Difference between revisions of "1992 AIME Problems/Problem 15"
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== 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>. | 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 == | ||
− | 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. | + | 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 | + | 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 01:54, 2 October 2020
Contents
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.