Difference between revisions of "2012 AIME II Problems/Problem 10"
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After a bit of experimenting, we let <math>n=l^2+s, s < 2n+1</math>. We claim that I (the integer part of <math>x</math>) = <math>l</math> . (Prove it yourself using contradiction !) so now we get that <math>x=l+\frac{s}{l}</math>. This implies that solutions exist iff <math>s<l</math>, or for all natural numbers of the form <math>l^2+s</math> where <math>s<l</math>. | After a bit of experimenting, we let <math>n=l^2+s, s < 2n+1</math>. We claim that I (the integer part of <math>x</math>) = <math>l</math> . (Prove it yourself using contradiction !) so now we get that <math>x=l+\frac{s}{l}</math>. This implies that solutions exist iff <math>s<l</math>, or for all natural numbers of the form <math>l^2+s</math> where <math>s<l</math>. | ||
Hence, 1 solution exists for <math>l=1</math>! 2 for <math>l=2</math> and so on. Therefore our final answer is <math>31+30+\dots+1= \boxed{496}</math> | Hence, 1 solution exists for <math>l=1</math>! 2 for <math>l=2</math> and so on. Therefore our final answer is <math>31+30+\dots+1= \boxed{496}</math> | ||
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+ | === Video Solution === | ||
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+ | [https://youtu.be/GqiNJuOsl1w?si=yd68PahrwevAa6fn 2012 AIME II #10] | ||
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+ | ~MathProblemSolvingSkills.com | ||
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== See Also == | == See Also == |
Latest revision as of 15:37, 24 July 2024
Contents
[hide]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
Solution 1
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 are nonnegative integers and (essentially, is a mixed number). Then,
Here it is sufficient for to be an integer. We can use casework to find values of based on the value of :
nothing because n is positive
The pattern continues up to . Note that if , then . However if , the largest possible is , in which is still less than . Therefore the number of positive integers for 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
Solution 3
Bounding gives . Thus there are a total of possible values for , for each value of . Checking, we see , so there are such values for .
Solution 4
After a bit of experimenting, we let . We claim that I (the integer part of ) = . (Prove it yourself using contradiction !) so now we get that . This implies that solutions exist iff , or for all natural numbers of the form where . Hence, 1 solution exists for ! 2 for and so on. Therefore our final answer is
Video Solution
~MathProblemSolvingSkills.com
See Also
2009 AIME I Problems/Problem 6
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 | ||
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