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

Home > Just Math
Geometric Integers (Posted on 2006-11-09) Difficulty: 2 of 5
The number 201 is divided by a positive integer N. It is observed that the quotient, remainder and divisor (that is, N itself), but not necessarily in this order, are in geometric sequence.

What can N be?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution (thanks to Charlie) | Comment 4 of 6 |
(In reply to computer solution by Charlie)

Good find, Charlie!  Not sure how I missed that, it seems so simple...

Still not sure why it gave me such a headache the first time around.  We are looking for three integers of the form a, ax, and ax^2.  These can be combined in three ways:

a(ax) + ax^2 = 201

a(ax^2) + ax = 201

ax(ax^2) + a = 201

Of those three it can be shown that only the last would work, since the lefthand side of the first two equations will always produce an even number.

So a is our remainder, and it must be odd.  Since ax and ax^2 are integers, x is some fraction h/k where h and k are integers and both k and k^2 divide a. 

Ignoring the case where a = k = 1, the first possible value of a is 9, and k is 3.  Plugging these values in we get:

9 + 3h^3 = 201
      3h^3 = 192
        h^3 = 64
            h = 4

So then x is 4/3, and our three values are:

Remainder = a = 9
Divisor = ax = 9*4/3 = 12
Quotient = ax^2 = 9*16/9 = 16

  Posted by tomarken on 2006-11-09 15:50:51

Please log in:
Remember me:
Sign up! | Forgot password

Search body:
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

Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information