Fun Math Problem?

Question

Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers?

So far none correct.

Not Easily Fooled is correct with an answer of 4 and 13.

Answer 1

not enough information.plz reframe question as there r many possibilities

Answer 2

X=3; Y=4

Answer 3

If x and y are prime numbers, then the value xy tells you what the two numbers are. For instance, xy=21, then x=3 and y=7. Since P can't tell what the two numbers are by looking at the product, then either x or y or both is not prime.

What are the cases when S could determine what the two numbers are, given 1 < x < y ? 2+3=5 is one such case. 2+4=6 is another. In both cases, P could tell by the product what x and y are. So, the answer is something larger than that.

(2,5) are prime, so that's not it. The next possibility is (2,6).

x+y=8 and xy=12 both have multiple answers x+y=8 could be (2,6) or (3,5) while xy=12 could be (2,6) or (3,4). Since both are ambiguous, it holds that neither could know at first.

If x+y=8, S would not know. And if xy=12, P would not know. By knowing that the numbers are not prime, S would then know the answer was (2,6).

Answer 4

x=48
y=49

x+y=97
xy=2352

there are two solutions for 2352, so P is unable to know the solution.

there is no unique solution for any of the combinations adding up to 97, so S knows that P cannot know the solution.

once s tells p that she knew that p could not determine the numbers, p knows that of the other possible solutions for 2352, only 42*56 has a sum below 100 (42+56=98). All other possible xy solutions have x+y values above 100. If the values were 42,56, then S could not have known for sure that P cannot determine the solution because there are multiple unique possible solutions for x+y values of 98 (11,87; 19,79; 31,67; 37,61).

Since there are only two possible solutions for 2352 that meet the x+y,100 constraint, once P knows that S knows that P could not have known the solution, it narrows down the possibilites from the two solutions of 42,56 and 48,49 down to the only possible solution of 48,49.

Before this point, S only knows that the sum is 97, and that there are no unique solutions for 97. Once S knows that P can get the solution from knowing that S knows that P doesnt know the solution, S knows that the components of the 97 must be 48 and 49 because there are no other components that have no unique solutions that would also give P enough information to positively narrow down the solutions based only on S's statement that she knows that P cannot determine the solution.

i would imagine that there could be other solutions, but the trick here is to have numbers large enough that the one extra piece of information (x+y<100) comes into play. this narrows down the possibilities substantially. for smaller numbers, there are multiple solutions so p cannot know the solution just by knowing that s knew that p could not determine the solution.

Answer 5

The correct values are 4 and 13, with a sum of 17 and a product of 52. It took about two hours manipulating a spreadsheet to narrow it down to this choice,

Basically, you have to determine which sums would have no components that can be uniquely factored. These are 11, 17, 23, 27, 29, 35, 37, 41, 47 and 53.

Next, you have to determine which products can only be derived from components of one and only one of these sums. There are 86 different products which meet this test.

Finally, you have to determine which of the 86 is the ONLY match for one of the sums, because if there are multiple matches, S cannot make his final statement.

Steve's answer (above) is incorrect. If it were 2,6, the sum would be 8. 5 and 3 also sum to 8, so S would not know that P could not determine them prior to his first statement.

Answer 6

It's easy to see that all the answers that are given so far are wrong. Right now I'm considering the numbers (3,32). It holds up to the first three statements, but it takes a while to check the fourth one. Kinda a pain, not seeing an easier way right now.


To the person who thinks "there are too many possibilities," please give some of these.

Answer 7

exciting difficulty! properly 7*2 = 14 legs for the females. 7*7 = 40 9 (the variety of cats) * 4 = 196 cat legs There are 196 more effective legs for the little cats. There are 406 legs assuming that between the females is illegally utilising the bus (or that no bus motive force is on the bus). There are 408 such because the bus motive force.

Answer 8

I keep coming back to the conclusion that more information needs to be given about the relationship between S and P. I will be curious to see the answer to this problem!

Answer 9

51 and 49

Answer 10

1 AND100