The worlds first riddle!

[size=8]A2K THREAD
37.5 (21.75 girls and 15.75 boys)[/size]

Mark:


A2K THREAD
37.5 (21.75 girls and 15.75 boys)



We can set up 2 equations:

.58*g = .42*b + 6
g = b + 6
Substituting the second equation into the first we get:

.58b + 3.48 = .42b + 6
.16b = 2.52
b = 15.75

Since we can't have decimal portions of people, we can truncate it to 15.
So, there are 15 boys and 21 girls for a total of 36 students.

Due to rounding of the percentages there are two more answers.
16 boys + 22 girls = 38 students and 17 boys + 23 girls = 40 students
So, all three answers are correct !!!




An A2K pilot overheard a woman say that she had missed her flight. The pilot said "I can give you a ride and I'll charge you nothing because you won't even be taking me out of my way."

The woman was very suspicious of this offer especially when she said "But you don't even know where I'm going !!" The pilot said "It doesn't matter. It will not be out of my way." The pilot was not lying so just where the heck was he going

[size=8]PILOT
to the airport[/size]

Mark:

PILOT
to the airport


I see nothing in the wording of the question to exclude that answer. However, I was looking for something along the lines of:

He was going to the opposite side of the Earth. For example, if he were in Honolulu, Hawaii he was headed for Mozambique, Africa.

On the surface of the Earth, the shortest distance between 2 points is a "great circle route". So, when you are traveling to the exact opposite side of the Earth, there are an infinite number of great circle routes you can choose and one of which would take you over the destination where the woman wanted to go.




The California Chamber of Commerce has invited one student from each of four local high schools to attend a leadership luncheon. However, the Sherman student and the Hadley student had just broken up and they did not want to sit next to each other.

In how many different ways could the students from Clarke, Hadley, Sherman and Jackson be seated in a row so that the Hadley and Sherman students do not sit next to each other

[size=8]SILLY TEENAGERS
12[/size]

Mark:

SILLY TEENAGERS
12


C H J S
C S J H
J H C S
J S C H
H C J S
H C S J
H J C S
H J S C
S C J H
S C H J
S J C H
S J H C

(there are 4x3x2x1 = 24 different ways in which the four students could be seated if there are no constraints.)





Mr. and Mrs. Lopez have always loved animals and encouraged their children to care for and respect their pets. As a result, the Lopez household boasts three dogs and two cats. Each of the children is responsible for a pet. Can you match up the pet names, the breeds, and their owners

Clues:

Neither the Dachshund nor Fritzy is cared for by Ann.

Tom's pet and the Collie are both 2 years old.

The Manx Cat, the Tiger Cat, and the Dachshund have been given the names, Hans, Fritzy, and Otto (but not necessarily in that order), but none belongs to Sue or Tom.

The Tiger Cat, Jim's pet, and Hans -- who howls and barks to Paul Simon tunes -- all came as Christmas gifts.

Buffy and the beagle both enjoy television.

Sparky was an Easter gift.

Bill is the oldest Lopez child.

[size=8]PETS
Buffy, collie, Sue
Fritzy, manx cat, Jim
Hans, dachshund, Bill
Otto, tiger cat, Ann
Sparky, beagle, Tom[/size]

Mark:


PETS
Buffy, collie, Sue
Fritzy, manx cat, Jim
Hans, dachshund, Bill
Otto, tiger cat, Ann
Sparky, beagle, Tom



Well done buddy, perfect answer.




Listen up gang; a chessboard has had its upper left square and its lower right square "sawed off". Underneath that is a domino that is exactly the size of 2 of the chessboard squares. If you had 31 of these dominoes, would you be able to arrange them so that ALL 62 squares were covered

[size=8]CHESSBOARD
No. Each domino covers one black and one white square. The removed squares are the same color. There are 32 of one color and 30 of the other.[/size]

Right, you will need all your fingers and toes for this little poser; natural numbers a,b,c,d between 1 and a xillion (you better believe this is a big number) are chosen at random, with each of the xillion-to-the-fourth-power possible quadruples equally likely.

What is the probability that gcd(a,b)=gcd(c,d)

[size=8]XILLION
It's right around 40% for natural numbers up to the 100-250 range.

I'll guess 1/e, which is approximately 36.788%.[/size]

Mark:
XILLION
It's right around 40% for natural numbers up to the 100-250 range.

Stop right there!
It is exactly 40%

This number arises as zeta(4)/zeta(2)^2. For each n=1,2,3,..., the probability that gcd(a,b)=n is n^-2/zeta(2), so the probability that gcd(a,b)=gcd(c,d)=n is n^-4/zeta(2)^2; now sum over n.

Can the answer 2/5 be proved by elementary means?

