The numbers are 4 and 13.
As BB is initially confident that they (i.e. he and CC) don't know the numbers, we can conclude that -
0) The 2 numbers are bigger than 1
1) The sum must not be expressible as sum of two primes, otherwise BB could not have been sure in advance that CC did not know the numbers.
Remember Goldbach conjecture? For our problem with such small numbers, we can assume it is ture. So from 1) we get:
3) The sum is odd.
4) The sum - 2 is not a prime.
From 3), we get:
5) One of the numbers is even, the other is odd.
Therefore, the product can be factored as 2^n*m, where m is odd.
If m is not a prime, then m=xy and x and y are odd and bigger than 1 ( x could be = y).
Then, the 2 numbers could be (2^n*x, y) or (2^n, xy). CC can't be sure which pair are the right pair.
Therefore, m is a prime.
Now, since CC knows the answer, BB should know that the product of the 2 numbers must be in the form 2^n*p.
From 5), BB knows one number is 2^n and the other is p.
From 4), BB knows that n>1.
Since BB also got the answer, that means the sum = 2^n + p is unique for n.
Now, we can use these conditions to get the answers:
(a) The sum is odd.
(b) The sum - 2 is not a prime.
(c) The sum = 2^n + p (p is prime) is unique for n, n>1
Let's see when sum=11:
11=4+7=8+3. In the case, n=2 and 3. It is not unique. Therefore, 11 is not the right sum.
The next sum is 17, since 13-2 is a prime and 15-2 is a prime.
17=4+13. Therefore, 4 and 13 are the solution. (17=8+9 and 9 is not prime)
After going thru the numbers, the next pair of right answer is (4,61).
As BB is initially confident that they (i.e. he and CC) don't know the numbers, we can conclude that -
0) The 2 numbers are bigger than 1
1) The sum must not be expressible as sum of two primes, otherwise BB could not have been sure in advance that CC did not know the numbers.
Remember Goldbach conjecture? For our problem with such small numbers, we can assume it is ture. So from 1) we get:
3) The sum is odd.
4) The sum - 2 is not a prime.
From 3), we get:
5) One of the numbers is even, the other is odd.
Therefore, the product can be factored as 2^n*m, where m is odd.
If m is not a prime, then m=xy and x and y are odd and bigger than 1 ( x could be = y).
Then, the 2 numbers could be (2^n*x, y) or (2^n, xy). CC can't be sure which pair are the right pair.
Therefore, m is a prime.
Now, since CC knows the answer, BB should know that the product of the 2 numbers must be in the form 2^n*p.
From 5), BB knows one number is 2^n and the other is p.
From 4), BB knows that n>1.
Since BB also got the answer, that means the sum = 2^n + p is unique for n.
Now, we can use these conditions to get the answers:
(a) The sum is odd.
(b) The sum - 2 is not a prime.
(c) The sum = 2^n + p (p is prime) is unique for n, n>1
Let's see when sum=11:
11=4+7=8+3. In the case, n=2 and 3. It is not unique. Therefore, 11 is not the right sum.
The next sum is 17, since 13-2 is a prime and 15-2 is a prime.
17=4+13. Therefore, 4 and 13 are the solution. (17=8+9 and 9 is not prime)
After going thru the numbers, the next pair of right answer is (4,61).
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