Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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..."

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,576

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**

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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 = k-1 + k + k+1

n = k-2 + k-1 + k + k+1 + k+2

n = k-a +... + k-1 + k + k+1 + ... + k+a

For some integers k or a. But we find something weird happens when we add those up;

n = k-1 + k + k+1 = 3k

n = k-2 + k-1 + k + k+1 + k+2 = 5k

n = k-a +... + k-1 + 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 non-power 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..."

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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 k-a + ... + a-k = 0, and there must be at least two terms to the right of a-k since k >= 1 and (k+a) - (a-k) = 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..."

Offline

Pages: **1**