[frookenhauer]

I find myself resisting the urge to hunt you down like a dog for the pain you are causing me, STD, I think I'll start asking questions in the Cambridge area....

[Impirion]

Frookenhauer, on Feb 22 2009, 11:52 PM, said:

I find myself resisting the urge to hunt you down like a dog for the pain you are causing me, STD, I think I'll start asking questions in the Cambridge area....

I live next door to him

Edit: And yeah, I frequently have to restrain myself. Be satisfied you don't have to listen to his punning every day

This post has been edited by Impirion: 23 February 2009 - 12:19 AM

[frookenhauer]

Thank god for small favours...Would you mind twatting him round the head for me? I'll owe you a beer

[Impirion]

It's 4 and 13 I believe, although I can't claim the credit. The power of computers and particularly computer scientists won the day.

[Impirion]

Frookenhauer, on Feb 23 2009, 03:16 AM, said:

Thank god for small favours...Would you mind twatting him round the head for me? I'll owe you a beer

Sure, course! Anything for a beer

[Sir Thursday]

Impirion is correct. It is his task (or our mutual computer scientist friend's, should he choose to accept it) to propose a new riddle.

In case you're interested, here is the reasoning I used to verify this solution:

Alex would be told '17', which has possible pairs (2,15),(3,14),(4,13),(5,12),(6,11),(7,10) and (8,9). These give products of 30, 42, 52, 60, 66, 70 and 72. All of these have more than two possible product pairs, so Alex is justified in his initial statement that neither of them know the pair of numbers.

Now Ben has been told '52', which has possible factor pairs (2,26) and (4,13). Given Alex's statement, he knows that it can't be (2,26), because if Alex had been told '28' then one possible pair is (5,23), whereupon Ben would have been told '65' and got the answer immediately. So now Ben knows that the answer must be (4,13).

When Ben says that he knows the answer, that allows Alex to eliminate all but one of the possible numbers Ben could have been given.

-If Ben had been given 30, then he wouldn't have known the pair - it could have been (2,15) or (5,6).

-If Ben had been given 42, then he wouldn't have known the pair - it could have been (2,21) or (3,14).

-If Ben had been given 60, then he wouldn't have known the pair - it could have been (3,20) or (5,12).

-If Ben had been given 66, then he wouldn't have known the pair - it could have been (2,33) or (6,11).

-If Ben had been given 70, then he wouldn't have known the pair - it could have been (2,35) or (7,10)

-If Ben had been given 72, then he wouldn't have known the pair - it could have been (3,24) or (8,9).

-However, if Ben had been given 52, then he would know the pair. This is the only remaining possibility, so Alex can determine that the pair of integers must be (4,13). Ergo, this is the correct answer.


Sir Thursday

This post has been edited by Sir Thursday: 24 February 2009 - 02:36 AM

[Impirion]

Here we go, not too difficult, but takes logic to work out.

Five sailors survive a shipwreck and swim to a tiny island where there is nothing but a coconut tree and a monkey. The sailors gather all the coconuts and put them in a big pile under the tree. Exhausted, they agree to go to wait until the next morning to divide up the coconuts. At one o'clock in the morning, the first sailor wakes up. He realizes that he can't trust the others, and decides to take his share now. He divides the coconuts into five equal piles, but there is one coconut left over. He gives that coconut to the monkey, hides his coconuts (one of the five piles), and puts the rest of the coconuts (the other four piles) back under the tree.

At two o'clock, the second sailor wakes up. Not realizing that the first sailor has already taken his share, he too divides the coconuts up into five piles, leaving one coconut over which he gives to the monkey. He then hides his share (one of the five piles), and puts the remainder (the other


At three, four, and five o'clock in the morning, the third, fourth, and fifth sailors each wake up and carry out the same actions.

In the morning, all the sailors wake up, and try to look innocent. No one makes a remark about the diminished pile of coconuts, and no one decides to be honest and admit that they've already taken their share. Instead, they divide the pile up into five piles, for the sixth time, and find that there is yet again one coconut left over, which they give to the monkey.

