[MortenS]

This is a well known mathematical puzzle. I do know the answer, but I am struggling a bit with arriving at the answer.

The problem can be specified like this.

There are two unknown whole numbers, m and n, both greater than 1, and less than 100. One mathematician, Mr. Product is given the product of these two numbers, while another mathematician, Mr. Sum is given the sum of these two numbers.

The following conversation takes place:

Mr. Product: I do not know the numbers.
Mr. Sum: I knew you didn't knew the numbers.
Mr. Product: Now I know the numbers
Mr. Sum: Now I know the numbers, too.

What are the numbers?

From Mr. Product's statement "I do not know the numbers", we can deduce that the product is not the product of two primes. If it were, then Product would have been able to factorize the product into two primes, and would then know the two numbers.

From Mr. Sum's statement: "I knew you didn't knew the numbers" we can deduce that the sum must be an odd number, because every even number (at least for small numbers) can be written as the sum of two primes (Goldbach's Conjecture). The only way for S to know that Mr. Product doesn't know the numbers, is for the sum to be an odd number.

From Mr. Product's statement: "Now I know the numbers", we can deduce that the information that the sum is an odd number helps Mr. Product to find the correct two numbers.

From Mr. Sum's statement: "Now I know the numbers, too", we can deduce that the fact that Mr. Product now knows the numbers, is enough information for Mr. Sum to find the correct two numbers.

A summary of what we know:

P1: m and n are not both primes
P2: the sum m + n is an odd number, so one of the numbers is odd, the other number is even.

I am a bit stumped on how to proceed from here. There are still lot of number pairs left to eliminate until a unique solution presents itself. What other deductions can be made from the sentences that Mr. Product and Mr. Sum says?

[C1ay]

MortenS said:

This is a well known mathematical puzzle. I do know the answer

So you know the numbers are 4 and 13

[MortenS]

I do know what the answer is supposed to be yes, I just can't seem to arrive at it myself...4 and 13 do fulfill the criteria I have deduced so far, but as far as I can see, a lot of other pairs also fulfill those criteria, unless I can specify some other criteria that can eliminate them.

[C1ay]

If I remember this problem correctly I believe your wording is wrong. I think it is given that numbers X and Y are greater than 1 and their sum is less than 100. As you stated, one mathematician is given the product of the 2 numbers and another the sum. The conversation then goes as you said. This set of conditions will yield more clues about the numbers than you have stated. Do you see a difference now?

[MortenS]

I am pretty certain that the problem specifies the unknown numbers to be between 1 and 100, and that it says nothing regarding the product and the sum (at least the version of the problem that I have got). If the sum is less than 100, it has to be derived from the conversation somehow.

Here is one presentation of the problem (and very convenient, the link to the solution does not work)


On the other hand, if the numbers are between 1 and 100 (from 2 to 99), I think we can state that any prime factor of each of the numbers must be less than 50. If a prime factor of any of the numbers is greater than 50, mr. product could easily state that the prime factor itself, was one of the unknown numbers (otherwise one of the numbers would have to be greater than 100). This means that we can rule out the primes from 53 and up to 97 as m and n.

[C1ay]

Perhaps this discussion will help you.

[MortenS]

Some tips there yes, but there is this assumption that the sum is below 100, that is not part of the original problem. I cannot make that assumption, without any reason for it.


There are so many different versions of this puzzle, just take a look at some of the various variants here:

http://www.mathemati...gic_sum_product

The version I am interested in solving is the one I specified, perhaps with the added clarification to S's first statement:

P: I don't know the numbers.
S: I knew you didn't knew the numbes. I don't know the numbers either.
P: Now I know the numbers.
S: Now I know the two numbers too.

[C1ay]

MortenS said:

Some tips there yes, but there is this assumption that the sum is below 100, that is not part of the original problem. I cannot make that assumption, without any reason for it.

The solution still works but the sumlist is longer since it is not limited to sums less than 100. It still amounts to being the only set in the list where the sum is unique. For all lines where the sum appears more than once in the list, Dr. P could not with certainty which number pair it is.

Here's another page discussing the solution.

[maddog]

From what I've read on all of y'alls websites is that putting a boundry on the sum to a 100 is a special
case to that of having both within the range of 100. Yes, the clarification that Mr S admits he didn't
know either does change the problem a bit. The two programs found within the newsgroup thread
submitted by MortenS and the website submitted by C1ay both seem to solve the problem. I am still
struggling with why.

I am still having problems at eliminating sums that don't work. For example why do the sums of 9 & 13
not work ? Ahh...

Since you say that S must not be the sum of primes or Mr. P would know the answer?

So 2 + 7 = 9 & 2 + 11 = 13.

However, for 9: 4 + 5 = 9, 3 + 6 = 9; for 13: 3 + 10 = 4 + 9 = 5 + 8 = 6 + 7.

I guess Number Theory is not my bag... I am intrigued by the problem...

Maddog

[maddog]

I did work through this a bit more and concluded that the examples I used earlier won't work since these
could be written as the sum of two primes. I going to dive into the two programs and see if I can get
further. Maybe I can deduce more from the algorithm.

Maddog

[MortenS]

I have been thinking. Isn't it possible to rule out all numbers that form products consisting of 4 or more prime factors? P can only eliminate between 2 sets of numbers, when one of the sums is even and the other is odd. This will happen if there are exactly three prime factors in the product, and one of them is 2.

This should eliminate some other number pairs, should it not?