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The invisible square (Posted on 2003-08-24) Difficulty: 3 of 5
You are on an infinite Cartesian plane at the origin (0,0). For every integer pair of coordinates (n,m) there's a null-dimensional point (that is, the point has zero width and height).

Some of these are "visible" to you, but some others are "invisible". For example, the point (2,2) is not visible from the origin since it is "blocked" by (1,1). On the other hand, (3,5) is "visible" to you since there are no other points in the way.

Where can you build an "invisible" unit (1x1) square (all four vertices of which are "invisible" points) as near as possible to you - and the origin?

See The Solution Submitted by maskass    
Rating: 4.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 2 of 3 |
I hope I can post a solution to this and not spoil it for the rest of you... It's actually a cool problem :)

It has been my opinion that x or y needs 2 DIFFERENT factors. If its factors are all the same, then x, y and x, y+1 couldn't be both invisible. (Also, 1 doesn't count as a factor here.)

This excludes all powers of 2, 3... and prime numbers. (A space means no and an = means yes.)

1
2
3
4
5
=6
7
8
9
=10
11
=12
13
=14
=15
16
17
=18
19
=20
=21
=22
23
=24
25

The lowest numbers where x and x+1 are both yes is 14 and 15, and the lowest other set are 20 and 21.

It happens that these two pairs work.

(14,20) is blocked by (7,10)
(14,21) is blocked by (2,3)
(15,20) is blocked by (3,4)
(15,21) is blocked by (5,7)
  Posted by Gamer on 2003-08-24 09:14:39
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