Pigeon Hole Principle

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Pigeon Hole Principle

siddhantshah 1208361519|%e %b %Y, %H:%M %Z|agohover

I just read the Pigeon Hole Principle and have a few questions of which I could make out neither head nor tail :(

1. 17 ants are roaming over a chessboard of area 8x8 (64 sq. inches) then atleast 2 of them mustr be always closer than :

(a) 2 inches (b) inches (c) (d) none

2. Out of (m+1) given integers 2 of them can be chosen such that their :

(a) difference is divisible by m (b) sum is div. by m
(c) product is div. by m (d) none

3. Justify giving arguments that there are 2 people in Chandigarh having same date of birth
(donno if Chandigarh has any importance)

Re: Pigeon Hole Principle

madnessweasley 1208361663|%e %b %Y, %H:%M %Z|agohover

3. Obviously Chandigarh has more than 366 people and there are only 366 possible birthdates so QED :)

PS: I assume that you are talking only about the date and not the year too.

Re: Pigeon Hole Principle

madnessweasley 1208362084|%e %b %Y, %H:%M %Z|agohover

1. Divide the 8*8 chessboard into 16 symmetric segments i.e. into 16 2*2 squares. Therefore by PHP, there exists atleast 2 ants at a distance of atleast 2 inches

Re: Pigeon Hole Principle

madnessweasley 1208362149|%e %b %Y, %H:%M %Z|agohover

2. a as 2 integers leave the same residue mod m

Re: Pigeon Hole Principle

madnessweasley 1208363284|%e %b %Y, %H:%M %Z|agohover

i seem to find some flaws in my 1st question argument, any suggestions?

Re: Pigeon Hole Principle

hell_ever_ 1208363579|%e %b %Y, %H:%M %Z|agohover

i think its ok, except that it should be as the ants can be on opposite ends of the 2*2 square

Re: Pigeon Hole Principle

madnessweasley 1208363958|%e %b %Y, %H:%M %Z|agohover

but can you generate 17 such positions? im getting that there can only be 13 such posns such that all the ants are on diagonals of such squares :( am i very confused? can you draw me a paint diagram with your configuration of ants? :)