Difference between revisions of "1985 AIME Problems/Problem 10"
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== See also == | == See also == | ||
+ | * [[Floor function]] | ||
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* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] |
Revision as of 13:35, 6 May 2007
Problem
How many of the first 1000 positive integers can be expressed in the form
,
where is a real number, and denotes the greatest integer less than or equal to ?
Solution
We will be able to reach the same number of integers while ranges from 0 to 1 as we will when ranges from to for any integer . Since , the answer must be exactly 50 times the number of integers we will be able to reach as ranges from 0 to 1, including 1 but excluding 0.
As we change the value of , the value of our expression changes only when crosses rational number of the form , where is divisible by 2, 4, 6 or 8. Thus, we need only see what happens at the numbers of the form . This gives us 24 calculations to make; we summarize the results here:
Thus, we hit 12 of the first 20 integers and so we hit of the first 100.
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |