Difference between revisions of "2012 AIME II Problems/Problem 10"
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=== Solution 2=== | === 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> | 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=== | ||
+ | Bounding gives <math>x^2\le n<x^2+x</math>. Thus there are a total of <math>x</math> possible values for <math>n</math>, for each value of <math>x^2</math>. Checking, we see <math>31^2+31=992<1000</math>, so there are <cmath>1+2+3+...+31= \boxed{496}</cmath> such values for <math>n</math>. | ||
+ | |||
+ | === Solution 4=== | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | === Video Solution === | ||
+ | |||
+ | [https://youtu.be/GqiNJuOsl1w?si=yd68PahrwevAa6fn 2012 AIME II #10] | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
== See Also == | == 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}} | {{MAA Notice}} |
Latest revision as of 16:37, 24 July 2024
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
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 | ||
All AIME Problems and Solutions |
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