The Labyrinthians Exchange... Solved.... I Think :)

[HyToFry]

Okay the answer is 2,25
A = 2
B = 25
S = 27
P = 50
I'm not sure if its THE only possible answer yet, but if the address was 2,25 They both would have figured it out EXACTLY when the riddle says they did.

Sorry about the new topic, but i can't get emailed if I just reply

The Product of 2,25 is 50 and there is only 3 products that can deduct this
1*50 2*25 and 5*10

So when Paul says I don’t know the house, but I do know that the first number is smaller than the second, this is all he knows.

Meanwhile Sam knows the SUM is 27 and that neither number is 1.
So now she knows that the first is greater than the second
There are twelve possible SUMS that can add up to 27 based on the criteria.

2+25, 3+24, 4+23, 5+22, 6+21, 7+20, 8+19, 9+18, 10+17, 11+16, 12+15, and 13+14

The following shows all of these possibilities, and what other numbers can be multiplied to get the same product for each of these cases PROVING that Sam KNOWS that Paul can’t answer the riddle YET.

2 * 25 = 50
50 = 2*25, 5*10,

3 * 24 = 72
72 = 2*36, 3*24, 4*18, 6*12, 8*9,

4 * 23 = 92
92 = 2*46, 4*23,

5 * 22 = 110
110 = 2*55, 5*22, 10*11,

6 * 21 = 126
126 = 2*63, 3*42, 6*21, 7*18, 9*14,

7 * 20 = 140
140 = 2*70, 4*35, 5*28, 7*20, 10*14,

8 * 19 = 152
152 = 2*76, 4*38, 8*19,

9 * 18 = 162
162 = 2*81, 3*54, 6*27, 9*18,

10 * 17 = 170
170 = 2*85, 5*34, 10*17,

11 * 16 = 176
176 = 2*88, 4*44, 8*22, 11*16,

12 * 15 = 180
180 = 2*90, 3*60, 4*45, 5*36, 6*30, 9*20, 10*18, 12*15,

13 * 14 = 182
182 = 2*91, 7*26, 13*14,


Okay this establishes that Sam KNOWS FOR A FACT THAT Paul can’t possibly know based on the fact that NONE of the possible numbers have only one possible outcome .


So Paul Says “I still don’t know”, which he doesn’t because as far as he knows it can be either 2,25 or 5,10.

When Sam says “I knew that” Paul knows that it is 2,25 because:
5,10 has a sum of 15 The only numbers that fit the criteria with a sum of 15 are:
2+13, 3+12, 4+11, 5+10, 6+9, 7+8

2*13 = 26, AND IT’S THE ONLY THING THAT IS TOO
So if the address is 5,10 Sam would know the sum was 15, and she wouldn’t KNOW that Paul Didn’t know because for all she knew it was 2,13 and if it was, Paul WOULD know the answer

NOW FOR THE LONG PART, PROVING HOW SAM FIGURED IT OUT…. I HOPE YOU CAN FOLLOW.

To make this short, I will only prove that 2,25 is the only two numbers that will add up to 27 where the product of the two numbers only has one set of sums that would allow Sam to know FOR SURE that Paul Couldn’t know the answer

2 * 25 = 50
2*25(True), 5*10(False),

3 * 24 = 72
2*36(True), 3*24(True), 4*18(False), 6*12(False), 8*9(True),

4 * 23 = 92
2*46(True), 4*23(True),

5 * 22 = 110
2*55(True), 5*22(True), 10*11(False),

6 * 21 = 126
2*63(True), 3*42(False), 6*21(True), 7*18(False), 9*14(True),

7 * 20 = 140
2*70(True), 4*35(False), 5*28(False), 7*20(True), 10*14(False),

8 * 19 = 152
2*76(True), 4*38(True), 8*19(True),

9 * 18 = 162
2*81(True), 3*54(True), 6*27(False), 9*18(True),

10 * 17 = 170
2*85(True), 5*34(False), 10*17(True),

11 * 16 = 176
2*88(True), 4*44(True), 8*22(True), 11*16(True),

12 * 15 = 180
2*90(True), 3*60(True), 4*45(False), 5*36(True), 6*30(True), 9*20(True), 10*18(True), 12*15(True),

13 * 14 = 182
2*91(True), 7*26(False), 13*14(True)


Any of the combinations with (True) would allow Sam to know FOR SURE that Paul Couldn’t know.

