Help me solve a RIDDLE, please

[Steph]

Yes Stratego, you are right, you cannot use this theorem for the largest number, but still it helps to eliminate a lot of cases.

 

[Scuffer]

I used a more lateral (i.e. my theoretical maths is nothing special, but my ability to find solutions is pretty good), and searched Google for "I know the numbers now" math. And what do you now, some clever chap has written some code to solve it.

http://www.32768.com/bill/weblog/2002_05.shtml

No idea if this will do want you need Stratego, but it might give someone some ideas

 

[Arminius]

The Product can't be a number that has only 4 factors. (such as [35: 1,5,7,35] or [25: 1,5,5,25] So that eliminates some numbers. I'm working on the problem. Between 2-25, I've narrowed it down a bunch: enough? We'll see.

 

[Steph]

4 and 13 is the answer
Just in case you missed it the first time

 

[Arminius]

I'm pretty sure it isn't. And if it is I apologize in advance. But, let me beat my head against the wall. It's gotten under my skin...

 

[Syterion]

They don't tell each other the number they got.

This is a variation of the better known "daughter's age" problem:
P: hey! how have you been?
S: i got married and i have three daughters now
P: really? how old are they?
S: well, the product of their ages is 72, and the sum of their ages is the same as the number on that building over there..
P: right, ok ... oh wait ... hmm, i still don't know
S: oh sorry, the oldest one just started to play the piano
P: wonderful! my oldest is the same age!
(even though we don't know the number on the building, we can use the fact that the professor knows to figure out what the ages are. This problem should be solved the same way)

Actually the reason it is worded like that is to tell you that there is an oldest, so it can't be 6,6,2, or any with two people as oldest. That problem is unsolvable.

I don't know how to solve the problem you gave, but if Steph is right Icongratulate him

 

[Steph]

I solved it long ago, when I was 18 in high school. I took some time because they guy who asked made an error and used a range of [2,20], and then there's no solution

 

[Steph]

About the other one
72=2*2*2*3*3

As there are 3 daughters, the possibilities are
4*2*9
4*6*3
8*3*3
6*6*2
Assuming there are no twins, it means we can't have 6*6*2 or 8*3*3
So to possibilities : 4*2*9 or 4*6*3.

Then you can't decided between them if you don't see the number on the building...

 

[stratego]

4 and 13 is the answer
Just in case you missed it the first time

thank you..

 

[samildanach]

Funny part is they have tried through computers to extend the limit from [2,100] to [2, many millions] and there is still a unique answer : 4 and 13.

It's not that suprising. I went to the high numbers so as to cut out all the low primes and all the other numbers below 80. I plumped on 90 and 90 because it holds for the first part of the riddle.
P doesn't know the numbers because it can either be 81 and 100 or 90 and 90 to give a product of 8100. Of course, it falls down because all the other numbers between 80 and 100 when multiplied give discrete products. So S solves it straight away because 81 and 100 add up to 181.
The higher the product the less numbers there are to multiply in a closed set. P gets the answer straight away because it is product of two primes or because there are only two numbers large enough to give that product within the particular set. If P can't get it then S solves it straight away.

I still contend my answer is right if S is a Professor of Divinty

good job Steph