# Two mathematicians..

#### triggernum5

Joined May 4, 2008
216
If you've seen this before, or find a solution on the net, kindly refrain from posting it until honest attempts have failed..

Two numbers (x and y) between 2 and 100 are selected.. These numbers are unknown to the mathematicians.. The sum (x+y) is given to one mathematician, while the product (x*y) is given to the second mathematician..
The second mathematician looks at the product he was given and says to the other "Hmm, I have no idea what your sum could be.."
The first mathematician replies "Thats no news to me, I already knew you could not deduce the sum I was given.."
And to that the second mathematician replies "Ah ha, now I know what your sum is!"
And the first mathematician responds, "And likewise I have now deduced your product.."

What are the two numbers x and y?

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#### triggernum5

Joined May 4, 2008
216
If you're stuck at how to aproach this, think about sums/products they could have been given instead, that could give them insight to the other mathematician's number..

#### Dave

Joined Nov 17, 2003
6,970
I remember this from years ago. Is it something to do with prime numbers? (Sorry if that ruins the trick!)

Dave

#### triggernum5

Joined May 4, 2008
216
Far from ruins the trick.. I was time displaced from this problem.. I remembered the numbers, and the basic logic of it, and it still took a couple hours of pondering to fully make sense of it in my head..
Primes, and what I call 'almost primes' like 15 are key to eliminating impossible combinations..

#### triggernum5

Joined May 4, 2008
216
I'll start by pointing out that for this to work, both mathematicians assertions rely on absolute confidence that neither one of them overlooked anything, or made any error of any kind..
I mentioned thinking about sums/products that could give insight, and 'almost primes' such as 15.. Aside from 1 and itself, 15 only has 2 factors 3 and 5.. Therefore if the second mathematician was given 15 as his product, he would know that the first mathematician's sum was 8, and his very first assertion would be false..

#### bloguetronica

Joined Apr 27, 2007
1,372
If both numbers are 2, the sum is four and the product is 4. In that case it is easy to deduce the numbers.

If we have a 2 and a 3. The sum is 5 (easy to deduce), and the product is 6 (as well).

I think I see a pattern here... If the product is the result of two primes, you can deduce the sum, because primes are not divisible by other numbers except one and themselves, and thus there is only one combination of multiplier and multiplied (actually two if you consider the order).

Now discovering the product from the sum is difficult. For example, if the sum is 6, it can be 2 + 4 or 3 + 3. So, the only possibility I see is that the other mathematician says that he discovered the sum. So the first mathematician would think "Ok, he discovered my sum...how can he be so sure... Ah ah, they must be primes, or else he wouldn't be so sure about that!". I think that narrows the possibilities.
Adding to that, if the mathematician knows that the numbers are primes (because the other mathematician suggested it), if the sum is an odd number, he automatically knows that the number 2 is involved and can deduce the other number. If the sum is an even number, he must try to figure what primes are involved, and that can be difficult.

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#### triggernum5

Joined May 4, 2008
216
You are on the right track.. The numbers are unique btw if I didn't clarify that..
One hint I will give, is that each mathematician will need to run 'a bunch' of possible calculations through their heads quickly, and flawlessly to make their assertions, and they have the advantage of atleast knowing the sum/product..
To save grief with wasted calculations I'll narrow it down and say that neither number is greater than 25..

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#### triggernum5

Joined May 4, 2008
216
"he must try to figure what primes are involved, and that can be difficult."

Its better to say he needs to figure out what non-primes are involved.. Maybe not though, depends on your thought process..

#### bloguetronica

Joined Apr 27, 2007
1,372
You are on the right track.. The numbers are unique btw if I didn't clarify that..
One hint I will give, is that each mathematician will need to run 'a bunch' of possible calculations through their heads quickly, and flawlessly to make their assertions, and they have the advantage of atleast knowing the sum/product..
To save grief with wasted calculations I'll narrow it down and say that neither number is greater than 20..
As far as I can see, there are still plenty of combinations that satisfy the conditions. However, we have to exclude obvious combinations like 2 and 2 or 2 and 3 that would led the mathematician to discover the product without any tip. And we have to exclude obvious combinations that would led the other mathematician to discover the sum, obviously. Thus, combinations like 3 and 3 are excluded.
I'm guessing that the product has to be an odd number and the sum has to be even. Combinations with 2 are to be discarded, because their product can easily lead to the numbers case the other is a prime, or lead to no conclusion if the other number is not prime. Combinations with 5 seem to be quite obvious too, for the same reason (since the products between 5 and odd numbers, like the most of the primes, end with a 5). Combinations with 3 are a little harder, but can be discovered easily. Combinations of equal numbers should be discarded too, because they are easy to break.

Of course I'm supporting my theory on the dialogues. I would pick the largest primes possible to make the calculations difficult. Two different numbers like 19 and 17.