I Have already done the math on all of them, be my guest and do it yourself if you like .

Or just right a program to do it, (that’s what I did)


Since 2,25 is THE ONLY ONE that has only one group of numbers that do this , Sam knows that it has to be 2,25, or Paul Never would have figured out what the answer was based off knowing that Sam knew that he didn’t know…..

If anybody can explain it better, be my guest, I’m too tired to do anything more





[This message has been edited by HyToFry (edited 03-08-2000).]

[HyToFry]

I've been doing some thinking on my 2,25 solution, and although it is still possible, Sam would have to be able to do calculations at about 20 per second, SO where as this is possible, i don't think the riddle is solved unless it is explained how she deducted it so fast.
Remember they only had "a few hours" till the party.

Are we to assume that these two are infinatley smart?


After I posted the text aboce, i did some more thinking, and realized.

NUMBER 1 it never says how much time passed between them saying things.

NUMBER 2 it never says they figured out the answer before the party started, so they could have been there for days calculateing



[This message has been edited by HyToFry (edited 03-08-2000).]

[araya]

Hey, what's this, some kind of parade? Time for some rain..

(2,25) works for most of the conversation but it fails on the last statement.

S knows the sum is 27 but the numbers could be 2,25 or 4,23 or 5,22 - maybe more (I stopped looking).

From S's point of view, if the numbers were 4,23 the conversation would go the exact same way. P wouldn't know the numbers to start off (P = 92 = 4*23 = 2*46) but after stating that, P would know (because 2+46=48 which is sum of two primes and can be ruled out).

There must be something wrong with your program. The "real" answer is 4,13 although other possibilities include (4,61), (16,73) and (64,73) - none of which I have checked (except for 4,13 of course).

[Ghost Post]

Sorry HyToFry, I think I found a hole!

Lets say S=27 as you say. Sam (I feel I can call her that now, because I know her so well!) knows Paul can't know the address...fine!

Consider the possibility of P=92:
From Paul's point;
P=92, options for address: (2,46) & (4,23)
When Sam tells him she knew he couldn't get the address, this will tell him the address is (4,23) because if it was (2,46) Sam couldn't know that Paul couldn't get the address [the address could be (31,17) for S=48 and therefore Sam couldn't know for definite] Therefore, if S=27, P could equal 92, and the address could be (4,23)...another option, which means Sam can't then get the address after Paul got it.

If you take S=27, Sam won't know if the Product is 50, 92 or maybe even 162, 176 or 180 [I haven't carried through because it wasn't neccessary]

By showing that for S=27 there are more than one possible option for P, this tells us that Sam couldn't get the address after Paul, and therefore S can't = 27....!

Sorry for your trouble.

[Ghost Post]

(4,61), (16,73), (64,73) are not possibilities here because of the current size of the village!

If (4,61) was the address, => P=244. The only other possible address for P=244 is (2,122) which is not an address in the village. Therefore, Paul would get the address straight away. [unless of course he also forgot the size of the town!]

The same occurs for the possibility of:

(16,73):
P=1168 => other addresses are (4,292) & (2,584)... not in the village.

&(64,73):
P=4672 => other addresses are (16,292), (8,584), (4,1168), (2,2336)... also not in the village.

(4,13) remains the only address.

[Worm]

Explaining is always the tough part.

1. B/c S. knew that P. couldn't solve at first, the sum must be a number that can't be formed from 2 prime numbers (11, 17, 23, 27,29, etc.).

2. B/c P. could now solve with the possible sums reduced, the actual product could not be common to two sums. e.g. 30 can't be the product because it is common to 11 (6,5) and 17 (15,2), so P. couldn't solve in this case.

