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.
my interpretation of the problem is that for the multiplication of x and y the point (x,y) is used. thus we are looking for 3 points
(a,n/a) (b,n/b) (c,n/c)
such that a,b and c are all unique divisors of a certain positive integer n.
now the equation from the first point to the second point is given by
y=(n/ab)x(n/b)+(n/a)
so for the 3 to be colinear the 3rd one must be on this line thus
n/c=(n/ab)c(n/b)+(n/a)
1/c=(c/ab)(1/b)+(1/a)
1=(c^2/ab)(c/b)+(c/a)
ab=c^2ac+bc
ab+ac=c^2+bc
a(b+c)=c(c+b)
a=c
but a can not equal c thus according to this intepretation of the problem it is impossible

Posted by Daniel
on 20060707 16:32:07 