Two mathematicians..

Thread Starter

triggernum5

Joined May 4, 2008
216
If your strategy is tight then there actually aren't that many calculations..
Can you simply explain how it could be possible for the second mathematician to deduce the sum using the info already stated?
 

Caveman

Joined Apr 15, 2008
471
Mathematician 2 can figure out the answer because only one pair of all of the pairs of numbers that add to his sum has a product that has only one factorization in which the sum of the factors has the property that the product of all pairs of numbers that add to this sum have multiple factorizations. :eek:
 

Caveman

Joined Apr 15, 2008
471
A little more clearly:

First, to reduce typing: M1 = mathematician 1, M2 = mathematician 2.
SUM = the sum given, PRODUCT = the product given.
Now let's define some shorthand:
P2S(x) is the set of possible sums that M2 could have if M1 is given a product x. Basically, you factor x and for every factoring add the results.
S2P(x) is the set of possible products that M1 could have if M2 is given a sum x.

So they can each predict what the other may have. Note that if you look at P2S(y) for every y in S2P(x), you will generally get x plus some other numbers that aren't correct, so you can't generally get very far with this.

But these guys are talking, so they are giving more information. The first piece of information is from M2. He says that he knows that M1 cannot figure out his sum. I'll call this R1(x), and it is true if P2S(y) has more than one element for every y in S2P(x). M2 is stating simply that R1(SUM) is true.

M1 picks up on this and does his own math. When he announces that he has the answer, he gives M2 some information. I'll call this R2(x), which is true if R1(y) is true for exactly one y in P2S (x). M1 was stating that R2(PRODUCT) is true.

Now when M2 states that he has the answer, he is giving us some information. That the puzzle is solvable. ;)
He is saying that if he looks at all x in S2P(SUM) that for only one x is R2(x) true.
 

Thread Starter

triggernum5

Joined May 4, 2008
216
Wow, thats impressive for a man who names himself after a stereotypically stupid homonid:).. I'll bring it up to the final step in a bit incase anybody wants to crunch some numbers.. I'm currently a few drinks past sober, so if you jumped an assumption I may not have caught it, but that sounds pretty smack on, and as far as I know, a description like that of this problem doesn't exist on the internet.. There are solutions, with descriptions, but most are cheater solutions since they seeem to deduce the numbers, yet only state parts of the conditions..

" Now when M2 states that he has the answer, he is giving us some information. That the puzzle is solvable. ;)
He is saying that if he looks at all x in S2P(SUM) that for only one x is R2(x) true."

Final part of the question:
How does the second mathematician's claim that he has deduced the first's sum allow the first mathematician to deduce the product?
 

Caveman

Joined Apr 15, 2008
471
Couple of finer points.
1. I think I may have swapped mathematicians, so in my posting M1 has the product, while M2 has the sum.
2. In either case the mathematician with the product figures out X and Y first. With this he can calcuate the sum.
3. Then the mathematician with the sum figures out X and Y, and can calculate the product.
4. I made one mistake in the answer I gave. Technically, when the mathematician with the product finds the answer, it doesn't necessarily mean the puzzle can be solved. It just means that he can solve it. We have it harder. We have to prove that for only one product out of all of possible products can this situation arise. It is not proven, but is an assumption of it being called a puzzle.

As to how to solve it as a puzzle (ie. I'm not one of the mathematicians), I would go about it differently. Basically, I would need to reduce the possible numbers first by looking at the statements, then look at ways to remove possible combinations.
 

Thread Starter

triggernum5

Joined May 4, 2008
216
The fact that the 2nd mathematician's deducing the sum, allowed the first to deduce the sum means that 'with the numbers (sum/product) they were given' the problem is solvable.. It says nothing more than that as far as I can determine..
If you can elaborate correctly on your point (3) then I think we can consider the problem solved.. I personally really hit a wall here.. Its almost as simple as it seems from this point, but not quite..
Actually, my difficulty may have arose because I knew the answer prior to working out the solution.. I had jumped the gun on the logic..
Essentially this is the conversation so far:
Second Mathematician: "My product has too many factors for me to determine your sum.."
First Mathematician: "I ran all possible combinations, and already determined your product had too many factors to determine my sum.."
Second mathematician: "Oh yea, well in that case, Based on the product I'm looking at, I've also already determined you could only have one possible sum that guarantees its secrecy from me, and therefore I know what it is.."
How does the first mathematician deduce the product?
 

Caveman

Joined Apr 15, 2008
471
It's not quite like that (and this is exactly what I said earlier):
Second mathematician: My product has too many factors for me to determine your sum.
First mathematician: I ran all possible combinations with my sum, and already determined your product had to many factors to determine my sum.."
Second mathematician: Oh yea, well in that case, based on the factors of the product I'm looking at, only the sum of one of these factor pairs would guarantee its secrecy from me, so these factors are the number.
First mathematician: Looking at the products of all possible pairs of numbers that have my sum, I see again that they all have multiple factorizations, but I also notice that only one of them has the property you just used to find your number. Of course, knowing the product and sum now, I can tell you the numbers.
 

Thread Starter

triggernum5

Joined May 4, 2008
216
"... Looking at the products of all possible pairs of numbers that have my sum, I see again that they all have multiple factorizations, I also notice that only one of them has the property you just used to find your number. Of course, knowing the product and sum now, I can tell you the numbers. "

Your ammendments to my version of the conversation are concise clarifications of points I felt were implied.. But this part may be backwards progress, or perhaps jumping to a conclusion..
The property the second mathematician used to deduce the sum was the fact that only one sum of two of its factors which multiply to itself could not be expressed as the sum of two primes..

