1. A week is seven days.

They met on sunday and left. Then after a week they no longer had to come every day.

So does this mean that the beginning of their 'week' was monday and the the last day was monday?

It is either six or seven people.

All the people looked at one of the members that had a mark. That person didn't come back. They did this for one weeks time. On the eighth day they didn't have to return. I am going to go with seven people.

2. yes beta
youre way off

and justin as i said in one of my prev messages no one speaks they just think it out

and as nvoigt will tell you it is possible
read the prev mesgs for more hints

3. They wouldn't have to speak.

Imagine a room that has five people and a moderator.

The moderator says that some in the room have marks...

Each person cannot see their own mark and cannot tell others if they have a mark.

Each person that has one has to leave.

*****

Ok... one and two have a mark.

Without communicating individually at all, if two, three, four, and five all look at one for a while... one should figure that he has a mark and leave. Then three, four, and five look at two until he leaves.

So... I guess I really suck at this. Please email me the answer before I start killing my neighbors.

betazep@hawaii.rr.com

4. ok so betazap gives up as well
ill mail you the answer(or atleast a part) beta

any more people with any ideas???

5. Look, I just don't think there is enough info provided. There are currently an infinite number of possibilities. The key to this is, if there is no way to know you have the mark, and no way for the others to let you know or keep you from attending, then there is no longer a solution. If the others can do anything but speak, then they jailed the sinners the first day, then took the next six days off to congratulate themselves.

Maybe I missed this, but can the priest say anything about it, like "there are still sinners" or "there are no longer sinners". The mathematical possibilities without any limitations are infinite. The mathematical possibilities with perfectly strict limitations are 0. So either no one was a sinner, or the riddle has some important missing information.. Consider that I give up and PM me, unless there is some limitation that was neglected earlier that you can tell everyone.

If there is a solution with the given info, then please PM and explain to me why 0 sinners does not meet the requirements. Thanks.

6. Here is an outline of a proof.

among those of you who have gathered here are some sinners
there is at least one sinner.

Base case.

I assume this all would happen simutanously.

day one passes
1.1
If there is one sinner left by process of elimination
not seeing any other sinners would know that
they are the sinner and leave.

day two passes
1.2
If there are two sinners left by process of elimination
they are able to tell that they are sinners.
This is because sinner #1 would see the sinner #2.
sinner #1 would be able to reason
case 1: I'm not a sinner.
sinner #2 would then would have left by 1.1
case 2: I'm a sinner
only possible case
They would both leave.

Generalizing it is possible to see that if a sinner sees
two other sinners who havn't left due to 1.k-1 then
they would be able to reason by 1.k-1 that they are
sinners. Therefore my rough guess is 7 sinners.

7. 1/2 of the people are sinners.

8. I'll try to do a better proof of this as it's worded
pretty bad and leaves out some stuft.

9. This is still an outline.

1.1
There is at least one sinner.

Let N = Number of sinners
Let n = number of days past

if N=1 and n=0 then that one sinner would see no other
sinners. By 1.1 he would reason that he was the sinner.

if N=2 and n=1 then both sinners would realize that
when n=0 no one left. But both sinners would see each other
by process of elimination both would leave.

Generalizing
if N=k and n=N-1 then N sinners would realize that
when n=N-2 no one left. So the N sinners would by the proccess
of elimination leave.

after a week none of the people with the mark were left
I'm assumming that on the 7 day when n=6 all the sinners leave.
Solving for N in n=N-1 gives N=7.

10. I get it now... I even had to think about it after I saw the solution. It makes sense now, though.

11. ## WELL DONE NICK

HI NICK
WELL DONE YOUVE GOT THE ANSWER

im in a hurry now but ill post the complete answer tonight

jv

12. Ah Ha! There was missing info if that is the solution. This assumes we know how many sinners there are to start with. If there are two sinners, and they don't know how many, then the one looks at the other one and thinks, "oh good, that person is the sinner". But they can't tell each other, so they just both comming to church. That solution does work if and only if there are a fixed and given number of sinners.

Granted I didn't read the original riddle carefully enough the first time I posted, but I'm pretty sure I'm right on this. Just because they know there are still sinners, by the fact they keep coming to church, doesn't mean they know how many sinners there are.

13. GRRRR JUSTIN

YOU STILL DONT GET IT OK ILL EXPLAIN

imagine and i mean imagine that there is just one sinner(this is hypothetical)

so all other people in the church see his mark and hope that he is the only marked one

now the sinner doesnt see a mark and knows that atleast one person is marked so s/he knows that it has to be him/herself

and so stops coming from monday

now imagine there are two sinners then they each see 1 person with the mark on sunday so they both come back on monday

now A wonders why B came back it could only be if b had also seen a mark so it must be A who also has a mark and in the same way B also figures it out that s/he has a mark

so they both stop coming on tuesday

and this can now be extended to seven days

i cant write down the whole proof obviously but mathematical inductio helps me there

it says if
p(1) is true and p(k) implies p(k+1) then p(n) generilastion is true

so 7 is the soln

14. I don't believe it is logical to solve this backwards. There couldn't be one or two people with the mark, because it took seven days. So if there are three or more with the mark to start with, then A sees B and C. B sees A and C. C sees A and B. D, who does not have the mark, sees that A, B, and C have the mark. A assumes B doesn't leave because B sees C (and what, three others?). So on and so forth until D, who is COMPLETELY innocent, decides to leave because he is also paranoid. Everyone else stays for another three or four days until the sinners are sick of going to church and ditch. This then, is the only reason all the sinners are gone by the seventh day, and there is no way to tell how many there were to start with.

You can't get rid of one per day with several people marked because there is no reason A should leave over B over C over the next four sinners. It is a stalemate. Put yourself in the shoes of one who doesn't know if you are a sinner or not and seeing several sinners return every day. Do you assume that makes you one?

Furthermore, if with geometry we can prove that a point is an infinitely small but existent amount of space, and that in the measure we call an inch, there are an infinite amount of said points, then how is movement possible? Must you not pass through all points? How long would it take you to travel through an infinite distance? It takes me about .01 seconds usually. Does this mean I skip space like in animation? Am I running at 4000FPS?!?!??!

15. I'm with Justin... either:

1 - the priest has to say how many sinners there are to start with, in which case they all realise the next day according to your logic, ie each sinner sees only 6 other sinners... or

2 - each day they go to church the priest has to say whether there are any sinners left or not... without this communication there is absolutely no reason why the last sinner would leave.