hard math riddle: two numbers between 1 and 99

[VSam]

Hello Ronald,

you are right. Actual problem here is, the man with the sum would have known the numbers right away when the man with product said he didn't know the numbers.

[Loki]

the answer is 2 and 2.

why? let me explain.
The product is 2X2=4 and that would make the sum 2+2=4
There is nothing in this word problem that says each number has to be individual and clearly the sum and product of these equate to the same answer. Therefore, the person with the sum and the person of the product would end up with the same answers. Look at how it is written and you'll see that 2 and 2 is the answer.



a man comes up with two numbers between 1 and 99 (meaning not 1 or 99 but from 2 to 98). The sum of these two numbers is under 99. He tells one person the sum of these numbers and another person the product of these numbers. The person with the product comes up to the person with the sum and says "i don't know the two numbers". The person with the sum replies with "i know that you don't know the two numbers". Immediately, the man with the product says "now I know the two numbers" and the man with the sum says "now I know the two numbers too". What are the two numbers? There is no word play involved.. it is a real problem with a real answer. Feel free to find the answer any way possible.

[rnast]

It is not 2 and 2, Product guy would know his number immediately since 2 and 2 are the only valid factors for 4. ((2,2),(4,1)) and "1" is not a valid choice.


[quote=Loki;40073]the answer is 2 and 2.

why? let me explain.
The product is 2X2=4 and that would make the sum 2+2=4
There is nothing in this word problem that says each number has to be individual and clearly the sum and product of these equate to the same answer. Therefore, the person with the sum and the person of the product would end up with the same answers. Look at how it is written and you'll see that 2 and 2 is the answer.

[Licia]

In my opinion it’s 25 and 50

As it was said, If the product guy don’t know the solution, Then
numbers both can’t be prime simultaneously (but one of them still can)

Аs the sum guy knows that product guy can’t know the solution, then there sum must be odd
(if the sum will be even then numbers can be prime and there is an opportunity that P guy knows the answer)

If the sum is odd then one of the numbers must be odd (let it be a) and the other is even (call it b)
Then the product will be a*b or 2*a*(b/2)
The P guy must know the separate numbers. But there can be two pairs from the last equation.
He can choose pair 2*a and b/2, or a and b. And there is only one point when these pairs will be the same thing – when a=b/2. Then b=2a and their sum will be
a+b=3a<99
a<33
And we already know, that it can’t be even (because 2a is even already)
So there is an odd row from 3 to 31
3 5 7 9 11 13 15 17 19 21 23 25 27 29 31

a can’t be prime Because if it will be than P guy will know the solution without asking S guy
so it leaves
9 15 21 25 27 for a
And
18 30 42 50 54 for b

And only the pair 25 and 50 has the product (1250=2*5*5*5*5) that can be factorize in only one way (2*5*5) and 5*5 that fit for all conditions

Am I wrong?

[parseljc]

I think the riddle is stated wrong. There is no need to tell us that the sum is less than 99, even if it is. Giving away too much info!

Plus, I think it goes like this:

Mr. P: "I don't know the numbers."
Mrs. S: "Hmmm, I don't either."
Mr. P: "Now I know the numbers."
Mrs. S: "Oh, now I know them, too!"

Update: I was wrong, the one I heard went like this:
Mr. P: "I don't know the numbers."
Mrs. S: "I already knew you couldn't know them."
Mr. P: "Now I know the numbers."
Mrs. S: "Oh, now I know them, too!"


Which is basically the same as what the one here posted. But telling us the sum is less than 99 is definitely an unnecessary hint.

[parseljc]

rbode got it right. this riddle is not solved by whipping up a quick formula. It requires brute force.

First, create a list of all your Possible pairs and write them out with their sums and products as well. I believe there are (97*97 + 97)/2 = 4753 such pairs, right? (I am ignoring the "sum < 100" hint.)

2,2,4,4
2,3,5,6
2,4,6,8
2,5,7,10
2,6,8,12
...
3,3,6,9
3,4,7,12
...
4,4,8,16
4,5,9,20
...
etc.


Step 1) We know it is not a pair of primes, or Mr. P would know them immediately from the product.
So eliminate all those from the possibilities, but keep them in a new list of "Prime Pairs" for future reference.

Step 2) We know it is not a pair whose sum could also come from a pair of primes, or S would not be sure that P doesn't know. So check the list of Prime Pairs and get a listing of all the sums. Now eliminate any pairs in the Possibilities that have a sum in that list. So here for example, you lose 3,4 because it has the same sum as the prime pair 2,5.

Step 3) Now, we know that with this information and knowing the product, Mr. P knows the answer, so therefore the product must only have one pair of possible multiplicants whereby their sum can not be expressed as the sum of a pair of primes. So find the list of products that were eliminated in Step 2. Now find all the remaining Possibilities that share one of those same products and only have one pair left that makes that product. These are the only products Mr. P could have.

Step 4) If there is more than one possibility left, your last step is to go through all the Possibilities eliminated in Step 3, and check their sums. Now find all the remaining Possibilities that share one of those same sums and only have one pair left that makes that sum. These are the only sums Mrs. S could have.

Hopefully at this point you are left with only 1 possible pair, and that is your solution. (4,13)

Math Forum Discussions

Wish I could find that Matlab code...

PS - I just realized that since you have an upper bound, there are other certain Products of non-Prime-Pairs that can be eliminated as possibilities because there is no other way to factor them with the given constraints. So they can be considered "Pseudo-Prime Pairs" for the purposes of this problem. For example, (95, 97) would give S=192 and P=9215 and there is only one way to factor this with two numbers between 1 and 99, so P would know right away. In fact, this might make a drastic cut in the possibilities left after Step 1, giving you approximately the same benefit of the "sum < 100" hint, (but still forcing you to logically deduce it). The best way to program this is to just eliminate all the possibilities who have unique products. That gets all the Prime Pairs and Pseudo Prime Pairs.

[rnast]

Excellent work parseljc!

Obviously this is a problem that spans the decades. Now someone just needs to email anabolik back in 2002:

"if you solve this, please email me at Yahoo!..."

Hopefully anabolik wasn't on a short deadline or anything. This is like math archeology. Drudging up ancient society's math problems and solving them. Hmmm... i'm seeing a TV show in the making.

Originally Posted by parseljc

rbode got it right. this riddle is not solved by whipping up a quick formula. It requires brute force.
...

Hopefully at this point you are left with only 1 possible pair, and that is your solution. (4,13)

Math Forum Discussions