The Labyrinthians' Exchange

[Griffin]

This is a very cool logic problem. In fact, I think it was the problem proposed by Araya in his very first post.

[Quailman]

I, too recognized this aas a problem that had appeared in the old forum. I couldn't figure it out then either.

ps: Griffin, are you ouini?

[throsby]

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).

[mithrandir]

18=7+11

[araya]

and the old forum comes back to haunt the minotaur. hehe.. did anybody else notice that the year is now 19100 according to the old forum?

[Gidon]

Hi

I looked at the old thread aswell, and wrote a program to generate the numbers. I correctly get the numbers 4,13, but also get the numbers (4,55) (4,61) (3,64) (4,67) (4,73) (3,76) (4,79) (4,83) (4,89)

any comments? the way the puzzle was phrased i hoped to find a unique sollution. the program is in PERL (61 readable lines), and if anyone wants it leave your email and request in this thread.

Cheers

Gidon

[Griffin]

Quailman - No. Why do you ask?

Gidon - I've only tried the first one (4,55), but there doesn't seem to be anything wrong with it. I remember one place I saw this puzzle, there was a requirement that the numbers be between 2 and 50, so you may be right.

[Ghost Post]

Here's the problem with (4,55)

If it were 4 & 55, Samantha knows the sum is 59. She can't discount the possibility that the numbers might be 6 & 53, in which case Paul would know from the product (6*53=318) that the numbers are 6 and 53. But she claims to know that Paul could not know the numbers.

Note: 318 = 6*53 = 3*106 = 2*159, but 106 and 159 are both over 100. If Paul knew the product to be 318, he would know the numbers are 6 & 53.

[Ghost Post]

The point is, Paul not knowing the numbers tells us not only that the product is not the product of two primes (thus that the product can be represented as the product of two integers greater than one in more than one way - like 318, for instance), but also that it can be represented as the product of two numbers in more than one way, where both numbers are in the range 2 to 100 (unlike 318).

[This message has been edited by extro... (edited 02-29-2000).]

[Wonko the Sane]

This may not be topical to the puzzle, but I figured I would raise that there is a minor typo in the phrasing of the problem. It should be "sharp-witted" not "sharp-itted". Might want to revise that sometimes. =) Sorry...writer's syndrome.

[kitakaze]

I'm having a lot of trouble getting beyond the logic of Paul's statement that he knows where the party is.

I have worked out that the only possible sums are
11,17,23,27,29,35,37,41,47,53

but I'm having trouble narrowing it down further. I need to get my head around this!

Since I know that a possible answer is 14,3 it really helps.

Kaze

[Ghost Post]

I only saw this problem today. I had a go, and failed. Then, without looking for it, I stumbled across the following page, which is about a full a description of the problem as you could ask for. Spooky, eh?
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/logic_sum_product

[Ghost Post]

I think the solution should be: 2-6;
I will try to explain the reasons in a following mail.

[HyToFry]

I've gotten 2-9 clear to where paul knows the answer, but not where sam does... YET

If the answer were 2-9 THEN
Paul would have gotten 18 for his product, and the first cant be greater than the second, this leaves 3 choices
1,18 ; 2,9 and 3,6 thats it

next Sam says I can only remember the sum and that neither is 1

She would have known the sum to be 11

this only leaves four possible answers for her
1. 2,9 which means paul would know 18 which is also possible with 3,6

2. 3,8 which means paul would know 24, which is also possible with 4,6

3. 4,7 which means paul would know 28 which is also possible with 2 and 14

and 4. 5,6 which is 30, also possible with 2,15

SO AT THIS POINT SAMANTHA KNOWS THAT TELLING PAUL THAT NEITHER NUMBER WAS 1 DIDN'T HELP HIM FIND THE ANSWER, in all four cases there is AT LEAST two possible sets of numbers that paul could use. So paul still dosn't know, because he knows it can be 2,9 or 3,6

NOW if Sam added up 3,6 she would get nine; nine is only possible with 2,7 3,6 and 4,5.

