Difference between revisions of "2012 AIME II Problems/Problem 10"
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== Solution == | == Solution == | ||
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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. | 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. | ||
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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 | + | 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 \cdot 32}{2} = \boxed{496}.</math> |
− | == 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> | ||
Revision as of 12:35, 6 August 2020
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 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.