3. Taking out all the "common" products leaves a lot of "uncommon" products. So even though P. can solve, S. might be lost. Say the sum were 11, S. would know the possible products are 18, 24, 28, 30. 30 is the only common product here so she has to pick from 18, 24, 28. S. couldn't solve in that case.

4. S. actually can solve, which means the sum has only one uncommon product. This is only true for the sum of 17, which has one uncommon product of 52 (4,13).

5. The address is 4,13...QED

[HyToFry]

I have already proved 4,13 CANT be it though

I to thought, for a good part of the afternoon, that 4,13 was the solution, HOWEVER.
4*13 = 52
Also 2*26 = 52

Now 2+26 = 28
The only numbers that add up to 28 are:

N1 * N2 = Product, Possible Products
2*26=52, 2
3*25=75, 2
4*24=96, 5
5*23=115, 1
6*22=132, 5
7*21=147, 2
8*20=160, 5
9*19=171, 2
10*18=180, 8
11*17=187, 1
12*16=192, 6
13*15=195, 3
14*14=196, 4

ALL of these numbers have AT LEAST 2 possible outcomes (you'll have to take my word on this as i'm too lazy to write it out) EXCEPT,
5*23 and 11*17, now if this were the perfect world, the game would be over, because 2*26 is the only other way to get 52, but unfortunately, if the numbers were 2*26, sam would still know that paul couldn't know, either way 2*26, or 4*13 she would know that he couldn't know
BECAUSE

5*23 = 115 the only other possible is 1,115, which is not even in the town, so if paul knew the answer was 115, that only leaves one possiblity, 5,23.

11*17 = 187 SAME OUTCOME

So once again we run into the problem:
PAUL WOULDN'T BE ABLE TO FIGURE IT OUT, He would know the product is 52, and because either way 4,13 and 2,26 Sam would KNOW that Paul Coultn't know, Paul wouldn't be able to figure out the number by knowing that sam knew that he couldn't know (incedently, he would already know that she would know that )

Sorry

MAN THIS IS MIND BREAKING!!!!!!!!!!

[Worm]

Don't be sorry...

28 is not a possible sum.

Like you said, 23 + 5 = 28 and 11 + 17 = 28.

23, 5, 11, & 17 are all prime. For this reason, 115 and 187 (the corresponding products) only have 2 factors (not including the identity factors 1*115 and 1*187 since they are excluded by Samantha's statement that there are no 1's). If Paul knows the product is 115 or 187 he could solve it with just the knowledge that there are no 1's.

What's needed are sums that cannot be derived from prime numbers. e.g. 11. The possible ways to arrive at a sum of 11 are:

9 + 2 = 11
8 + 3 = 11 (Note that there is one
7 + 4 = 11 non-prime number in each
6 + 5 = 11 equation)

("10 + 1" is excluded by the no 1's rule)

The corresponding products are:

9 * 2 = 18
8 * 3 = 24
7 * 4 = 28
6 * 5 = 30

18 can also be factored into 6 & 3, 24 into 6 & 4 or 12 & 2, 28 into 14 & 2, 30 into 6 & 5 or 15 & 2 or 10 & 3. Because there is more than one pair of factors for each product, Paul cannot solve if Samantha's sum is 11. The same is true for sums 17, 23, 27, 29, 35, 37, 41. There are more possible sums, but these give you the idea.

[HyToFry]

Araya You Said "
From S's point of view, if the numbers were 4,23 the conversation would go the exact same way. P wouldn't know the numbers to start off (P = 92 = 4*23 = 2*46) but after stating that, P would know (because 2+46=48 which is sum of two primes and can be ruled out).
"


This is because you are thinking wrong, and if you think about it, you will figure it out, if I tried to explain it, it would sound something like this;
"Although S cannot be the sum of two primes, this isn't to say that Product of one of the (sum)roots of S cannot have a (product)root that has a Sum which has a root of two prime numbers"

To make it simpler to prove, i will prove 4,13 wrong using the same logic (however 4,13 is not ruled out because of this, it is ruled out for other reasons)

(I will use your EXACT text, replaceing numbers accordingly)

S knows the sum is 17 but the numbers could be 2,15 or 7,10 or 8,9 - maybe more (I stopped looking).

