All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
A real p puzzle (Posted on 2007-10-02) Difficulty: 2 of 5
Determine all possible positive real p satisfying 18[p] + 35{p} = 673, where [y] denotes the greatest integer ≤ y and {y} = y - [y]

See The Solution Submitted by K Sengupta    
Rating: 1.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Calculus....... | Comment 1 of 5

Hi!

I start with replace {p} = p-[p] and so I find a new equation

35p=673+17[p].

Because the right expresion is a integer and positive number the left is the same and so 35p = k where k is a positive and integer number.

So I replace p with k/35

From the new relation k = 673+17[k/35] I find that

[k/35] = (k-673)/17 and after this I use the properties of the [y] function ( y-1<[y]<=y) and I find two inequalities.

k/35 - 1 < (k-673)/17 <= k/35.

After resolve this I found

1275<k<=1308 but [k/35] can be only 36 or 37 so the answer is

(1) p = 1285/35

(2) p=1302/35

 

 


  Posted by Chesca Ciprian on 2007-10-02 13:35:45
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information