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^2-ac+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
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Posted by Daniel
on 2006-07-07 16:32:07 |
another solution