From S's point of view, if the numbers were 7,10 the conversation would go the exact same way. P wouldn't know the numbers to start off (P = 70 = 2*35 = 5*14) but after stating that, P would know (because 5+14=19 which is sum of two primes and can be ruled out).

Although your correct that 2,46 could be ruled out, there are other combinations that CAN'T Giving ALL of Sams Root AT LEAST 2 of the roots Paul would have, that would let Sam know that Paul couldn't know the answer, EXCEPT 2,55, which ONLY has 2, 25 and 5,10

if the roots were 5,10 Sam Wouldn't know that Paul Couldn't know, so now Sam knows that it has to be 2,25 BECAUSE all of the other products of her roots have at least two sets of roots that tell Sam that Paul Can't know, so if it was any of these, Paul wouldn't know, but he DOES, so SHE DOES

you guys ever hear of thinking outside of the box?, I feel like we're having to think ourselves right back into the box on this one

I still say 2,25 is the answer, if my Definition doesn't explain it, maybe this will.

(This is the full story)

Paul says "I still don't know the answer"
Sam says "I knew that"

(Paul sees a house decorated with "Welcome

Paul and Sam" banners on it)
Paul Says "Okay now i know"

(Sam now sees the house too)

Sam Says "Okay so do I"

HAHAHHAHAHHAHA LMAO, Man I need to get out more

[HyToFry]

Worm Wrote
Don't be sorry...
28 is not a possible sum.

If this were true worm, than Paul would know the answer after 1 turn

if 28 is not a possible sum
and 2*26 and 4*13 are the only possible products of 52 then
Paul would know it was 4,13 because 2,26 is not a possible sum

So once again, sorry

Oh and BTW 28 is a possible sum, which is to say, if the sum was 28, Sam would KNOW for a Fact that paul couldn't know the answer.



23, 5, 11, & 17 are all prime. For this reason, 115 and 187 (the corresponding products) only have 2 factors (not including the identity factors 1*115 and 1*187 since they are excluded by Samantha's statement that there are no 1's). If Paul knows the product is 115 or 187

[HyToFry]

Something else i noticed when i was examining this problem, the product would have to be less than the maximum of the second number, (for any number that would alow Sam to know that Paul Couldn't know) so so a product of >100 wouldn't work and can always be ruled out by both paul and sam anyways, THERE IS ALOT OF TIME INVOLED TO PROVE THIS TO YOURSELF, SO IF YOU WANT TO GO RIGHT AHEAD .

So far these are all the criteria i have come up with

1 < x <= y
and x + y is not a prime number
And x * y < max(y)

if the numbers do not meet the third criteria, then paul knows the answer in the first place

A good example would be if P = 115 because 5*23 = 115 and 1*115 = 115 and 1,115 is not a possible answer (out of town), Paul would know from the start the answer was 5,23.

I haven't done much reasearch on this, however 2,25 meets all the criteria required, also ALL of the products of the sumroots of 2,25 have at least two sumroots that meet thees requirements except 2,25, the only other possible sumroots it could have is 5,10 and since 5,10 dosn't meat these requirements, it can be ruled out.

If you find a hole PLEASE let me know.
thanx Chaz

[worm]

okay hytofry, listen up

First I have to give credit where it's due. Paul wouldn't have needed the no 1's rule to solve if the product was 115 or 187. The 1, 115 and 1,187 possibilities are ruled out by the size of the town. Good catch. Doesn't affect 4,13 at all but attention to details is admirable albeit occasionally annoying.

Now, just to humor you, let's say that the sum is 28. Samantha knows that 5,23 and 11,17 are possible addresses in this case. I just agreed with you in the above paragraph that either of these would give Paul easy solutions w/o Samantha's help. Therefore, if the sum is 28 Samantha does not and cannot know (before he tells her), that Paul can't solve it.

