Number theory in 21 minutes
> Why is that?
Well for one all prime numbers are odd. And we all know given n (an odd number) n - 1 is even. Providing n > 1 :) (depending on your definition of zero's oddness).
Other than that, it's just a pattern. You could of course replace 24 with any other number that 24 is divisible by (der).
Now all you have to do is prove it inductively -- just to make sure :)
Number theory in 21 minutes
Quote:
Originally Posted by
zacs7
Well for one all prime numbers are odd.
Wrong. 2 is an even prime number.
Quote:
And we all know given n (an odd number) n - 1 is even.
Right. So?
Greets,
Philip
Number theory in 21 minutes
Quote:
Originally Posted by Snafuist
Wrong. 2 is an even prime number.
I think zacs7 forgot the phrase "within the given range" or "greater than 3".
Number theory in 21 minutes
> 2 is an even prime number.
And you specified a range > 3. Last time I checked 2 < 3.
> Right. So?
What do you mean so? It's a key property of the pattern.
Number theory in 21 minutes
Quote:
Originally Posted by zacs7
What do you mean so? It's a key property of the pattern.
Right, but honestly I do not see how it proves the property. Could you elaborate?
Number theory in 21 minutes
I never said it proved it :)
Number theory in 21 minutes
Take n where
p = 2*n + 1
p^2 - 1 = (2*n + 1)*(2*n + 1) - 1 = 4*n*n + 2*2*n + 1 - 1
= 4*n*n + 4*n = 4*(n*n + n)
So n*n + n = n*(n+1) must be dividable by 6.
If n is not dividable by 2, then n+1 is. So now we need to proof n*(n+1) is dividable by 3.
If n is not dividable by 3, and n+1 is not dividable by 3, then n+2 must be. Then:
(n+2)*2 = 2*n + 4
must be dividable by 3. And thus
2*n + 4 - 3 = 2*n + 1
must be dividable by 3. However, p = 2*n + 1, and p is a prime number so can't be dividable by 3.
q.e.d.
Number theory in 21 minutes
Right, EVOEx!
Your proof is technically perfect, but it hides the fundamental ideas, so I will present a simpler version:
From school, we remember that p^2 - 1 == (p-1)(p+1)
We want to show that (p-1)(p+1) is dividable by 24, i.e. it's dividable by 2^3 and 3 (the prime factors of 24).
Now have a look at the three consecutive numbers (p-1), p and (p+1). We know that p is a prime > 3, so (p-1) and (p+1) are even. And what's more, if you have two consecutive even numbers, one of them is also dividable by 4. Thus, (p-1)*(p+1) is dividable by 2*4.
Another simple fact about numbers is that of three consecutive integers, one of them is dividable by 3. This is not p (as p is a prime greater than 3), so it must be either (p-1) or (p+1).
Hence, (p-1)*(p+1) is dividable by 2*4 and by 3, i.e. 2*3*4, i.e. 24.
Greets,
Philip
Number theory in 21 minutes
Heh. That really was a lot easier. ;)
Riddle #2: one night in the UAE
This is taken from my exam in "system architecture":
Consider the number abcabc, where a, b and c are arbitrary digits. abcabc is always dividable by 13:
123123 = 13 * 9471
597597 = 13 * 45969
666666 = 13 * 51282
Why is that?
Greets,
Philip
Riddle #2: one night in the UAE
Code:
a_number_in_the_form_of(abcabc) =
= c + 10 * b + 100 * a + 1000 * c + 10000 * b + 100000 * a =
= 1001 * c + 10010 * b + 100100 * a =
= 1001 * (c + 10 * b + 100 * a) =
= 13 * 77 * (c + 10 * b + 100 * a)
Riddle #2: one night in the UAE
Right, anon!
Or to put it in simpler terms:
abcabc = 1001*abc, and 1001 is dividable by 13.
Where do you guys get those byzantine explanations? :P
Greets,
Philip
Riddle #2: one night in the UAE
Quote:
Originally Posted by
Snafuist
Right, anon!
Or to put it in simpler terms:
abcabc = 1001*abc, and 1001 is dividable by 13.
Where do you guys get those byzantine explanations? :P
Greets,
Philip
Actually, that was the proof I wanted to provide before I read the rest of the posts. However, after typing it in my calculator, I found that "10001" wasn't dividable by 13. I always make those kinds of stupid mistakes....
Riddle #2: one night in the UAE
Quote:
I always make those kinds of stupid mistakes....
Well, you're not alone. My inability to perform mental arithmetic is famous by now...
Your brother,
Philip