Fun questions :)

[/0]

From the binomial theorem:







xD

[kamil9876]

wowowow niceee.

There's an easy way to prove it if n is odd. Since in that case you can just pair up the terms and use . I then proved it for n+1 using induction via .



Using this result, generalise

into:


where is the sum of all the cardinalities of intersections of all combinations of k sets.

i.e for :

[TrueTears]

A snooker table measuring 8 metres by 4m with 4 pockets measuring 0.5m and placed at diagonal slants in all 4 corners contains 10 balls each with a diameter of 0.25m placed at the following coordinates

White ball: (2,1)

Red balls: (1,5), (2,5), (3,5), (1,6), (2,6), (3,6), (1,7), (2,7), (3,7)

The white ball is then shot at a particular angle from 0 to 360 degrees (0 being north, and going clockwise).
[a ball is 'potted' if at least half of the ball is in area of the 'pocket']

Assuming the balls travel indefinitely, what exact angle(s) should you choose to ensure that all the balls are potted the quickest?

[This question is not for the light hearted]

[TrueTears]

This one's out there for kamil

A sphere 5.5 centimetres in diameter is filled with 1cm diameter hemi-spheres. What is the theoretical maximum amount of spheres that can be crammed into the mega-sphere?

[dejan91]

What level maths is this just out of interest? Uni maths?

[TrueTears]

Hmm I don't think you need to use any maths beyond year 12 to do most of these questions, it's more about elegance/simplicity xD

[dejan91]

Wow and I can't even touch this stuff...count me as awkwardly complex then

[TrueTears]

Fun one:

Prove that all natural numbers can be written as a sum of numbers of the form where and are non-negative integers and none of the summands divides another.

Most likely failed attempted to prove by induction. :uglystupid2:

Let the statement be true for all numbers which are smaller than n.

Scenario 1: n is even

n can be expressed as

This follows both the conditions above

Scenario 2: n is odd

n can be expressed as where b is largest number such that

is even which implies that every summand in there will be even. Thus none of these will divide .

What remains to prove is will divide none of these. If divides a summand it has to be of the form .

Consider



Which shows that the claim that was the largest power of such that is false.

[kamil9876]

Excellent!! Different to my solution but very nice

will post mine soon if you want.

[TrueTears]

Excellent!! Different to my solution but very nice

will post mine soon if you want.

Wow I thought I screwed up somewhere heh but awesome question nonetheless!

Can't wait to see your solution

[kamil9876]

1.)Prove that if one chooses more than n numbers from the set then two of them are
relatively prime.

2.)Prove that if one chooses more than numbers from the set , then one number is a multiple of another. Can this be avoided with exactly numbers?

[/0]

1) By the pigeon-hole principle, if you choose more than numbers, you will have at least 1 pair of consecutive numbers.
Let these numbers be and . If they have a common divisor d, then

     ,     

Then , so must divide 1. But this is only possible if . So the consecutive numbers are relatively prime!

(I'll leave the other to someone else)

[TrueTears]

1) Nice Pigeon holing

Another method:

By partitioning as .

There are n parts one part must contain two of the numbers two of the numbers are adjacent.

Since adjacent numbers are always relatively prime, QED.

[kamil9876]

Here is a cool fact I discovered today when playing around:

Consider I have the following rectangular array:

0   0   0   0
1   1   1   1
2   2   2   2
3   3   3   3
4   4   4   4....


Suppose I have to choose one number from each column. I then have to take the sum of the numbers and then divide it by 5 and take the remainder.

Prove that:

a.) the number of combinations that yield a remainder of 0, equals the number of of combinations that yield a remainder of 1, equals the number of combinations that yield a remainder of 2.... etc.

b.) prove that the number of combinations that yield a remainder of 0 is: where is the number of columns.

You can independantly prove a and use it to prove b. Or independantly prove b and then use it to prove a. Both ways are equally beautifull

[zzdfa]

re. the above question

There are 5^(n-1) ways to pick the first n-1 columns, and for each of these 5^(n-1) combinations, there are 5 choices for the last number, and each choice will give you a different result, from {0,1,2,3,4}.

example for clarity:

0   0   0   0
1   1   1   1
2   2   2   2
3   3   3   3
4   4   4   4

3+1+2 gives remainder 1.


6+0 ->remainder 1
6+1-> remainder 2
..
..
..
6+4 -> remainder 0


more questions :

1) There's a n x n square grid of people seated. They are told to find a new seat, but they are only allowed to move up, down, left or right one spot. Prove that if n is odd, it is impossible for them to all find a new seat.

2) How many ways are there to color n identical spheres with k colors? Not all colors have to be used.

3) A number p is chosen uniformly randomly from [0,1]. An unfair coin is constructed with a probability p of getting a heads. The coin is tossed n times. What is the probability of getting k heads?

4) On a 13x13 lattice, 53 dots are colored. Prove that amongst them there are 4 dots that form a rectangle.
(this has been annoying me... it seems like an easy pigeonhole, and probably is, but I can't get it  )