"I also notice that only one of them has the property you just used to find your number. "

Hint: Using the actual numbers in the solution this statement is 100% incorrect..
^^
Huge hint..
 

Caveman

Joined Apr 15, 2008
471
The property the second mathematician used to deduce the sum was the fact that only one sum of two of its factors which multiply to itself could not be expressed as the sum of two primes..
The second mathematician took his product, got all factor pairs, added each pair together. Now he has a bunch of possible sums that the first mathematician could have. He looks at each of these numbers to see if any of them can be expressed as the sum of two primes. The only way he can figure it out is if only one of these possible sums has this property.

The first mathematician then takes his sum, gets all possible pairs, multiplies them together. Now he has a bunch of possible products that the second mathematician could have. He looks at each of these product, gets all factor pairs, and adds them together. He looks at each of the sums to see if any of them can be expressed as the sum of two primes. He is looking for a possible product that has only one sum with this property. Because this is a puzzle with only one solution, only one product has this property.
 

Thread Starter

triggernum5

Joined May 4, 2008
216
"He is looking for a possible product that has only one sum with this property."

Thats the nugget I was looking for..

The numbers are 13 and 4.. The nugget was important because the sum 17 yeild products 14*3 = 42, and 13*4 = 52..
Before the second mathematician announces he has deduced the sum, the first cannot narrow it down beyond these two possibilities.. But 52 yields only one possible sum of its factors that cannot be expressed as the sum of two primes (13+4=17).. 42 on the otherhand yields two possible sums that cannot be expressed as the sum of two primes (14+3=17, 21+2=23)..
Therefore the first mathematician deduced the product because he knew that if the product was 42, the second mathematician could not have deduced the sum..
Thats pretty impressive Caveman!
 

Caveman

Joined Apr 15, 2008
471
From the point of the mathematician with the sum of 17:
Sum is 17, so all possible products the other guy could have are:
XY X*Y
2 + 15 => 2*15 = 30
3 + 14 => 3*14 = 42
4 + 13 => 4*13 = 52
5 + 12 => 5*12 = 60
6 + 11 => 6*11 = 66
7 + 10 => 7*10 = 70
8 + 9 => 8*9 = 72

Now I have to assume the logic that he would use, so:
If he had a product of 30, he would look at all of his possible factors.
Then he would look at the possible sums that I could have from that.
Then he would look at each sum, and determine the possibility of getting the same sum by adding two primes.
For his product, he found that exactly one of his factorizations had this property.

30
--
2*15 => 2+15 = 17 = 4+13 = 5+12 = 7 + 10, so 17 works.
3*10 => 3+10 = 13
5*6 => 5+6 = 11 = 9+2 = 8+3 = 7+4 = 6+5, so 11 works.

There are two possibilities that work, so his product sum is not 30.
----
42
2*21 => 2+21 = 23 = 19+4 = 17+6 = 13+10, so it works
3*14 => 3+14 = 17, works (above)
6*7
At least two possibilities, so his product is not 42
---
52
2*26 => 2+26 = 28 = 23+5
4*13 => 4+13 = 17, works (above)
Has only one possibility... but just to be sure...
---
60
2*30
3*20 => 3+20 = 23, works (above)
4*15
5*12 => 5+12 = 17, works (above)
6*10
At least two possibilities, so the sum is not 60
---
66
2*33 => 2+33 = 35 = 31+4 = 29+6 = 23+12 = 19+16, so it works
3*22
6*11 => 6+11 = 17 works (above)
Two possibilities, so the sum is not 66
---
70
2*35 => 2+35 = 37 = 31+6 = 29+8 = 23+14 = 19+18, so it works
5*14
7*10 => 7+10 = 17 works (above)
Two possibilities, so the sum is not 70
---
72
2*36
3*24 => 3+24 = 27 = 23+4 = 19+8 = 17+10 = 13+14, so it works
4*18
6*12
8*9 => 8+9 = 17 works (above)
Two possibilities, so the sum is not 72

The answer must be 4*13 = 52

As involved as this mathematicians logic is, it is even harder to find the information if you don't know his sum. The logic is the same, but there are a lot more numbers to look into. There are other ways though...
 

bloguetronica

Joined Apr 27, 2007
1,372
...
The answer must be 4*13 = 52
...
It seems that you are following a wrong logic here. Given the product of 52, you can have two different sums: 2 + 26 or 4 + 13. Thus it is impossible to be certain about the sum. Keep in mind that this is the mathematician who first discovered the numbers and gave the tip. He had to be sure that the numbers were those and only those.

It seems that you are assuming that the sum must be a prime, but, unless you explain me why, that doesn't make sense.
 

Thread Starter

triggernum5

Joined May 4, 2008
216
No, as far as I can tell, Caveman is right on the money.. The second mathematician already knew that the sum was not (26+2=28), because it that was the case, the first mathematician could not righteously make his initial claim:
"Thats no news to me, I already knew you could not deduce my sum.."
If his sum was 28, he would have to accept the possibility that the numbers could be 23 and 5, which are two primes..
If those were the numbers, the second mathematician would have looked at his product and said "115 only has 2 factors, so your sum is 23 + 5 = 28..
This is the part I love.. The follow-up debates.. Yep, I'm a massochist..:)
 
Hi,

It can't be solved
first: you did not specified if the numbers are integers.

If the number are integers, there is not way on mathematician can be sure of each others numbers.

each on has one equation with one unknown
the solution to the problem requires two equations.

I may be wrong!

Let me know where my deduction is wrong.

Myidismine,
 
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