The worlds first riddle!

CUBIC EQUATION
6, 12, 12
p(11)=5
p(13)=7

Assuming the roots were distinct prolonged finding a solution.

Mark:
DIE
Player 1: 6/11
Player 2: 5/11



The first player wins 6 out of 11 times, the second 5 out of 11 times.
Let the probability that the first player wins be p. If the first player does not win on the first roll, which happens with probability 5/6, the game looks just as if the second player had started the game, so the probability that the second player wins is 5/6 p. Since someone has to win the (silly) game,
p + 5/6 p = 1,
and p = 6/11.

The problem was also solved by summing an appropriate infinite series, and, of course, by computer simulation.


Mark:
DRUGS
83.898%
99.947%

Of the 5000 employees in the company 5%, or 250 people, will be drug users, while the remaining 4750 will be non-drug users. The test will yield a positive result for 99% of the drug users, which amounts to 247.5 people (however you want to interpret that), as it will for 1% of the non-drug users, which amount to 47.5 people. The total number of positive results will be 247.5 + 47.5 = 295, of which only 247.5, or about 83.9%, are actually drug users.

Using the logic above, to get a test that will correctly identify 99% of drug users, the test would have to be 99.947% accurate.
The number of employees in the company is irrelevant to the problem.


Mark:
CUBIC EQUATION
6, 12, 12
p(11)=5
p(13)=7


Damn, darn and dang. I thought that one had slipped the net. Good answer.

Let the roots be a, b, and 2a. Since the largest is twice the smallest, all the roots are positive. Write the polynomial as p(x) = (x-a)(x-b)(x-2a). Since p(B) = (B-a)(B-b)(B-2a) is prime, B-b and B-2a must equal 1. Therefore b = 2a = B-1. Similarly, since p(A) = (A-a)(A-b)(A-2a) is prime, A = b-1 = B-2. Given that one of the numbers is thirteen, the possibilities for B, p(B), A, p(A) are 13,7,11,5; 25,13,23,11; 15,8,13,6; 29,15,27,13. Only the first one fulfils all the requirements, yielding the roots 6,12,12 and the polynomial
p(x) = (x-6)(x-12)2




Given a standard deck of cards, what is the fewest number of cards you would have to draw in order to be certain that you have three of a kind, that is, three twos, or three kings, etc

What is the fewest number of cards you would have to draw in order for you to have a 50% chance of having three of a kind




Ignoring the weight, a package will be accepted for shipping by the United Parcel Service if its length plus girth is no larger than 130 inches.

(The length is the longest dimension while the girth is twice the sum of the other two dimensions.)

What is the size of the square window of smallest area through which you can pass all acceptable packages

CARDS
a) 13*2+1 = 27
b) I'll have to think about this

WINDOW
side=130/(3*sqrt(2)) in.
area=938.888... sq. in.

CARDS (b)
13 only gets you to 49%, but 14 puts you up to 58%.

Mark:

CARDS
a) 13*2+1 = 27
b) 13 only gets you to 49%, but 14 puts you up to 58%.


With 27 cards you are guaranteed a three-of-a-kind. Twenty-six cards could consist of 13 pairs of cards, but the next card drawn will necessarily complete a trio.

The second part of the problem is more difficult. You will have less that a 50% chance to have a trio with 13 cards, and more than a 50% chance with 14 cards.


Mark:
WINDOW
side=130/(3*sqrt(2)) in.
area=938.888... sq. in.

We need consider only packages whose length equals one of the other two dimensions, since longer packages will have a smaller girth. So consider a package whose dimensions are a, a, b with a + 2(a + b) = 3a + b = 130, and b  a. Hence b = (130 - 2a)/3. There are two basic ways in which a package can be pushed through a square window -- with sides parallel to the sides of the window, or with sides parallel to the diagonal of the window.

In the first case, the length of one side of the square is a, and the diagonal has length d = 2 a. In the second case, the perimeter of the box is twice the diagonal of the square, so d = a + b = (130 + b)/3. In the first case, the diameter decreases as b increases, while in the second case the diameter increases with increasing b. We will find our desired result when these two expressions for the diameter are equal. Solving gives b = (130 (2 - 1))/(1 + 2 2 ) and the length of the side of the window would be a = (130 (22 -1)/7 = 33.9565...




A fun ice breaker at a family function is the following game: Take a group of people, and have them stand in a small circle, facing into the center. Each person in the group reaches out with their hands and, at random, takes hold of the first two hands they encounter.

