Fun questions :)

[Over9000]

22. Mao, over9000, /0, TrueTears and kamil play a game in which each person is a frog or ugly duckling. Frogs' statements are always false while ugly ducklings' statements are always true.
Mao says that over9000 is a ugly duckling.
/0 says that TrueTears is a frog
kamil says that Mao is not a frog
over9000 says that /0 is not a ugly duckling
TrueTears says that kamil and Mao are different kinds of animals
How many frogs are there?


http://img19.imageshack.us/img19/1332/image0024s.jpg ...... Part 1
http://img8.imageshack.us/img8/2856/image0025f.jpg .......Part 2
http://img4.imageshack.us/img4/6641/image0027u.jpg ......Part 3

Sorry for the whiteouted part, its just meant to be false

[TrueTears]

Haha, interesting method. That is correct.

(Would have been interesting if you happened to be the ugly duckling )

[TrueTears]

NEW QUESTIONS POSTED.

ENJOY~!

[Over9000]

26. On a tiny, remote island where the death sentence still exists, a man can be granted mercy after receiving a death sentence in the following way: He is given 18 white balls and 6 black balls. He must divide them among three boxes, with at least one ball in each box. The blindfolded, he must choose a box at random, and then a ball from within this box. He receives mercy only if the chosen ball is white. The probability of the man receiving mercy (provided he has distributed the balls in the most favourable manner) is?

18 white balls, 6 black balls
Put a white ball in one box, another white ball in the other, then dump the rest of the balls in the last box.
Therefore last box has 18-2=16 white balls and 6 black balls

Pr(white total) (pr white in box with 1 white ball only) (pr white in box with 1 white ball only) (pr white in box with 16 white balls and 6 black balls)
                   
                   
                   
                   
                   

[TrueTears]

That is correct over9000.

[TrueTears]

28.

Hence :


However which is larger than



Subbing this back in yields:





Subbing this back in yields:



Fun one.

[kamil9876]

20.

56 cubes. I'll explain later coz i dont have time now, just put this up to reserve this question

explanation:

consider the corners of the cube. The centre of the sphere, coincides with a corner. Hence we will make this the origin of an x,y,z box. Because of the symmetry of this situation we can just focus on the "quadrant" where x,y and z are positive and just find how many cubes satisfy the condition there and just multiply the result by 8.

How to find a valid cube:

Lemma: If we ensure that the corner of a unit cube that is furthest away from the origin is inside the sphere, then every point of that unit cube is inside the sphere.

Every cube in our quadrant can be identified by it's furthest most corner from origin (we'll call this "the corner"). The only possible such corners have x,y,z values of either 1,2 or 3. (not 0, since if x y or z are zero, then it isn't a "furthest away corner")
The Corners must satisfy this following inequality:




now , and can only be 1,4 or 9. Hence we can turn the question into how many different ways can u add those numbers three times to get a result less than or equal to 9:

1+1+1  --- 1 way
1+4+4  --- 3 ways
4+1+1  --- 3 ways

total 7 ways and hence 7 different corners and hence 7 different cubes and hence 7*8=56 unit cubes altogether.

[TrueTears]

20.

56 cubes. I'll explain later coz i dont have time now, just put this up to reserve this question

lols okay, I'm interested 

[TrueTears]

some 7 new questions added, found em in some old chinese books and translated em

[kamil9876]

34:




so the fraction must be natural (it can't be zero thank god). In order for that to happen, must be a factor of 24.

24=2*2*2*3

so for n-7 u can choose some number that has a combination of any of those factors, including the combination of no factors, meaning the number 1. You can say:



a can be 0,1,2 or 3. b can be 1 or 0. Hence the total number of possibilities is 2*4=8

[TrueTears]

Elegant and rigorous solution kamil.

[TrueTears]

Q 25.(as requested by a certain someone )



Now Area = Area + Area - Area



solving for x yields x = 2

[Over9000]

I praise you TT, you are truly a maths machine. I just know when im older, I will see you competing against terrence tao 
_{How the fuck did you do that question you freak}

[TrueTears]

A few new questions courtesy of /0

[kamil9876]

23.)
V- sum of vertex numbers
F- sum of face numbers
Each face number is added 4 times since each face number appears in 4 different vertex number calculations. hence:

V=4F
F is the sum of 6 different integers, and there exists an infinite ammount of prime numbers that can be written as the sum of 6 different integers. Here is a proof:

add 5 different numbers(call this sum S), by the infinity of primes we know there is some prime greater than 2S+1, so just add on the appropriate number that will make ur sum that prime.

hence there exists two V values that are of the form:




Where and are two different primes. And so only 4 is divisible by both.
Hence 4 is the largest such number divisible for all possibly face numbers.