Home > Just Math
| Going Greatest With 2007 (Posted on 2007-11-04) |
 |
Determine the total number of real y with 1 ≤ y ≤ 2007 satisfying this equation:
[y/2] + [y/3] + [y/8] + [y/9] = 77y/72
where [p] denotes the greatest integer ≤ p.
|
|
Submitted by K Sengupta
|
|
Rating: 2.7500 (4 votes)
|
|
|
Solution:
|
(Hide)
|
We observe that y/2 + y/3 + y/8 + y/9 = 77y/72
Thus, the given equality is possible only when:
y/2 - [y/2] + y/3 - [y/3] + y/8 - [y/8] + y/9 - [y/9] = 0 ……(*)
But, p> = [p], for any real p.
Accordingly, the relationship (*) holds if:
y/2 = [y/2] , y/3 = [y/3], y/8 = [y/8] and [y/9] = y/9
Therefore, y must be divisible by LCM(2, 3, 8, 9) = 72
Now, there are precisely 27 positive integers less than or equal to 2007, which are divisible by 72.
Consequently, the required total number of real y is 27.
|
Comments: (
You must be logged in to post comments.)
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|
