Difference between revisions of "2012 AIME II Problems/Problem 10"
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The pattern continues up to <math>a = 31</math>. Note that if <math>a = 32</math>, then <math>n > 1000</math>. However if <math>a = 31</math>, the largest possible x is <math>31 + 30/31</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for n is equal to <math>1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}</math> | The pattern continues up to <math>a = 31</math>. Note that if <math>a = 32</math>, then <math>n > 1000</math>. However if <math>a = 31</math>, the largest possible x is <math>31 + 30/31</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for n is equal to <math>1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}</math> | ||
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+ | == Solution 2== | ||
+ | Notice that <math>x\lfloor x\rfloor</math> is continuous over the region <math>x \in [k, k+1)</math> for any integer <math>k</math>. Therefore, it takes all values in the range <math>[k\lfloor k\rfloor, (k+1)\lfloor k\rfloor) = [k^2, (k+1)k)</math> over that interval. Note that if <math>k>32</math> then <math>k^2 > 1000</math> and if <math>k=31</math>, the maximum value attained is <math>31*32 < 1000</math>. It follows that the answer is <math> \sum_{k=1}^{31} (k+1)k-k^2 = \boxed{496}.</math> | ||
== See Also == | == See Also == | ||
{{AIME box|year=2012|n=II|num-b=9|num-a=11}} | {{AIME box|year=2012|n=II|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:58, 4 March 2014
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
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 |
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