2*7 is 14 and its the only possible way to get 14 there is no other way. so if sam knew it was 9 (3,6) she wouldn't KNOW that paul couldn't know the answer. SO when she says i knew that, paul knows that the answer is 2,9 because if it was 3,6 sam wouldn't KNOW that he didn't know.... Well i hope you've followed me so far, and i hope i'm right, i'm so bloody confused now I'm tempted to delete this post... just kidden... even IF i'm wrong, at least i tried

I'm still trying to figure out how Sam figured out at this point that if paul knew that she knew that he couldn't know then she figured out the answer, but MAN I'm TRYING, this will DEFINATELY be the hard part with lots of numbers and stuff

[HyToFry]

Sorry guys i think i found a hole in my own theory.... 4,7 acts the same way

[HyToFry]

2,6 wont work, 2,6 sums up to 8
so does 3,5 which has a product of 15
15 is only possible 2 ways where A<=B
1,15
and 3,5
so sam would not KNOW that paul didn't know the answer when she said that neither number was 1, because for all she knew it may have been 3,5 and paul could have known the answer at that point
2,6 WONT WORK...
neither will 2,9

[Ghost Post]

SORRY
Thank you HyToFri, my 2-6 is not a solution.
(Samantha said "I knew that", which refers, as HyToFri noticed, to the fact that Samantha WAS SURE that Paul could not know the numbers; instead I reasoned as she said "I do not yet know the numbers", that would have been better for us)

forgive me! (English is not my first language!)

I must restart all my previous reasoningand maybe I will write again.

[Ghost Post]

The last message I posted was nearly, but not quite there. I made a stupid mathematical blunder which threw me, but that’s all sorted out now! (and I deleted the message!)

To rectify the situation, here’s my understanding of the puzzle and solution:

1) If Samantha knows that Paul can’t solve the address, S (the sum) must be so that all its root pairs [numbers added to give the sum] multiply to give products which ALL have more than one set of factors [numbers multiplied together to give a product ] For this to happen, S is a number which doesn’t have roots of the form (prime, prime), because if there is a set of roots (prime, prime) they are the exclusive factors of their product , and so Samantha can’t be sure Paul can’t get the address.
[Most numbers have a set of prime roots, e.g. 16: (3,13), 18: (11,7)… some don't, like 27. Prime numbers, [with the exception of some] don’t [7 for example has roots (2,5), both primes]. These, are therefore possible ‘S’s].

Next;

2) If Paul can get the address after Samantha tells him she knew he couldn’t get it, P (the product) must be so that there is only ONE set of factors which add to give a possible ‘S’; the others being dismissed so that Paul can get the address. For this to happen, P is the product of a square and a prime, and the address is of the form (square, prime) or vice versa.

For example, 112 has the following factors: (2,56), (4,28), (7,16) & (8,14) The only set that gives a possible ‘S’ is (7,16) = 23, and they are of the form (prime, square) [23 is a possible ‘S’ because the product of ALL its root pairs have more than one set of factors; as is required above]

However,

3) Lastly, if Sam can get the address after Paul gets it, S must have only ONE set of roots whose P is in this form. For example, for S=23, (4,19) & (7,16) are both possible addresses!


Considering these criterion, the address is (4,13), P=52, and S=17.

I hope this makes sense.



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

[HyToFry]

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!!!!!!!!!!

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

[Ghost Post]

What HyToFry said is true. In fact, it's true for S = all numbers. The only set of roots whose product won't have more than one answer is in the form (prime, prime) and, obviously here Paul would be able to know the address. e.g.; S=12: (5,7) is a possible address, but if it was, Paul would know the address so it can't be the address! Samantha can rule out that option, and all others carry multiple possibilities, so Paul can't get the answer!

By that thinking, S = any number!!!!

But, if any number can be S, Paul couldn't get the address after Samantha told him she knew he couldn't get it. Yet he does!

In order for this to be true, the above scenarios musn't be applicable. For (4,13) they're not, and yet all the rest can be true! It has to be the address.

Interestingly, you can assume that all you can deduce, they too can deduce, unless ofcourse they don't have to because of the extra information they have that we don't!!

[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.

