b******y
发帖数: 139 | 1 A colleague asked me this question today:
Prove that all two digit even numbers (between 10 and 98) can be expressed
the summation of two primes. | s**c
发帖数: 1247 | 2 enumerate就行了吧
10到98，不算很多
【在 b******y 的大作中提到】
: A colleague asked me this question today:
: Prove that all two digit even numbers (between 10 and 98) can be expressed
: the summation of two primes.
| b******y
发帖数: 139 | 3 Of course your solution works. But obviously the point here is to test
analytical ability.
【在 s**c 的大作中提到】
: enumerate就行了吧
: 10到98，不算很多
| r*******y
发帖数: 290 | 4 so what's your solution
【在 b******y 的大作中提到】
: Of course your solution works. But obviously the point here is to test
: analytical ability.
| b******y
发帖数: 139 | 5 I was expecting someone to come up a solution, I just got one.
My idea is that any number n between 10 and 98 can be written as one of the
following three forms:
n = (n-3) + 3;
n = (n-5) + 5;
n = (n-7) + 7;
Now it is enough to prove at least one number s in S ={n - k, k=3, 5, 7} is
prime. By further simplification, it is enough to prove that there is at
least one s in S such that none of 3, 5 and 7 divides s. This s is prime.
【在 r*******y 的大作中提到】
: so what's your solution
| ad
发帖数: 110 | 6 This is wrong for 98
98-3=95, 98-5=93, 98-7=91, none is prime
Actually, the assumption behind this method is that for every six
consecutive natural numbers within 100, there is at least one prime number
(because the gap between 3 and 7 is 4, plus the gap between two consecutive
even numbers, which is 2). This is true for all cases except 89...97.
But this is interesting thought, though. I believe it can be made perfect by
just adding one other element to your S, say, n-19? However, it will look
【在 b******y 的大作中提到】
: I was expecting someone to come up a solution, I just got one.
: My idea is that any number n between 10 and 98 can be written as one of the
: following three forms:
: n = (n-3) + 3;
: n = (n-5) + 5;
: n = (n-7) + 7;
: Now it is enough to prove at least one number s in S ={n - k, k=3, 5, 7} is
: prime. By further simplification, it is enough to prove that there is at
: least one s in S such that none of 3, 5 and 7 divides s. This s is prime.
| b******y
发帖数: 139 | 7 Thanks for pointing out the misktake, I found one error in my proof.
You misunderstood one thing in my previous mail, that is the assumption you
mentioned. If such an assumption is made, I wouldn't call it a proof, it
just uses one conjecture to prove another conjecture.
I changed a little in my previous proof and include more details here, |
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