[HyToFry]

Sorry the third criteria is only needed in order for paul to be able to find out the answer when sam says that she knew that he didn't know. If the third criteria is not met, than ALL possible numbers that sam would know from pauls point of view, would allow sam to know that paul couldn't know the answer yet.

There is only one set of roots that sum up to 27 that have a product with only one set of roots that that meet all three criteria, all other roots of 27 either don't meet these criteria requirments, or have two or more roots that do, makeing it impossible for paul to know the answer based off of knowing that sam knows (because there is still two possibilities)

3*24 product of 72 which has the roots 2,36 - 3,24 - and 8,9

and 4,23 product of 92 which has the roots 2,46 - and 4,23

hope this helps you see the way to the party, which is incedently at 2,25

[HyToFry]

WORM THIS IS FOR YOU

[HyToFry]

Araya, sorry about what i said earlier, you were right in what you said, however still this dosen't prove 2,25 wrong, because Paul doesn't know that Sams Sum can't add up to be a prime number (at first) he would only know this after sam says I knew that, so Sam can still assume that it might be 4,23, until Paul says that he figures it out, then she could rule this out, and probably all others in her set EXCEPT 2,25 which would give her the answer.

[This message has been edited by HyToFry (edited 03-09-2000).]

[worm]

HyToFry

Let me ask you a question.

At the end, how does Samantha know that its 2,25? If the sum were 27, how does Samantha know it's not 7,20 or 11,16 or 12,15? Is she just a good guesser?

[HyToFry]

Because ALL of the other possible roots of 27 have products which have at least 2 possible sums that would allow Sam to know that Paul Couldn't know, so she knows that it is 2,25 because if it wasn't Paul Wouldn't know the answer

Worm, why don't you register your name?

[HyToFry]

ALL RIGHT I'M A DEAD ALIEN

[worm]

Never mind the question I asked you, HyToFry. Your questions simply amaze me. It's like trying to explain why 2 + 2 =4 to a kid who is pretty darn sure the answer is actually 5.

[HyToFry]

bradan, this is for the "hole" you found in my theory...... DAMN THIS IS FUN GUYS I LOVE THIS

quote:

---

Lets say S=27 as you say. Sam (I feel I can call her that now, because I know her so well!) knows Paul can't know the address...fine!

Consider the possibility of P=92:
From Paul's point;
P=92, options for address: (2,46) & (4,23)
When Sam tells him she knew he couldn't get the address, this will tell him the address is (4,23) because if it was (2,46) Sam couldn't know that Paul couldn't get the address [the address could be (31,17) for S=48 and therefore Sam couldn't know for definite] Therefore, if S=27, P could equal 92, and the address could be (4,23)...another option, which means Sam can't then get the address after Paul got it.

---

If the nubmer was 17,31 that would mean that the only other possible answer would be 1,527 which would give paul the answer in the first place , so in that case Sam KNOWS for a FACT that it cant be 17,31 and would rule it out of the set. any other numbers in the set act the same way, or have multiple numbers that paul could use to get the answer

[HyToFry]

What the hell do you mean? 2+2 = 78 EVERYONE KNOWS THAT... just kiddin

You Still have yet to prove that Paul would know that it was 4,13 and not 2,26 BOTH of which would allow Sam to KNOW that Paul Couldn't know.

Still waiting

[Ghost Post]

The answer is 2,8 (product=16, sum=10)

Paul wouldn't know from product. It could be 2,8 or 4,4.

Sam wouldn't know from sum. It could be 2,8 or 3,7 or 4,6 or 5,5, but she eliminates 3,7 and 5,5 from the fact that Paul does not know the numbers. That leaves 2,8 and 4,6.

Paul still doesn't know. If it were 4,4, Sam would know the sum to be 8, and would have been considering 2,6 and 4,4 as possibilities- i.e. more than one possibility. So Paul can't eliminate the 4,4 possibility yet.

Sam knows that Paul didn't know. It can't be 4,4 then. If it were 4,4, Sam would have to consider that the numbers are 3,5, and that Paul would know them from the product 15.

Paul now knows it couldn't be 4,4, and so is 2,8.

