Difference between revisions of "2012 AIME II Problems/Problem 10"
Arrowhead575 (talk | contribs) (→Solution 3) |
m (→Solution 3) |
||
Line 30: | Line 30: | ||
=== Solution 3=== | === 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+...+ | + | 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=== | === Solution 4=== |
Revision as of 13:52, 7 May 2023
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.