An old puzzle about two numbers
 

Two integers are chosen. It is known than both of them are greater than 1 and less than 100. It's also known that Mersenne was informed of their sum, while Fermat was informed of their product. So, the following conversation between them took place:

Fermat: Of course you know that I'm not able to find the numbers chosen...
Mersenne: You are right. And I can't find them too...
Fermat: But I've just found them!

What numbers were chosen, and how did Mersenne and Fermat draw their conclusions?

Of course both of them were honest :smile:

I assume that only one pair of numbers have a product which
can be formed in more than one way and the sum of the two numbers
is the same for more than one case. Does this make sense?
I'll try it in Basic.

David

Lets say the two numbers are a, b, with a<b.
The first statement means the two numbers a,b are not both prime, their product is not the cube of a prime, and for at least two k|a*b with 1<k<100, a*b/k is < 100. I.e., a*b wasn't 2*31*59, as that would only allow a=59, b=62.
The second statement means that the sum Mersenne was given can be written in more than one way as a+b with such a,b. Also, the sum isn't 198 or 197.

It may be possible to solve it from there with an exhaustive search.

Alex

[Spoiler]
The first numbers that spring to mind are 4 and 3. Giving the product 12 and the sum 7.

12 can be expressed as 4*3 or 6*2 so Fermat could not define them.
7 can be expressed as 4+3 or 5+2 likewise for Mersenne.
The statement did say the numbers where greater than 1 and other numbers would not give such a simple addition.

[/Spoiler]

I have not thought this completely through but it seems to give a quick solution.

Regards
Patrick

But Fermat said he *knew* that Mersenne knows that Fermat can't figure out the numbers. If Fermat had been given the product 12, he would have known that Mersenne must have been given the sum 7 or 8. With the sum 7, Mersenne could not have known for certain that Fermat was stuck, as it might have admitted a=2, b=5, which Fermat could have solved...

Nice little mind twister.

Alex

So I guess the sum a+b was an odd number, as any even number can (conjecturally) be written as the sum of two primes, and a+b-2 must not be prime.

Alex

Ah! I see what you mean. I thought it was too easy.

Regards
Patrick

[b]akruppa[/b] is on the right way :)

[Spoiler]
The two numbers are 9 and 21. Product 189 - Sum 30.
[/Spoiler]

Taking Alex's lead, I used Excel and filtered out:
All Evens.
All primes.
All products of two primes.
All numbers where n-2 was prime.

I then factorized the remaining numbers and created a permutation of all the ways you can sum the numbers. There where 2 products that had the same sum with one being a cube. This is all Fermat needed to know to solve this.

Regards
Patrick

The answer (9, 21) is wrong :)

I saw this puzzle on the net and the following line is included:

Both are unequal to 1 and the sum of them is less than 100.

This is different than XYYXF's statement. I don't know if this is supposed to be part of this puzzle or not.