Sam (recall she was last considering 2,8 and 4,6 from knowing the sum of 10) now knows it is not 4,6 (because 4*6 = 2*12 = 3*8).

A note about the 4,13 solution: 4,13 is a solution to a variant of this puzzle. The puzzle as stated is has the following statements being made:
1) P says I don't know.
2) S says I don't know.
3) P says I don't know.
4) S says I knew that you didn't.
5) P says I now know.
6) S says I now know.

Leave out statements 1 and 2, and the solution is 4,13. But With statements 1 and 2, it's quite different.

[HyToFry]

Sorry Extro... I already gave this a run for its money

quote:

---

The answer is 2,8 (product=16, sum=10)
Paul wouldn't know from product. It could be 2,8 or 4,4.

Sam wouldn't know from sum. It could be 2,8 or 3,7 or 4,6 or 5,5, but she eliminates 3,7 and 5,5 from the fact that Paul does not know the numbers.

---

At this point Paul might think it was 3,7 OR 1,21 and 5,5 or 1,25 respectively..., because Sam dosen't tell him that neither number is one until AFTER, so if it were 2,8 Sam WOULDN'T KNOW that paul Couldn't Know......

Your thinking ahead of the puzzle, you have to wait for Sam to tell him that it can't be 1 for your statement to be true.

It's 2,25 why can't you people just believe me? :b

(I hope)

I'm still waiting for someone to either prove me wrong, or prove 4,13 right?


[This message has been edited by HyToFry (edited 03-09-2000).]

[HyToFry]

I have relooked at throsby's criteria that would produce a winning number, and although he was close, he was slightly off....

quote:

---

a.b cannot be a prime (ruled out by a and b both greater than 1).
a.b cannot be the product of two primes, or else P would know the address.
a+b cannot be the sum of two primes, or else S could not know that P did not know the address.
However, as the fact S knew that P did not, P can then work out that a+b is not the sum of two primes. This means that a.b is a number that has only one set of two factors that add to a number which is not the sum of two primes.
My back of the envelope calculations give me 18 as the first a.b to satisfy this condition, and 2.9 as the address. I have not ruled out all other addresses (I will leave that to people who are either not at work, or much faster at this than I).

---

There is an exception to rule "a+b cannot be the sum of two primes, or else S could not know that P did not know the address." a+b CAN be the sum of two primes if A*B = a number that is greater than the MAX(B), because the other number of the two primes would be 1 and A*B which is outside of the town, and P would still know the answer from the start....

I hope this clarifys my point that the answer MUST be 2,25.
A = 2, B = 25, S = 27, P = 50

[HyToFry]

Worm Wrote:

quote:

---

Explaining is always the tough part.

1. B/c S. knew that P. couldn't solve at first, the sum must be a number that can't be formed from 2 prime numbers (11, 17, 23, 27,29, etc.).

---

Your assumeing that P knew neither of the numbers could be one, which He didn't, he only knew that B > A, he didn't know that B and A > 1 until after S told him, so S cannot rule out primes as the answer because for all S knew it was the prime numbers, take s = 9
A = 2 and B = 7, both primes
S knows that A <= B but dosn't know if paul is thinking it could be 2,7 OR 1,9, so when she says neither number is 1, that might help Paul to know that it was 2,7

Oh and i still think 2+2 = 17.453445434

[Ghost Post]

Missed that about Paul not knowing the numbers were greater than 1 until Sam said so. With that taken into account, I get 24 different solutions:
(2,25) (2,27) (2,42) (2,44) (2,54) (2,58)
(2,82) (3,25) (3,35) (3,45) (3,51) (3,69)
(3,74) (3,86) (4,74) (4,82) (5,35) (5,58)
(6,74) (7,58) (7,62) (10,58) (11,46) (13,34)

I sure hope that the "official" solution doesn't dismiss the subtle variations between this puzzle and the more common variations (such as where 4,13 is the answer).

[worm]

I think I see the gap that separates us HyToFry.

In his first statement, Paul did not say that he couldn't figure out the address. He said he had forgotten it, but remembered the product. He doesn't actually tell Samantha that he can't figure it out until after she says she knows the sum and that there are no 1's. But she knew it before he told her, b/c she knew that the sum could not be derived from two prime numbers.

