1992 AIME Problems/Problem 15: Difference between revisions
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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>. | ||
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) = | 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{ | 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 20:02, 12 June 2008
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 positiive integers less than 
are not factorial tails?
Solution
The number of zeros at the end of is 
.
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 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 | ||
