Difference between revisions of "2012 AIME II Problems/Problem 10"
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<math>a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}</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{ | + | <math>a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}</math> |
Revision as of 02:48, 24 February 2013
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
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 a,b,c are nonnegative integers and
(essentially, x 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
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 |