Difference between revisions of "1992 AIME Problems/Problem 15"
m |
m (→Problem) |
||
Line 1: | Line 1: | ||
== 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 | + | 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 == |
Revision as of 16:41, 4 March 2012
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
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 |