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Addition of consecutive numbers
By Geoff Rutherford (T1869) on Thursday, January 6, 2000 - 04:12 pm:

If you attempt to take all the numbers on a 100 square and see which can be made by adding consecutive numbers eg. 3+4=7 or 1+2+3=6 The only numbers you can't do are
2,4,8,16,32,64 etc

Can anyone explain this for me?
and does this particular pattern have a special name?

By Michael Doré (P904) on Thursday, January 6, 2000 - 07:54 pm:

I'm not sure but I think that the following works: [It does! - The Editor]

If a number can be written like

(r + 1) + (r + 2) + (r + 3) + ... + (n - 1) + n

then this is equal to n(n+1)/2 - r(r+1)/2 using the result: sum of first n integers = n(n+1)/2.

So if x is a number that can be written like that then

x = n(n+1)/2 - r(r+1)/2 for integers n and r which aren't negative, and where n is at least 2 more than r.

We can re-write x as (n-r)(n+r+1)/2. This isn't obvious, but if you multiply out you'll see it works.

If we now let q = n - r then q will also be a non-negative integer and: 2x = q(2r+q+1).

The right hand side is an odd number (greater than 1) multiplied by an even number: if q is even, then 2r+q will also be even and 2r+q+1 will be odd.

Therefore the left hand side (2x) must have an odd factor greater than 1. Therefore x has an odd factor greater than 1. Therefore it is impossible to write any of the powers of 2 (20, 21, 22...) like this. What's more 1,2,4,8,... are the only ones that do not work - because it is always possible to factor an even number into q and q + an arbitary odd number as long as it has an odd factor.

Therefore the powers of 2 (1, 2, 2×2, 2×2×2, 2×2×2×2 etc) are the only ones left.

Hope this is right,

Michael