on. Evaluating this gives us 
. - Spacesam(edited by srisainandan6)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
