Difference between revisions of "2012 AIME II Problems/Problem 10"
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== Solution == | == Solution == | ||
+ | === Solution 1 === | ||
+ | We know that <math>x</math> cannot be irrational because the product of a rational number and an irrational number is irrational (but <math>n</math> is an integer). Therefore <math>x</math> is rational. | ||
− | == See | + | |
+ | Let <math>x = a + \frac{b}{c}</math> where <math>a,b,c</math> are nonnegative integers and <math>0 \le b < c</math> (essentially, <math>x</math> is a mixed number). Then, | ||
+ | <cmath>n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}</cmath> | ||
+ | |||
+ | Here it is sufficient for <math>\frac{ab}{c}</math> to be an integer. We can use casework to find values of <math>n</math> based on the value of <math>a</math>: | ||
+ | |||
+ | <math>a = 0 \implies</math> nothing because n is positive | ||
+ | |||
+ | <math>a = 1 \implies \frac{b}{c} = \frac{0}{1}</math> | ||
+ | |||
+ | <math>a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}</math> | ||
+ | |||
+ | <math>a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}</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 <math>x</math> is <math>31 + \frac{30}{31}</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for <math>n</math> is equal to <math>1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.</math> | ||
+ | |||
+ | === 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+1\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 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.</math> | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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> | ||
+ | == See Also == | ||
+ | [[2009 AIME I Problems/Problem 6|2009 AIME I Problems/Problem 6]] | ||
{{AIME box|year=2012|n=II|num-b=9|num-a=11}} | {{AIME box|year=2012|n=II|num-b=9|num-a=11}} | ||
+ | [[Category:Intermediate Number Theory Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 04:40, 22 December 2020
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
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
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 | ||
All AIME Problems and Solutions |
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