At this point everyone's arms will appear quite knotted. Then without letting go of their hands, they try to untangle themselves to form a large circle. This will almost certainly require crawling under legs, stepping over arms, and other twisting and turning. (It is permitted to have your hand "spin" in your partner's hand as long as you do not lose contact.)

Starting small, let's take a group of three people. Is it always possible for them to un-knot their arms
If not, how many different knots can they form



Is it always possible to choose five points in 3-dimensional space so that the line through any two of them is perpendicular to the plane through the remaining three

If so, how; if not, why not.



Consider a square with sides of length 1. Choose a point at random in the interior of the square (by throwing a dart, perhaps).

What is the probability that the point you selected is closer to the center of the square than it is to one of the sides of the square

WINDOW
I don't deserve smiles. I got it wrong.

DART
.219 (the region is bounded by four parabolas)

KNOTS
Assuming nobody is stuck holding his own hand, there are eight configurations. None of them are knots.

FIVE POINTS
If it's "ever" possible, then it's always possible
I need more time on this.

2004 can be expressed as the sum of x different positive integers, who all contain the same digits:

For instance: 2004 = 725 + 752 + 527
or
2004 = 617 + 671 + 716
2004 = 509 + 590 + 905

Can you make 2003, 2005, 2006 in this way?

Whim

x>3 for 2003, 2005, 2006
If x=4, then sum of the digits in each number is
8 or 17 for 2003
13 for 2005
11 for 2006

LANGUAGE
A
AAAAAAAAAAAAA (rule 1 twice)
ZZZAZZZ (rule 2 six times)
ZZZA (rule 4 - nothing follows the second ZZZ)
AA (rule 4)
Z (rule 2)

WHIM
2003 = 127+172+271+712+721 (only solution)
2005 = 265+526+562+652 (only solution)
2006 can't be done

Mark:
LANGUAGE
A
AAAAAAAAAAAAA (rule 1 twice)
ZZZAZZZ (rule 2 six times)
ZZZA (rule 4 - nothing follows the second ZZZ)
AA (rule 4)
Z (rule 2)


I thought this question was most difficult, and open to various interpretations. The question was: "Is the word "Z" in your dictionary?"
I saw it this way.

You can reach the word Z starting with A using all four rules, but you cannot reach it using only the first three rules.

A > AAA > AAAAAAAAAAAAA > AAAAZZZAAA > AAAAAAAAAA > AZAAZAZ >Z. In fact, every word is in the dictionary and can be made using only rules 1, 2 and 4.

For the second part, note that the initial word has an odd number of A's, namely one A, and that rules 1, 2, and 3 transform a word with an odd number of A's into another word with an odd number of A's. Since the word Z has an even number of A's, namely zero A's, it can never be reached starting from A.




This was sent to me, I have not yet started on the answer, see if you can answer before I do

It's your birthday and you want to have a big birthday party. You invite 100 people to your fête. Each of them invites 99 people, each of whom invite 98 people, each of whom invite 97 people, and so on...

Assuming that no one is invited by more than one person, and that everyone who is invited attends your gala, how many people will be there to wish you a happy birthday



Just a bit of fun.

A hair salon has three hair stylists. One of them is an excellent stylist, one is an average stylist, and, unfortunately, one is a very poor stylist. Every morning, the three stylists style each other's hair -- of course, no one styles her own hair, each one works on the hair of one of the other two ladies, and who styles whom is selected at random.

One day you walk into the shop to get your hair styled. Which stylist should you select in order to look your very best

Suppose you are satisfied enough to return to the shop a second time. Whom should you select now

BIRTHDAY
Let N be the number of people you invite. Then the total number of invitees is
N! * sum(i=0 to N-1, 1/i!)
which approaches
N! * e
as N approaches infinity

HAIR
First visit: Pick the one with the worst hair. Note who has the best hair and vow never to pick her.

Second visit: Pick the one with the worst hair unless she's the one you vowed never to pick. In that case stick to your original pick. Again, note who has the best hair and vow never to pick her.

There are only two possible configurations for who cuts whose hair. Eventually, you will have seen both, and will have identified the best stylist. The worst case is you pick the average stylist on the first visit. You will have average cuts until you see the second configuration.