This can be regarded as a form of a question attributed to Wagstaff in problem B48 of R.K.Guy's Unsolved Problems in Number Theory (Springer, 1981): ``Wagstaff asked for an elementary proof (e.g., without using properties of the Riemann zeta-function) that [the product of (p^2+1)/(p^2-1) over all primes equals 5/2].''




[A problem I remember in the 1988 Qualifying Exam for Harvard]

Can you prove that a finite group of 8k elements (k=1,2,3,...) must have at least five conjugacy classes


Ps, Don't worry if you find the question too difficult, you can still qualify for Yale if you can write your name with fewer than three mistakes.

[size=8]CONJUGACY CLASSES
OK, I'll do the heavy lifting and get us 80% of the way there.

Here are some givens about conjugacy classes (CCs):

(1) Each element in a group belongs to exactly one class.
(2) The identity element is always in its own class.
(3) The conjugacy class orders of all classes must be integral factors of the group order of the group.

To account for all of the CCs, we need factors of 8k that sum to 8k.

From (1), we know that there is a CC with order 1.
Since 8k is an even number, there must be at least one more CC with odd order. That order must be less than 8k/2 - 1 since 8k/2 - 1 doesn't divide 8k.
Therefore, the sum of the orders of these first two CCs is less than 8k/2.

Let the third CC have the largest possible order: 8k/2. 8k/2 plus a number that is less than 8k/2 (from the first two CCs) doesn't sum to 8k.
Therefore, a fourth CC is required to get to 8k.

Examples:
k=1: CCs could have order 1, 1, 2, 4
k=2: CCs could have order 1, 1, 2, 4, 8
k=3: CCs could have order 1, 3, 8, 12

Now that I've done the bulk of the work, I'll let the mop up crew get us that last 20% of the way there. :wink: [/size]

Mark wrote:

CONJUGACY CLASSES
OK, I'll do the heavy lifting and get us 80% of the way there.

Here are some givens about conjugacy classes (CCs):

(1) Each element in a group belongs to exactly one class.
(2) The identity element is always in its own class.
(3) The conjugacy class orders of all classes must be integral factors of the group order of the group.

To account for all of the CCs, we need factors of 8k that sum to 8k.

From (1), we know that there is a CC with order 1.
Since 8k is an even number, there must be at least one more CC with odd order. That order must be less than 8k/2 - 1 since 8k/2 - 1 doesn't divide 8k.
Therefore, the sum of the orders of these first two CCs is less than 8k/2.

Let the third CC have the largest possible order: 8k/2. 8k/2 plus a number that is less than 8k/2 (from the first two CCs) doesn't sum to 8k.
Therefore, a fourth CC is required to get to 8k.

Examples:
k=1: CCs could have order 1, 1, 2, 4
k=2: CCs could have order 1, 1, 2, 4, 8
k=3: CCs could have order 1, 3, 8, 12

Now that I've done the bulk of the work, I'll let the mop up crew get us that last 20% of the way there.

I think you have 99% of it. Great work!


Apologies for overlooking an answer from the day before.

Mark:

CHESSBOARD
No. Each domino covers one black and one white square. The removed squares are the same color. There are 32 of one color and 30 of the other.





Ok! Let's slow things down a tad. Using each of digits 2, 5, 7 once, and only basic arithmetic operations, can you write a formula for 181

Yippee, it's the weekend at last, let's take it easy with:



For this little puzzle you will need:

• six cups
• a pile of 100 or so paper clips (or coins, or match sticks or any objects that you have a lot of)
• a six-sided die

Put 10 clips in each cup and the extras in a pile. The idea is to move clips from one cup to the next according to the following rules:

1. Start with cup #1. Roll the die and see what number you get. Move that many clips from cup #1 to cup #2.

2. Roll the die again and see what number comes up. Move that number of clips from cup #2 to cup #3.

3. Continue this way down the line. When you get to cup #6 move the clips to pile of extras since there's no 7th cup.

4. Go back to the first cup and start again with step 1 above.

One more rule: Cup #1 never runs out of clips. If you need more for a move to cup #2 than you have in cup #1, just get what you need from the pile of extras.

What happens after you've done this for a while

[size=8]CUPS
I did it 5000 times (took all weekend*) and all I noticed was that the clips would accumulate in one or two cups and the accumulation would cycle through the cups.

*not really - I wrote a program[/size]

"Using each of digits 2, 5, 7 once, and only basic arithmetic operations, can you write a formula for 181?"

Using only +, -, *, /, the answer is no (at least according to the program I wrote).

[size=8]CIRCLES
I have a hunch that it is independent of r and r', but I have to prove it. I'll be back.[/size]

Whilst we wait for Mark to come up with an enlightening answer, consider this:

Using each of digits 1, 2, 3 once, and only basic arithmetic operations, can you write a formula for 19

I am serious, I think it can be done.