mathematical puzzle, maybe!?

In his phone call, A indicates that for any decomposition S = u + v of his sum S, the product uv has several decomposition namely uv = u'v'. If this were not the case A could not be sure that B has no way to crack.

B looks at all all decomposition P = ts. From his answer, we know that only one sum s + t satisfies this requirement.

Now any even S would have a decomposition as the sum of 2 odd prime numbers yielding an immediate solution for B. So that's excluded and S must be odd.

For the same reason S cannot be of the form p + 2 with p prime. so S is of the form c + 2 with c composite odd.

So the first possibilities for S are 11,17,23,27,29,35,37, 41, 47, 51, 53, 57, 59, 65, 67, 71 etc call this list L

It is the task of whoever divised this problem to show that it has only one solution. Ours is just to find one.

To be admissible, the sum S must be such that of all products P_1, P_2, P_3 associated to it, only one has the property that allows B to conclude.

A product P is "good" if of all the sums are out of L except just one, which will be selected by B.

A sum is excluded as soon as it is associated to at least 2 "good" products since in this case A would not be able to finish.

S = 11 ?

P_1 = 18 2*9 --->11 6*3 ---> 9 since only 11 is in L, P_1 is good

P_2 = 24 3*8 --->11 other products would give even sums. So P_2 is good and so S is not equal to 11.

S = 17

P_1 = 30 2*15 ---> 17 6*5 ---> 11 11 and 17 are int the list P_1 is bad

P_2 = 42 3*14 --->17 6*7 ---> 13, 21*2 ---> 23 17,23 in L, P_2 bad

P_3 = 52 4*13 ---> 17 so P_3 is good

P_4 = 60 5*12 ---> 17 15*4---> 19 3*20 ---> 23 17,23 in L, P_4 bad

P_5 = 66 6*11---> 17, 2*33 ---> 35 so P_5 is bad.

P_6 = 70 7*10---> 17 2*35 ---> 37 so P_6 is bad

P_7 = 72 8*9 --->17. 24*3 ---> 27 so P_7 is bad.

Finally S = 17 and P = 52 so (x,y) = (4,13)

Let's summarize. B wonders if P = 52 corresponds to 2*26 or 4*13. When he hears that A knows for sure he is has not found x and y, then he knows

S is not equal to 28 coz if that were the case A could not be sure that P is not 11*17 = 187 in which case B would be done. So P knows that S = 17.

A knowing that can eliminate all the bad products and finds 52 for himself.

One of the answer is 1 and 8. The sum is 9 and the prod is 8

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Here's the logic.

A and B don't know the numbers. That means B doesn't have prime number as product. A says that B can't have his number with just his prod because A knows this. If the product is prime number, then it's definite that the sum is p+1. Using this statement, we get if the sum isn't p+1, it's definite that the product isn't prime number. If the sum is p+1, A will still get confused whether B knows or not.

So the sum isn't p+1. Now B also knows this from A's statement. And B gets the number. That means the product B has only has two possibilities:

- one or more possibilities resulting in sum in p+1 form

- exactly one possibility resulting in other form of sum

That means B has a composite number with those natures.

Now A found out the number. That means, with the sum A has, there is only one product with those natures. It seems likely that the more sum A has, such possible product with such nature will increase, but I haven't managed to prove it or set the limit yet. So far I only searched the composite numbers from 1 to 10.

x=2 and y=2

Hawking Radiation is not due to tunneling. It arises because of vacuum fluctuations outside the event horizon. We know these occur because they can be measured in a lab - the Casimir Effect is a famous example. In a vacuum fluctuation, particle-antiparticle pairs arise spontaneously but have life times that mean dEdT <= h/2pi. Normally the particle pairs are unobservable directly (they are virtual), but at the event horizon it is possible for one of the pair to be captured by the black hole. The main objection to the idea is its impact on entropy. In effect, a know particle falling into the event horizon tells us something about the entropy of the black hole (conveys information) and this should not be possible. Your friends is totally wrong re the LHC - he clearly has no clue. The supposed risk of the LHC is that it creates densities large enough to create a Schwartzchild metric (black hole) locally, but this would be infinitely tiny. The same math that shows this could happen also shows it would evaporate almost instantly - hence no risk.

x = 2, y = 2 ANSWER

A was given 2 + 2 = 4

B was given 2*2 = 4

x=2, y=2 BAH! xD

in z^2 -(x+y)z +xy=0 the roots are (x+y) & xy

God bless you.

hmmmmmmmmmm