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

Home > Numbers
Multiplication Table (Posted on 2006-07-07) Difficulty: 3 of 5
Imagine a multiplication table (like the one below, except it continues on forever):

   1   2   3   4     
 +---+---+---+---+---
1| 1 | 2 | 3 | 4 |...
 +---+---+---+---+---
2| 2 | 4 | 6 | 8 |...
 +---+---+---+---+---
3| 3 | 6 | 9 | 12|...
 +---+---+---+---+---
4| 4 | 8 | 12| 16|...
 +---+---+---+---+---
 |...|...|...|...|...

Find three of the same number in a straight line somewhere within the table. If this is not possible, show why not.

No Solution Yet Submitted by tomarken    
Rating: 3.0000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution A better answer | Comment 5 of 12 |
Since the problem asks for NUMBERS and not DIGITS, let's suppose there are three squares, in a straight line, with the same value.

WLOG we can suppose the middle one to be (x,y), one of the others (x+a,y-b), and the third (x-ac,y+bc); a and b must be positive integers, but c could be a rational. Neither a nor b nor c can be zero.

Since xy=(x+a)(y-b), -bx+ay=ab, and since xy=(x-ac)(y+bc), bcx-acy=abc^2. Multiplying the first by c and adding, abc+abc^2=0. Since c isn't zero, c=-1.

Now we need (x+a)(y-b)=(x-a)(y+b), but this leads to ay-bx=bx-ay, so ay=bx, so y=ax/b.

Substituting, we get to a^2=b^2, so a=b, and y=x. But x.x can never be equal to (x+a)(x-a) unless a=0, so the problem is impossible.

  Posted by Old Original Oskar! on 2006-07-07 14:34:10
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