Difference between revisions of "2012 AIME II Problems/Problem 10"
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== Problem 10 == | == Problem 10 == | ||
Find the number of positive integers <math>n</math> less than <math>1000</math> for which there exists a positive real number <math>x</math> such that <math>n=x\lfloor x \rfloor</math>. | Find the number of positive integers <math>n</math> less than <math>1000</math> for which there exists a positive real number <math>x</math> such that <math>n=x\lfloor x \rfloor</math>. |
Revision as of 17:01, 9 August 2018
Contents
Problem 10
Find the number of positive integers less than
for which there exists a positive real number
such that
.
Note: is the greatest integer less than or equal to
.
Solution
We know that cannot be irrational because the product of a rational number and an irrational number is irrational (but
is an integer). Therefore
is rational.
Let where a,b,c are nonnegative integers and
(essentially, x is a mixed number). Then,
Here it is sufficient for
to be an integer. We can use casework to find values of n based on the value of a:
nothing because n is positive
The pattern continues up to . Note that if
, then
. However if
, the largest possible x is
, in which
is still less than
. Therefore the number of positive integers for n is equal to
Solution 2
Notice that is continuous over the region
for any integer
. Therefore, it takes all values in the range
over that interval. Note that if
then
and if
, the maximum value attained is
. It follows that the answer is
See Also
2012 AIME II (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.