Mark:
BIRTHDAY
Let N be the number of people you invite. Then the total number of invitees is
N! * sum(i=0 to N-1, 1/i!)
which approaches
N! * e
as N approaches infinity

At first glance, it would appear so. However, I am not convinced it is so high. I will need a few days.



Mark:
HAIR
First visit: Pick the one with the worst hair. Note who has the best hair and vow never to pick her.

Second visit: Pick the one with the worst hair unless she's the one you vowed never to pick. In that case stick to your original pick. Again, note who has the best hair and vow never to pick her.

There are only two possible configurations for who cuts whose hair. Eventually, you will have seen both, and will have identified the best stylist. The worst case is you pick the average stylist on the first visit. You will have average cuts until you see the second configuration.

No arguments from me.



You may be familiar with the puzzle known as the Towers of Hanoi. There are three posts and n disks of distinct radii each with a hole in the middle. At the start the disks are stacked at one of the posts in order of descending size, with the disk of smallest radius on top.

The problem is to transfer the disks from one post to another by moving one disk at a time from one post onto another one and never stacking a larger disk on a smaller one.

I ask you to look at the Tower puzzle with four posts.
What is the fewest number of moves required to transfer eight disks from one post to another if you have four posts to work with

TOWERS OF HANOI
I get 33.

Algorithm for N disks:
1) Move some number of disks (N-X) to another post.
2) Move the remaining X disks to a different post (only three posts available now; so this is like the original problem).
3) Move the disks from step 1 onto the disks from step 2 (four posts are available again).

The recursive formula for N disks is
F4(N) = F4(N-X) + F3(X) + F4(N-X) = 2*F4(N-X) + F3(X)
where
F4(i) = the number of moves required with 4 posts and i disks
F3(i) = the number of moves required with 3 posts and i disks
X = ceiling((sqrt(8*N+1) - 1) / 2)
ceiling(i) = smallest integer greater than or equal to i

Another formula is
F4(N) = F4(N-1) + 2^floor((sqrt(8*N-7) - 1) / 2)
floor(i) = largest integer less than or equal to i

I'm unable to come up with a simpler formula.

F4 grows much slower than F3. The first 20 values of F4 are:
1, 3, 5, 9, 13, 17, 25, 33, 41, 49, 65, 81, 97, 113, 129, 161, 193, 225, 257, 289
F3(20) = 1,048,575

Mark:
TOWERS OF HANOI
I get 33.

You set a new standard in perspective.



The fewest number of moves is 33.
Let's label the pegs 1 through 4 with the disks initially on disk 1 and denote by Fn the minimal number of moves required to transfer n disks with four pegs and by Tn the minimal number of moves with only three pegs. It's a straightforward exercise in induction to show that Tn = 2n -
1.

One possible strategy is the following. First moving k of the disks to peg 2 using four pegs, then move the remaining n-k of the disks to peg 3 using only three pegs then the original stack of k pegs now on peg 2 onto the stack on peg 4 using all four pegs. The number of moves is then
Fn = Fk + (2n-k - 1) + Fn-k .
You then find the minimum value of this for k between 1 and n - 1. Using the following table of Fn allows you to determine the value of F8.
n….. Fn
1….. 1
2 …..3
3 …..5
4 …..9
5 …13
6….17
7 …25
8 …33

The formula for Fn :

Fn = (n - s (s - 1)/2 - 1) 2s + 1, where s = ë (Ö(8n + 1) - 1)/2 û,
and ë x û means the nearest integer less than or equal to x.



Ten potato chips are arranged in the shape of a triangle. You start by eating the top chip, numbered 1. Being hungry, you want to eat as many chips as possible but may only do so by following this rule:
You may move a chip to a blank spot in the triangle by jumping over another chip; you then get to eat the chip that you jumped over, creating an additional blank spot.

For example, you can move 6 to 1 and eat 3, leaving 3 and 6 now blank.

Is it possible to eat all but one chip

POTATO CHIPS
I believe the best you can do is leave 3 and that the final configuration must be equivalent (via rotation and/or flip) to 1, 4, 8.

Way off!!! The answer is approximately 2.5*10^158

Look at the first few terms:
100 + 100*99 + 100*99*98
These total 980,200, they're only 3% of the terms, AND they're growing. The last term is 100!

Draw the pictures (inverted trees) for N equals 1 through 5. Note the total at each level (just for that level). There is a definite pattern that makes perfect sense.

PALINDROMIAL
If the order of p(x) is N, then the coefficients of p(x) are the Nth row of Pascal's triangle.