I Will post later with the Proof... ITS BEEN FUN... ALOT OF FUN GUYS

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

[HyToFry]

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

OH BTW this is for the minotaur, LETS GET SOME MORE LOGIC PUZZLES, i'm gettin boared with these Math ones They take me too long

Oh and one more thing, WHERE DID YOU GET THE DECK OF CARDS IN THE SKINNEY PROBLEM, I TOO WOULD LIKE A DECK OF THESE SNAZZY CARDS



[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 this, 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).]

[Ghost Post]

HyToFry, don't say anybody, but I think there is a little problem with your proof in the 2,25 solution, in the rows:

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


2*46 is False, because 2+46=48 and 48 (from the Sam point of view) admit the 5-43 possibility,which would have been a solution for Paul.

THIS PUZZLE IS LIKE A VIRUS!
(I can hardly succeed in reading my own reasoning after a while)

Of course, to find a solution, if one exists, we are supposing that:
1) paul and samantha are perfectly quick witted, but more important: PAUL KNOWS THAT SAMANTHA IS QUICKWITTED AND SAM KNOWS THAT ALSO PAUL IS QUICKWITTED;
2) Paul and Sam are short memoried, BUT THEY HAVE ENOUGH MEMORY TO REMEMBER THEIR WORDS AND THEIR REASONING FOR THE TIME OF THEIR DIALOGUE.

[HyToFry]

If P and S know the size of the town than S can eliminate the possibility of 5,43 as 5,43 only has two possible answers

5,43 and 1,215 (if P knew a Product of 215, he would have been able to figure out that 5,43 is the only possible answer because 1,215 is out side of the town...

[Rwilkinson]

Whoa! I had a perfect solution, until I realised that my math teacher had messed it up when he asked us this morning. We spent the whole day working solutions to the _wrong problem!_ When I got online I saw the _correct_ problem, and am in the process of working a solution.

Also, if the old site is listing the year as 19100, it's because of JavaScript. it counts years as 'years since 1900', and so, the date string probably contains:
date= '19'+year
when it should be
date=year
if (year<1900) {
date=date+1900
}
Note: this is to fix IE5, where Microsoft tried to Y2K fix JavaScript, when it is already fine.

Robert Wilkinson.

[Ghost Post]

OK, I solved this puzzle and got 4,13. I can't say that this is absolutely the only solution, since I haven't written a program to check. Hopefully it could work as a test for other possible solutions in question.


First imagine the Town as an x,y graph.

Product (P) indicates that the address is north of the diagonal from the Southwest to the Northeast corners of town.

Sum (S) indicates it cannot be along the West edge of town.

In this remaining triangle of possibilities, S's solution set falls along an unknown 45 degree diagonal line from NW to SE.

The key is that when S says, "I knew that," it tells P that ALL possible solutions along the diagonal line have multiple factor pairs. That's how S KNEW that P could not have known with certainty what the answer was. These diagonal lines sum to non-prime odd numbers+2, such as 11,17,23,27,etc. Now we've reduced the possibilities to some select diagonal stripes.

Since saying, "I knew that," gave the answer to P, it means that the product of the address exists in exactly ONE of these diagonal lines.

However, there are many addresses which only exist in one of these lines. When P said he knew the answer, S was able to determine it as well. This can only be if the solution is the ONLY address on the diagonal which does not exist in any of the other diagonal lines.

Therefore, we need to find the one diagonal whose addresses ALL have multiple pairs of factors, and exacly ONE address whose product does not exist on any other such diagonals.

The only diagonal line which has exactly 1 address whose product does not exist on any other diagonals is the line described by a sum of 17.

This logic is very difficult to translate into words, but hopefully this description is complete enough to understand.

[Ghost Post]

The solution is 4,13.
(Don't ask me why, I am tired. I solved it last week, and I admit that I had to write down a program on my PC, because trying with paper and pen I made a mistake and jumped the right sum 17)

[Ghost Post]

Help. I'm trying to understand the answer by creating my own proof. I don't understand why 2-9 doesn't work. If it is because the sum is 11 and other addresses (3-8, 4-7, etc.) also sum to 11 then why does 4-13 work when other addresses that haven't beeen eliminated (6-11, 7-10, etc.) also sum to 17?