
 Ricky
 Moderator
Special numbers
A special number is a number which can be written as a sum of 2 or more consecutive numbers.
Prove that a number n is special iff it is not a power of 2.
I have a proof, but it took me 2 and a half pages, so I want to see if anyone can come up with something reasonable.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Special numbers
So for 14, it is 2+3+4+5 which is 3.5 times 4, 4 terms above, average is 3.5 . Hmm. So for 5, it's 2+3. For 3, it's 1+2. For 6, it's 1+2+3. For 7, it's 3+4. For 9, it's 4+5. For 10, it's 1+2+3+4. For 11, it's 5+6. For 12, it's 3+4+5. I don't see the pattern yet. Is there a pattern??
igloo myrtilles fourmis
 Ricky
 Moderator
Re: Special numbers
You have to be more abstract, John. Finding patterns is useful, but most of the time is hopeless. But abstract patterns are much easier to find.
If a number n is special, then one of the following must be true:
n = k n = k + (k + 1) n = k + (k+ 1) + (k + 2) n = k + (k+ 1) + (k + 2) + ...
Where k <= n.
So expanding on this, we get:
n = k or n = 2k + 1 or n = 3k + 6 or n = 4k + 10 or ...
And so on for some positive integer k. First thing to notice is that all the odds other than 1 are covered because every odd number can be expressed as 2k + 1. So all we have left are the evens.
We can continue that pattern, getting more and more numbers, but it will get us nowhere (or so I think). We won't ever be able to say we have all the numbers.
So lets do the same thing, but write numbers a different way:
n = k n = k1 + k + k+1 n = k2 + k1 + k + k+1 + k+2 n = ka +... + k1 + k + k+1 + ... + k+a
For some integers k or a. But we find something weird happens when we add those up;
n = k1 + k + k+1 = 3k n = k2 + k1 + k + k+1 + k+2 = 5k n = ka +... + k1 + k + k+1 + ... + k+a = ak
So from this fact, we can construct a proof that if n is divisble by an odd number, it must be special.
Now if you know a little (but very important) fact, you can figure out where to go from here. But assuming you don't, that little fact is:
Every number can be written as a product of primes.
The only numbers that can be written as solely a product of 2's is a power of 2. Thus, for any nonpower of two, there must be a prime in that product that is not 2. So it must be the case that...
... such a number is divisble by an odd number.
And that concludes that half of the proof (informally). The other half can be deduced by finding that any special number must contain a prime factor other than 2, and every power of two must contain only powers of 2, so they can't be equal.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Re: Special numbers
There might be a flaw in that proof, depending on exactly what it is that you're trying to prove.
Take 14 as an example. By your proof, 14 is a special number because it is divisible by 7 and so there are 7 consecutive numbers that add to give 14. However, those numbers are 1, 0, 1, 2, 3, 4 and 5. Are you allowed to use negative integers or 0 as part of the consecutive numbers?
Why did the vector cross the road? It wanted to be normal.
 Ricky
 Moderator
Re: Special numbers
Right, I realized this when I wrote it out on paper. However, you will notice that:
(1 + 0 + 1) + 2 + 3 + 4 + 5 = 2 + 3 + 4 + 5, and thus, 14 is special.
It takes a little fix, but you find that if k  a < 0, then ka + ... + ak = 0, and there must be at least two terms to the right of ak since k >= 1 and (k+a)  (ak) = 2k.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
