Hard maths problem for you (Warning: VERY hard!)

TheCenturion said:
It's entirely possible to do. I asked my head of maths today, and he said that he knew this particular problem. He knew 5 people who'd solved it (including himself) and all had used a different method. One of these people in fact had to write a program and left it overnight to figure out the answer :p
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I'd love to find out these other methods. The only one I am aware of, the only one my colleagues seem to be aware of and the only one I can find on the internet is a brute force method (which boils down to eliminating a large collection of pairs then examining the rest case by case).

Your maths teacher and his five friends seem to know more than the academics! :eek:
 
w11tho said:
Your maths teacher and his five friends seem to know more than the academics! :eek:
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Actually, maths teacher was one of the five, so it's maths teacher + only four others that he knows who have ever solved this. One of the people who has solved it has now gone to Trinity (Cambridge) to do maths. He was literally a maths god. He took his GCSE in year 10, and at the time of doing his GCSE, he already had 8 maths A level modules under his belt :eek:
 
vonhelmet said:
Google it. It's fairly simple, but it's not elegant. You won't read it and think "Oh, that's so obvious".
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Oh yeah I knew it wasn't one of these obvious things, but I just had to read how it's done after seeing the problem :)

I like simple-ish number things like this, nothing too taxing though.

Didn't sink in too well last night as was pretty tired so brain was hurting a bit anyway, but I think it made sense :p
 
TheCenturion said:
Actually, maths teacher was one of the five, so it's maths teacher + only four others that he knows who have ever solved this. One of the people who has solved it has now gone to Trinity (Cambridge) to do maths. He was literally a maths god. He took his GCSE in year 10, and at the time of doing his GCSE, he already had 8 maths A level modules under his belt :eek:
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Maybe you could get your friend to pop over to the maths department and tell me all these different methods!

To emphasize my point though: doing the problem is hard, but certainly within the reach of someone who is mathematically competent. But I find it very difficult to believe your maths teacher is aware of five different methods!
 
The solution is not a quick formula, it's brute force (to an extent).

You'll need to go through many-many calculations and create lists of numbers to cross out before you can get the result, so it's not down to math wizardry, just knowing how prime numbers work.

Something I want you to check for me though,

How about these two set of pairs, wouldn't they satisfy the conditions of the problem?
first pair is 6 & 8, second pair is 4,10.

6,8 product=48,sum=14, the same product is made out of 4,12 so paul can't tell which pair of numbers are the right one (no primes there either)
4,10 product=40, sum=14, the same product can be made out of 5,8 so paul can't tell which is the right one there either (no primes there either).

I'm sure there is something to exclude these two pairs according to the problem statement but what is the flaw?
 
kgi said:
I'm sure there is something to exclude these two pairs according to the problem statement but what is the flaw?
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There is a flaw. It comes from the line:

I know you don't know what the numbers are

Any even number can be made from the sum of two primes. When two primes multiply together to make a product, the product can only have four factors: itself, 1, and the two primes.

Therefore, if the sum of the numbers was even, then there was a chance that both numbers could be prime numbers. However, this is not the case as he says with 100% certainty that he knows that the other person cannot know the two numbers. Therefore, the sum must be odd :)
 
TheCenturion said:
There is a flaw. It comes from the line:

I know you don't know what the numbers are

Any even number can be made from the sum of two primes. When two primes multiply together to make a product, the product can only have four factors: itself, 1, and the two primes.

Therefore, if the sum of the numbers was even, then there was a chance that both numbers could be prime numbers. However, this is not the case as he says with 100% certainty that he knows that the other person cannot know the two numbers. Therefore, the sum must be odd :)
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I've thought about that but can't we assume the guy said he doesn't know because he isn't sure? As the product can be made of either primes or non-primes so "not sure" equaled "don't know"? Maybe I'm getting too philosophical about it.
 
kgi said:
I've thought about that but can't we assume the guy said he doesn't know because he isn't sure? As the product can be made of either primes or non-primes so "not sure" equaled "don't know"? Maybe I'm getting too philosophical about it.
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But the whole point is that if it is even, then there is a possibility that it could be made of two primes. However, the fact that it was said "I know you don't know what the two numbers are" means that the possibility could not have existed, therefore must be an odd number :)
 
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