Difference between revisions of "2012 AIME II Problems/Problem 10"
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| − | <math>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}</math> Here it is sufficient for <math>\frac{ab}{c}</math> to be an integer. We can use casework based on the value of a: | + | <math>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}</math> 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: |
Revision as of 18:11, 31 March 2012
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 x cannot be irrational because the product of a rational number and an irrational number is irrational (but n is an integer). Therefore x 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:
a = 0 --> nothing because n is positive
a = 1 --> b/c = 0/1
a = 2 --> b/c = 0/2, 1/2
a = 3 --> b/c = 0/3, 1/3, 2/3
The pattern continues up to a = 31. Note that if a = 32, then n > 1000. However if a = 31, the largest possible x is 31 + 30/31, in which n is still less than 1000. 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 | ||
