The cook in the noodle shop

After a tiresome day boiling vegetables, the cook sits down in the corner of his dingy kitchen for dinner. As is his quaint habit, he puts exactly 100 noodles in his soup and begins his ritual:

He extracts two noodle ends from the bowl, ties a knot and slips it back into the soup. He repeats this 100 times until all the ends are tied. Now, he reaches his hand into the soup.

What is the probability that he would extract a garland containing all the 100 noodles?

credits: tuku

A number born of 3 digits

ok, all this fence planning has prompted me to offer a simpler puzzle for those who would like to relax a little: what is the largest number that can be created out of 3 digits and any operators ? no googling please.

Fencing without fights...

A farmer, one Mr. S, owns an uncultivated square field of green grass surrounded on all sides by pastures owned by his friends. Annoyingly for Mr. S, the cows grazing in the surrounding pastures often travel across Mr. S 's land to one of the other pastures.

Mr. S decides to stop this nonsense and considers isolating his field with a fence, but then a brainwave strikes him: "Instead of making the field impassible for these cows, what if I simply fool them into thinking it's impassible? I could build the fencing in such a way that they cannot see the other pastures and therefore won't try to walk across my field!"

At first, Mr. S considers building two fences in the shape of an 'X' across his field, which would block the view from any side of the square field to any other, thereby preventing the cows from trying to cross while saving on the amount of fencing needed.

But then he realizes that he could accomplish the same goal with even less fencing...

Can you determine the minimum amount of fencing required to block the view from any side of Mr. S's field to any of the other three sides?

This one is a big guy and carries 2.0 points!

Can you sum me out?

The rules of this game are simple:

1. Numbers 1, 2, 3... 9 are arranged in a row. At every turn, each player chooses a number and adds it to all the previously chosen numbers.
2. The player who gets to 15, by summing any 3 of his chosen numbers, wins.

For example

Turn 1:
Me 5
You 9
Me 8
You 2
Me 3
You 4

You win (9+2+4=15)!

If you win, I will give you triple the money. If I win, I keep your money. If neither wins, the pot is split. To give you the advantage of winning, I will let you choose your turn - that is, you can choose to play first or second!

Would you want to play with me?

Clue: This problem can be reduced to a certain well-known game played by us all during boring classes!

The figures of the Hindus...

This one is from the book Liber Abaci, published in 1202. Authored by Leonardo Picano, as a compilation of all that he learned from his extensive travels, the book played a pivotal role in introducing Eastern knowledge into the (then) developing West. Because of the reputation of his father Bonacci (who was a state official), he sometimes took on the name of Leonardo Fibonacci (Fibonacci, from Latin Filius bonacci meaning the progeny of Bonacci). Eight hundred years later, only his last name stuck, and today he is known as Fibonacci.

The following puzzle from Fibonacci's book is a play on the "Hindu" decimal number system that he learnt in India:

"A group of people are seated in a row, and one of them is wearing a ring on a certain joint of a certain finger. One person, who knows the whereabouts of the ring, computes a number as follows: He finds the wearer's position in the row, multiplies it by x, adds 5, multiplies that by z, and adds p. Then, he adds a number indicating the finger, and multiplies this by q. Then, he adds a number indicating the joint. When this number is announced, it is easy to pinpoint the ring. "

Find x, z, p, q and tell me how to nail Frodo Baggins!

(*) The problem is not that hard. You can assume that the joints in each finger are numbered 1 to 4. Further, a particular solution will suffice but make sure that you state the crucial logic. As usual, the first solution gets a point. But there is a bonus for the most elegant solution!

Squarely solvable?

Consider N squares of sides 1, 2, 3...N. What is the minimum length of the square that will circumscribe all of these smaller squares?

If you find that to be too easy, extend it to a circle!

(No overlaps allowed)