What is the smallest amount of coconuts that there could have been in the original pile?

[Sir Thursday]

Impirion, on Feb 24 2009, 11:30 PM, said:

Here we go, not too difficult, but takes logic to work out.

Five sailors survive a shipwreck and swim to a tiny island where there is nothing but a coconut tree and a monkey. The sailors gather all the coconuts and put them in a big pile under the tree. Exhausted, they agree to go to wait until the next morning to divide up the coconuts. At one o'clock in the morning, the first sailor wakes up. He realizes that he can't trust the others, and decides to take his share now. He divides the coconuts into five equal piles, but there is one coconut left over. He gives that coconut to the monkey, hides his coconuts (one of the five piles), and puts the rest of the coconuts (the other four piles) back under the tree.

At two o'clock, the second sailor wakes up. Not realizing that the first sailor has already taken his share, he too divides the coconuts up into five piles, leaving one coconut over which he gives to the monkey. He then hides his share (one of the five piles), and puts the remainder (the other


At three, four, and five o'clock in the morning, the third, fourth, and fifth sailors each wake up and carry out the same actions.

In the morning, all the sailors wake up, and try to look innocent. No one makes a remark about the diminished pile of coconuts, and no one decides to be honest and admit that they've already taken their share. Instead, they divide the pile up into five piles, for the sixth time, and find that there is yet again one coconut left over, which they give to the monkey.

What is the smallest amount of coconuts that there could have been in the original pile?

The answer I got was 15621. Seems a bit large, but it definitely works. There may be smaller, but I don't see how to figure it out.

Sir Thursday

[Giles]

Yeah the answer i got seemed to be going into large numbers but i got bored before getting to an actual answer

[Mentalist]

assume there was 1 coconut in each pile at the end.

therefore, 6 when all 5 woke up.

31 when the fifth woke up.
156 when the fourth woke up
781 when the third woke up
3906 when the second woke up
19531 when the first woke up

that's what I get by doing 5x+1, where x is the number leftover each time.

so, 19531

[Giles]

Ment yours doesnt work. The pile when all 5 wake up must be a multiple of 4 as its comprised of 4 equal piles

[Mentalist]

bah
missed that part.

base number must be 16.

so, 16, when all wake up

81 when fifth wakes up
406 when fourth wakes up
2031 when the third wakes up
10156 when the second wakes up
50781 when the first wakes up

[Mentalist]

lol, it still doesn't work, b/c i'ts not times 5

gah, my brain's being moronic.

[Giles]

whenever someone wakes up excluding the first person the pile must consist of a multiple of 4 as it is made up of 4 piles, so yours still doesnt work

[Mentalist]

lol, the very idea behind my method's faulty

the 5x+1 would mean each succssive person took 80% of the coconuts for themselves. rather then 20%

[Mentalist]

each number except the first must also end in a 6.

my brain's incapable of going any further with this today, lol.

[Giles]

Thats a bloody huge tree

Gah it doesnt seem to be working when i try it any more, must have got my formula wrong )

This post has been edited by baudin: 25 February 2009 - 02:23 AM

[Mentalist]

wait, how does that work?
you divide that into 5 you get 258 with three left overs.
you can only have one leftover.

Excel has failed you....
now that I think about it, at every stage the number has to end in 6 so that when divided into 5, there's 1 leftover, and so that it's divisible by 4.

the only exception's the first time where it can end in 1.

i'm sure there's an equation that can be made but I can't figure it out now.

[Slow Ben]

[/quote]

The answer I got was 15621. Seems a bit large, but it definitely works. There may be smaller, but I don't see how to figure it out.

Sir Thursday
[/quote]

I dont know about you all, but i can't find a way to make the pile any smaller than this.

[Mentalist]

ok
so, every time you want to go back an hour, you have to add the quarter of the previous total. plus 1
so it'd be 16 +5, for example. of course, that doesn't work, b/c it's nor didisible by 4.

This post has been edited by Mentalist: 25 February 2009 - 02:34 AM