You trackin'?

[worm]

Sorry, Charlie, Paul DID know that 1 was out of the question when he told Samantha that he couldn't solve.

[HyToFry]

Yes accept that he says, "I still can't figure out the answer" after she tells him that is greater than 1, so we know at that point that P can't equal 5*23 or 11*17, because P WOULD be able to figure out the answer at this point, and it stated that both P and S were being Honest about the whole thing

2,25 works Man try it, i'm not sure about all of the other that extro posted

[This message has been edited by HyToFry (edited 03-09-2000).]

[HyToFry]

[Ghost Post]

I'm back with 4,13, based on worm pointing out that the first two statements don't say that Paul and Sam can't figure out the numbers, but only that they don't remember them. One could argue about what that means. Can you remember a number is equal to 2+2, and not remember what the number is? So:

1) P says - I don't remember, but the first was not greater than the second.
2) S says - I don't remember, but neither was 1.
3) P says - I can't figure them out.
4) S says - I knew that you couldn't.
5) P says - I now know the numbers.
6) S says - I now know the numbers.

So after statement 2, we (and they) know that the numbers p and q are such that p<=q, 2<=p<=100, and 2<=q<=100

[HyToFry]

I've said it once, and i'll say it again

Yes accept that he says, "I still can't figure out the answer" after she tells him that is greater than 1, so we know at that point that P can't equal 5*23 or 11*17, because P WOULD be able to figure out the answer at this point, and it stated that both P and S were being Honest about the whole thing
2,25 works Man try it, i'm not sure about all of the other that extro posted

[HyToFry]

The reason i know its not 4,13 is because when i was typing up my explanation of why it HAD to be 4,13 I proved that it couldn't be 4,13 I also typed up why it was 2,9 but later proved that wrong, nobody has presented any solid evidance that 2,25 is wrong though, or if the have, I have proved them wrong, so I'm gonna stay with 2,25 for the time being

[HyToFry]

extro... I have looked at your 2,27 and can prove it wrong

(2,25) - I can't prove the correct answer wrong

(2,27) - Both primes so P would know after S told him A and B > 1

I didn't have time to do any others, as they all meet my criteria for a winning number, but this dosen't mean that all of their roots meet the criteria

[Ghost Post]

HyToFry: 27 a prime? I'd like to see a proof.

[HyToFry]

RIGHT!!!!!!!!!! HEHE

My bad

[HyToFry]

Okay extro you win I have tried a few of the numbers that you posted and have come to the conclusions that ALL of them that I tried hold true for as far as i can tell

This means that either A the puzzle was stated incorectly ie the town should have been 100 by 100+ or B there CAN and IS multiple answers

Or C (posted later) Worm is correct with his theory that Paul Forgot the number, which doesn't mean that he couldn't have figured it out in steps 1 and 2 if it was the product of two primes, and also that Sam wouldn't have thought that Paul could have figured it out in steps one or two, so consider that S = 28, Sam would think that Paul might think the Products of 5,23 (115) could be either 5,23 or 1,115 and because for all she knew her telling him that it can't be 1 would have ruled out the 1,115 and Paul would have known the answer given the information that neither number was 1, she can't rule out the fact that Paul, A VERY QUICK WITTED MATH-MATITION, Wouldn't have been able to figure out that 1,115 is somewhere in Pleasentville the neihboring town, and so she wouldn't know that he couldn't know.
Also Paul would have to assume that Sam wouldn't count on him being smart enough to know that 1,115 is not a possible address.

But if this is the Case, i don't see how paul could possibly know that his products must not be the sum of two primes (unless he is absent-minded) , i mean come on it doesn't take logic to know that 1,115 isn't possible, just common sense , so she couldn't count on the fact that Paul might have just Remembered the address all of the suddon, and she shows up at 4,13 when the party was really over at 6,11 - where Paul is because he happend to remember the address.

(Think about guys it never says that Paul Didn't Just Remeber the address all of the suddon)

This sounds a little far fetched to me though.

