The worlds first riddle!

PIRATES
Here's a way to do it in 5 days for 9-16 pirates

Mark that is brilliant, and clean.

This is a bit of a ramble.
First note that the number of people who know any one piece of information can at most double on any given day. This means that the information cannot be distributed in fewer than 4 days, since 4 is the smallest positive integer, k, satisfying 11 < 2k.

There are various schemes for exchanging the information. Here is one possible one.
Think of the islands as being numbered from 1 to 11, as on a clock, so that 1 follows 11. On the first day, everyone sends his (I think all mutinous pirates are men, aren't they?) information to the next island, that is, numbered one greater; on the second day, everyone sends his two pieces of information two islands further on; on the third day, four islands further on; and on the fourth day, eight islands further on. For example, the pirate on island 8 would send his various pieces of information to islands 9, then 10, then 1, and finally 5.

Why does this work? Since the scheme is the same for all the participants, we only need to see that the first piece of information is universally distributed. Look at island i, and write i -1 as a sum of powers of 2. The powers appearing will tell you how the message got from island 1. For example, if i = 6, write 5 = 4 + 1; the message from 1 arrived at island 6 by going first to island 2 via a single island flight, followed by a four island flight.

Exactly the same procedure solves the problem for 12 mutineers.



I have a d-digit positive integer where
• each digit from 1 to d appears exactly once,
and in which
• each digit (except for the leftmost) differs from one further to the left by +1 or -1.

For example, if d = 3, the only numbers satisfying the conditions are, in increasing order, 123, 213, 231 and 321.


Can you find a formula in terms of d for the number of integers satisfying these conditions

(More difficult problem: Instead of the first condition, suppose I only tell you that the d digits are distinct. Now how many integers satisfy the conditions )

PIRATES
"Mark that is brilliant, and clean."

But apparently not optimal.

DIGITS (easier version)
I haven't proven it, but it appears to be 2^(d-1).
Also, the number of solutions that begin with digit 'c' appears to be C(d-1,c-1). If you let c run from 1 to d you get line (d-1) of Pascal's triangle which sums to 2^(d-1). In fact, looking at the solutions for d=5, they've got Pascal's triangle written all over them if you consider the number of solutions that start with a given digit, then the number of solutions that start with a given pair of digits, etc. It seems like it should be so easy to prove...

DIGITS (harder version)
The digits have to be consecutive or the second condition can't be met. Therefore, it's essentially the same as the easier version. The following assumes I answered the easier version correctly.

If zero is included and is allowed to be the leading digit, then the number of solutions would be (11-d)*2^(d-1).

If zero is included and is not allowed to be the leading digit, then the number of solutions would be [(11-d)*2^(d-1)]-1.

If zero is not included, then the number of solutions would be (10-d)*2^(d-1).

Glittering Points: Stalactites

Alive without breath: Sorry, I've read The Hobbit.

DIGITS (revisited)
The key is the position of the digit 1.

First, here are two observations that I'll state but not prove:

- If there are N digits before 1, then they must be the digits 2 through N-1.
- The digits to the right of 1 must be the largest digits and they must be in increasing order from left to right.

For any number of digits, there is exactly one solution with 1 in the first position:
123...d

When 1 is in position N>1, the number of solutions is equal to the total number of solutions for d=N-1. This is because the digits in the first N-1 positions must be 2 through N. They can be arranged in exactly the same number of ways as the digits 1 through N-1.
Therefore, the number of solutions with 1 in positions 1 through N is twice the number of solutions with 1 in positions 1 through N-1.

The sequence doubles and the first term is 1 for d=1. Therefore, the number of solutions for d digits is 2^(d-1).

DrewDad: Aka Bart, who drew (angry) dad said,

"Glittering Points: Stalactites

Alive without breath: Sorry, I've read The Hobbit."

That's the problem, people are too well read. :wink:



"But apparently not optimal."

Mark, just because I say it, dose not make it so. I do not judge replies that vary from mine, there are many ways to reach a valid conclusion.
It is more amazing that we agree almost all the time, and nobody has raised any objections to any result. How long can it continue?


The number of integers satisfying the conditions is 2d - 1. One method of solution is the following:

Look at the left most digit, call it L. The digits less than L must follow in decreasing order while those greater than L must follow in increasing order. If L = d, there are no digits larger than L; if L = d - 1, there is one digit larger than L and it can appear in any of d - 1 places; if L = d - 2, there are two digits larger than L and they take up two of the following d - 1 places; and so on. This gives the total number to be
C(d - 1,0) + C(d - 1,1) + C(d - 1,2) + ... + C(d - 1,d - 2) + C(d - 1,d - 1),
where C(d-1,k) is the number of ways to select a subset of size k from a set of size d - 1. The above sum is the desired number given above.



A priest and a pirate are shipwrecked on an island with a boat which is large enough to carry only one of them to the mainland. They decide to let Lady Luck decide which one of them will be rescued.

On the island they find two coins. Assuming that at least one of the coins is fair -- that is, gives heads or tails 50% of the time -- but not knowing which one that may be, how can they use the coins to decide fairly who gets off the island



(This is an opinion only sort of question.)

A cattle town in the old west comes under attack by rustlers.
The townspeople dispatch their fastest horse and rider to the nearest fort with a note seeking help. When the horse arrives at the fort, the rider has been killed and most of the note destroyed, except for the letters

EST

To make matters worse, there are two towns about the same distance from the fort, to the east is the town of Rhester, to the west lies Testchester. Because of lack of troops, the fort can only send help to one of the towns.

To which town would you send the troops



Die without me,
Never thank me.
Walk right through me,
Never feel me.
Always watching,
Never speaking.
Always lurking,
Never seen.

Answer:


Each morning I appear
To lie at your feet,
All day I will follow
No matter how fast you run,
Yet I nearly perish
In the midday sun.


Answer:

Hi Try,

Each morning I appear
To lie at your feet,
All day I will follow
No matter how fast you run,
Yet I nearly perish
In the midday sun.

are you shadowing me?

the answer is the same as the question, me thinks.

woops wrong forum

the answer is the same as the question, me thinks.

the answer is the same as the question, me thinks.

PIRATE/PRIEST
Flip both coins.
HH - pirate wins
TT - priest wins
HT or TH - flip again

EST
It would seem that Testchester would be twice as likely to have EST survive on the note than Rhester.

هلااااااااااا

simply thinking

the letters EST are present in " west Testchester"
more than "east Rhester"

so it has more probability to be the town

i have no other idea.

too late

Die without me: Air

what about 4 days.

the 1st

1----2
2----1
3----4
4----3
5---1
6----1
7----1
8----9
9----8
10---11
11----10

the 2nd

1---3
3---1
2---4
4---2
8---10
10---8
9---11
11---9
5---2
6---2
7---2

the 3rd

1---8
8--1
2---9
9---2
3---10
10---3
4----11
11----4
5---4
6---4
7---4

the 4th

4----11
1----5
2---6
3---7

right? or i didn't understand the riddle?

Hooter: The leash starts shortening when Hooter turns the corner, and is at its minimum when you are both 4 feet from the corner. At the moment when Hooter reaches the corner and turns, the leash is shortening at 6 feet per second.

Three lives: Water (water, vapor, ice)

Burden: a snail?

the river

1700 feet