bonanova

You have only one box. You have one key for each of your locks, but no keys for any of your friend's locks. Your friend has one key for each of her locks, but no keys for any of your locks.

Looking at the bright side, that still leaves quite a few candidates for friends ...

Why is there debate about this? The OP says four southern exposures. Only at the Geo NP is this the case.

Bumping this to the top so maybe the author will reply to the last question. Thanks ...!

Another quiet night at Morty's. Until Alex came in. I bought a bunch of factory reject dice the other day, he said. They're OK except the numbers are all wrong. Mostly they have extra two's and three's, but this one [he held one up, far enough away so the numbers could not be read] has all different numbers. Still, the numbers aren't 1-6. I have a wager for anyone here who thinks he's a genius. I'll roll the thing three times against that wall over there. The bottom and back die faces won't be visible, but I'll give you the sum of the four faces that are visible. As a bonus, I'll give you the sum of the top and front faces. I'll buy a pint for anyone who can tell me all six numbers on the die. If you try and can't figure it out, you'll buy me a pint. He rolled the die three times and called out the numbers: Roll #1 = 28 and 18 Roll #2 = 18 and 7 Roll #3 = 22 and 6 Jim thought for a while, then said, no way. There's too many possibilities. So did Ian and Jamie. Davey paused to scratch his beard and said, I'll try. Would you have taken the bet?

Kudos to writersblock [2nd post basically does it]

Sorry, I missed the question.

Hint: the answer is different if there are 10 killers from the answer if there are 11 killers.

You want to send a valuable object to a friend. You have a box to contain the object. The box has a locking ring which is more than large enough to have a lock attached. You and your friend have several locks with keys. But your friend does not have the key to any lock that you have, and vice versa. How do you do it? Note that you cannot send a key in an unlocked box, since it might be copied.

Hint 1: What does it mean to survive? Hint 2: Would the result be different if Killville were populated by 11 killers and 11 pacifists?

You live in Killville - a town populated by 10 killers and 10 pacifists. When a pacifist meets a pacifist, nothing happens. When a pacifist meets a killer, the pacifist is killed. When two killers meet, both die. Assume meetings always occur between exactly two persons and the pairs involved are completely random. Are your odds of survival better if you are a killer? or a pacifist? Or does it matter?

They ended up paying the hotel $25 and the boy $2. Of their original $30 payment, $25 stayed with the hotel, $3 came back to them, and $2 went to the boy.

Question -- are you allowed to look at the table, and use row/col sorts to make the table sorted??? If so, you can use sort_up and sort_down to manipulate numbers to their correct positions, much like solving the "15" puzzle. I'm assuming you're asking for a procedure to sort tables that could be programmed to sort any and all tables without human "guidance".

Clueless.....

Still no clue. Martini is being cruel....

letter #5

This has been posted in a slightly different form elsewhere in this forum. The usual answer is A much easier solution is

No. Your answer is the right one. After a little math, you can show that the final share has to be 1 less than a power of 2. A simple program to calculate the other shares does not give integral values for 15, 31, 63 or 127, but does for 255: -> 255 319 399 499 624 ---> 3121.

By sorting rows then columns [or v.v.] you get the smallest number to the upper left and the largest number to the bottom right, and all rows and columns sorted. Repeated sorts of rows and columns have no effect. Elements of the array are not sorted except within a row or a column. e.g. sort rows: then sort cols: 6 9 4 3 ....... 3 4 6 9 ....... 1 2 4 7 8 2 1 5 ....... 1 2 5 8 ....... 1 3 5 8 7 6 5 4 ....... 4 5 6 7 ....... 3 4 6 8 8 1 4 3 ....... 1 3 4 8 ....... 4 5 6 9 Sorting row1, col1, row2, ... col4, the result is sorted rows and cols, but not sorted table: 1 1 4 7 2 3 5 8 3 4 5 8 4 6 6 9 Once you reach a configuration where rows and cols are sorted, repeated sorting makes no change to the table. If that's all you can do, you're left with an unsorted table. To sort the table, you have to do something more than / different from sorting rows or cols. [a] sort diagonals or mess with its shape somehow.

Have a look... [url:6963e]http://www.travelmasti.com/international/malaysia/image/penang/map.jpg

You're right... there is no defensible best answer. Um... except [4] is defensible. Where does that leave us?

Not at all. Re-reading my post, I don't think I made the connection clear; tossed it in as a teaser, kind of ...