#### triggernum5

Joined May 4, 2008
216
Again, for this puzzle, the 2 numbers are unique.. I don't know if a solution exists using 2 identical numbers or not, but these are unique..
Think about the order the statements are made, and 'any' insight each statement can give based upon the claims before it.. Let me be the mathematician holding the sum, and imagine my sum is 23..
23= ->
21+2
20+3
19+4
18+5
17+6
16+7
15+8
14+9
13+10
12+11
(and then the same, but reversed)
None of those is a sum composed of 2 primes, so 23 is special in the sense that the first mathematician can indeed make the claim that he was positive his sum could not be deduced if he ran through all possible sums.. On the otherhand, if he was holding say 8 as his sum, he could not be sure his secret was safe because he'd be aware of the possibility that the other guy's product was 15 (in which case the sum would be deduced)
Like I said, both mathematicians have to run the numbers flawlessly, and have absolute confidence that the other mathematician is absolutely flawless.. If either one made any mistake then the logic collapses..

#### bloguetronica

Joined Apr 27, 2007
1,372
Like I said, both mathematicians have to run the numbers flawlessly, and have absolute confidence that the other mathematician is absolutely flawless.. If either one made any mistake then the logic collapses..
Exactly.

I just had a flash. The sum has to be a unique combination between two primes. If there is another combination of two primes giving the same sum result, the combination is to be discarded (uncertainty). 2 + 2, 2 + 3, 2 + 5, 3 + 3 and 3 + 5 are unique combinations, but too obvious. Besides, combinations with 2 and 5 are to be discarded, because the products they give are too obvious. The solution is to trial and error other combinations:
3 + 7 = 5 + 5
3 + 11 = 7 + 7
3 + 13 = 5 + 11
3 + 17 = 7 + 13
And so on...

I will do this on paper and I tell you when I find an unique combination.

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#### triggernum5

Joined May 4, 2008
216
Maybe I'm confused, but it seems like you're progressing backwards (or just misspoke)..

"I just had a flash. The sum has to be a unique combination between two primes. If there is another combination of two primes giving the same sum result, the combination is to be discarded (uncertainty)."

That statement is completely incorrect.. So incorrect, that its inverse statement is absolutely correct..

#### bloguetronica

Joined Apr 27, 2007
1,372
That statement is completely incorrect.. So incorrect, that its inverse statement is absolutely correct..
So, how the mathematician would be so sure about the product?

One says: Ahah, I've discovered your sum. The other thinks they must be primes (he only knows that the numbers are primes, but doesn't know the numbers). For him to be so sure as well, the sum has to be a unique combination of primes. The same sum cannot give two combinations, or else he wouldn't be sure.

Of course, I'm not ordering the trial and error tries by their sums. I think that might raise some confusion.

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#### bloguetronica

Joined Apr 27, 2007
1,372
I've found what numbers are: 5 and 7. I just took the logic I was following.

It seems we should consider the prime 5 , because after all it seems that the thinking of the mathematician that discovered the sum is not that fast. It took me a bit to try that. I was following a subjective assumption and precipitating myself.

#### triggernum5

Joined May 4, 2008
216
Those numbers are not correct..
Using those, the sum would be 12, and the product 35..
But here is what I mean my you being on the opposite path.. If the mathematician was holding 12 as his sum, then he would need to account for the possibility that the 2 numbers were infact 5 and 7.. If that was the case, the other mathematician would be holding 35 as the product, and 35 is one of those 'almost primes'.. The only combination that multiplies to 35 is 5 and 7, so if those were the numbers, the sum would obviously be deduced to equal 12..
'Because' 12 'can' be expressed as the sum of two primes, it needs to be eliminated as a possible sum if the statements are accurate..

#### studiot

Joined Nov 9, 2007
4,998
Cume, maybe this clue will help:

The second mathematician looks at the product he was given and says to the other "Hmm, I have no idea what your sum could be.."
The guy with the product is the first to speak.

If the number he was given can only be obtained in one way he would not have said the above quote

eg given 35 as the product, this can only be obtained from 5 x 7.

So he must have been given a number such as 60

which is either

4 x 15

or

5 x 12

But he doesn't know which and says so.

#### triggernum5

Joined May 4, 2008
216
I think steps one and two have been covered pretty well..
1. The product has more than two factors besides 1 and itself..
2. Its been determined that a set of numbers we are interested in is the set of numbers within parameters that cannot be expressed as the sum of 2 primes.. Remember that x<>y..
No2PrimeSums = [11, 17, 23, 27, 29, 35, ...]

I may have goofed in that set, I just ran it through my skull.. I'll check it over and expand it when I'm sharper..

Now, the next statement is made by the mathematician holding the product:
"Ah ha, now I know what your sum is!"
We need to figure out the logistics that make this actually possible..
(Hint: Its easier for him because he actually knew his product from the start, and calculation requirements to deduce the sum were reduced by that knowledge..)

#### bloguetronica

Joined Apr 27, 2007
1,372
I call it quits.

#### triggernum5

Joined May 4, 2008
216
You did figure out the major component of the question.. Well you had it, then got it turned around in your head, (same thing happened to me when I was solving it)..
I'll let this sit for a while..
Want me to PM you the solution?

#### Caveman

Joined Apr 15, 2008
471
Frankly, I figured out how to get the answer long ago, but just didn't want to work through all of the possibilities. It just takes so long to check them all. And I was using Excel.