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

Home > Numbers
Product of 3 boxes (Posted on 2020-10-14) Difficulty: 2 of 5
Determine all integers x satisfying [x/2][x/3][x/4]=x2. ([y] is the largest integer which is not larger than y)

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
No Subject | Comment 2 of 3 |
First off, x=0 is a trivial solution.  There cannot be negative solutions.

If we ignore the [.] function we get x^3/24 = x^2 which has solution x=24.  This is also a solution since each of x/2, x/3, x4 will not be rounded down.

The LHS of the equation is bounded above by x^3/24 and below by (x-2)(x-3)(x-4)/24.  These cross the x^2 at x=24 and x=32.216 respectively.

A quick table shows no other solutions in this range.  So the solution set is {0,24}

There is an additional non-integer solution at sqrt(882)=29.7 where the equation becomes 14*9*7=882 (I play slope game )

  Posted by Trudie Kirk on 2021-06-16 01:57:34
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 (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information