My guess is the problem is A (the puzzle was stated incorrectly)... hehe


[This message has been edited by HyToFry (edited 03-09-2000).]

[HyToFry]

My final answer for this riddle is 2,25.

I do however have new reasoning behind this, you see you people may not know this, but i live in Mathematica and my address is 2,24, now i live one house west of 2,25 and those darn kids kept me up all night with their loud music, and times tables. In fact it got to the point where i had to call my girlfriend (who incendently lives at 4,13) and stay the night at her house. (this is how I know that it wasn't at 4,13).

There was a party at 4,13 however I don't rememeber ANY one with the names Sam or Paul attending the party, (in Mathematica we also refer to all people by where they live, ie my name is not HyToFry it is 2,24-3 my Dad(2,24-1) and Mom(2,24-2) have lived here pretty much their whole life, and when I save up enough money, I'm moveing out of this God-Forsaken Town (the reason i hate it so bad, is because whenever there is a town meeting, i have to walk CLEAR down to 1,1 because there is never enough parking for everyone, they couldn't have town square at 50,50, NO thats 50,50-1's house.. oh sorry 50,50-1 in joe parker, he just moved here a month or so ago....

Well i hope this clears things up for you people .

[This message has been edited by HyToFry (edited 03-09-2000).]

[HyToFry]

Okay I think i should post this before someone else does, I would like to change my answer to 4,13 (which was my second choice untill I discovered the odd 5,23 and 11,17 problem, so i figured that minotaur was trying to trick us to thinking that the answer was 4,13 by limiting the playing field to 100,100 (AND I WASN'T ABOUT LET MINOTAUR TRICK ME AGAIN )

HOWEVER the problem with the problem is that IT NEVER SAYS THAT SAM AND PAUL DO OR DON'T KNOW THE SIZE OF THE PLAYING FIELD. I still think worms theory on Paul not being able to figure out the sum is kinda far fetched .

I am changing my answer to 4,13 ONLY because if Sam and Paul DO know the size of the town then there is 24 different addresses that can fit the criteria, and if they don't know the size of the town there is only 1 4,13 (my girlfriends house, no wonders i couldn't sleep) and I knew that we had to be able to answer the question, thus telling us that Sam and Paul DIDN't know the size of the town....... I WOULD LIKE TO GO ON PROTEST THAT THIS INFORMATION SHOULD BE GIVIN IN THE RIDDLE THOUGH

[Ghost Post]

HyToFry: I get 4,13 WITH the assumption that they know the size of the town.

Here's how. I wrote a program which considers possible solutions - initially 10000 possible solutions (I assume they both know both numbers <= 100.) - and eliminates whichever it can based on provided information.

Paul: I'm afraid I've forgotten the address. I can only remember the product of the two numbers, and that the first number wasn't greater than the second.

This just tells us Paul knows first*second, and that first<=second. So we eliminate all solutions with first>second.

Samantha: I've forgotten it too, but can only remember the sum of the two numbers, and that neither number was 1.

This just tells us that Samantha knows first+second, and that first>1 and second>1. So we eliminate all solutions with a 1 in them.

Paul: I can't figure out where the party is.

Now we know that there is more than one way to express the product as p*q while sticking to the above constraints (2<=p<=q<=100). So we can eliminate all pairs having a product that has a single valid factorization.

Samantha: I knew that.

This tells us that the sum is such that any two numbers in the range 2..100 that add up to that sum have a product with more than one valid factorization. So we eliminate all pairs having a sum that can be the sum of two numbers whose product has a single valid factorization.

Paul: OK, I know where the party is.

Now, from the pairs of numbers we have not yet eliminated, some have products with unique factorizations into pairs of numbers we have not yet eliminated. These we keep. All others we eliminate.

Samantha: OK, so do I.

Now, again from the pairs of numbers we have not yet eliminated, some have the same sum as other pairs, and the rest have unique sums. We keep the ones with unique sums.

You are left with 4,13


[This message has been edited by extro... (edited 03-09-2000).]