pouët.net

because their squareroots aren't counted twice?

[quote]and, only squares have odd number of divisors? why?[quote]
integers have "divisor pairs". if a is divisible by n then it's divisible by a/n. squares have n == a/n. that's pretty a primary school question in disguise of a rather more complex alogrithm.

ok Anes, you win. Then I will have to post a harder one :D

Because every division of n by a e I , a<SQRT(n) yields a second integer divisor of n: b=n/a . So the number of divisors is always even.. only iff a=sqrt(n), b=n/a=a and only a single divisor results and the total number of divisors becomes odd.

uh... second! :)

Indeed, its more like a middle school problem.

But in the end discrete maths is nothing but a lot of intricate, easy to understand problems ;)

On a similar level: What happens if you start with n=2 and only turn cards if they are face down?

you get all the numbers n st. "(n-1)! + 1 is divisible by n" :P

Get a 2d-evenly-spaced-grid with unlimited size (only shown a part here by space restrictions):


Pick up 5 points from the grid intersections:


Trace straight line segments from every point to each other:


If any line cuts by a grid intersection point (not being the end or origin line points), you loose (look at the WTF example). Else you win the game.

The question: could it be ever possible to win this game? Demostrate it.

*I forgot to draw a line in the example

How about placing 4 dots as a square and placing the 5th in the center?

gargaj, try to draw it and you will see

0 0
1 1
2 3
5 8
13 21

do you want a proof also? (:

ah ok it didn't work out :P

anes, in the line (1,1) to (13, 21) there is an intersection in (7, 11)

anes, you have lost a glöp for your chessmastering. Chessmasters has no errors :P

Here is my solution

(1 1)
(1 1)
(1 1)
(1 1)
(1 1)

check your rules.. ;)

sure spock. Well, now try it with different points

Sorry about my errors. (I'm not going to be chessmaster ever :D)

Well, essentially the problem boils down to this:

You have to sets of numbers, the x and the y coordinates. xi and yi

A line intersects an integer coordinate if |xi-xj| and |yi-yj| have a common divisor. (0 is dividable by any integer).

It is sufficient to look at pairs that are dividable by two. This is the case whenever xi and xj are both even or both odd. or in computerish: (xi^xj)&1=0 the same applies ro yi and yj.

Since there are only four possible combinations of even and odd coordinates the problem above can not be solved with five coordinates.

I am sure there is a more precise way to show this...

great Spock, you win :D Parity and pingeonhole :) Does it is a more beatiful problem or not?

Tomorrow I will post any other :D

spock for chessmaster :P

Now solve the above problem in four dimensional space with 17 dots!