Difference between revisions of "2012 AIME II Problems/Problem 10"
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<cmath>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}</cmath> | <cmath>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})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 n based on the value of a: | + | 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 = 0 \implies</math> nothing because n is positive | ||
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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 | + | 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=== | === Solution 2=== |
Revision as of 12:37, 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
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
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.