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Consecutive Sums (Posted on 2016-07-12) Difficulty: 3 of 5
In the list below I have tried to express each integer 1-9 as a sum of consecutive positive integers:

1 = ?
2 = ?
3 = 1+2
4 = ?
5 = 2+3
6 = 1+2+3
7 = 3+4
8 = ?
9 = 2+3+4

This suggest that there is a sum for every positive integer which is not a power of 2, and there is never a sum for powers of 2.

Prove or disprove my claims.

No Solution Yet Submitted by Brian Smith    
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more complete answer Comment 2 of 2 |
First, I apologize because I forgot to fill the heading of my precedent post. 

Re-reading the puzzle I see that my answer refers only to the fact that a positive integer which is a power of 2 can't be expressed as a sum of consecutive numbers. 

But the puzzle also mention the possible existence of a sum of consecutive numbers for every positive integer (except the powers of 2), and ask to prove or disprove this.

Well, I think that the proposition is true: it applies to every positive integer (except the powers of 2).
PD: I correct here the first version of this post (a wrong one)

In the precedent post I get a formula to express a positive integer (p) as a sum of consecutive integers (from n to n+a).

p=((a+1)(2n+a))/2        [1]

The numerator is always the product of an even factor and an odd factor (if "a" is even, the first factor is odd and the second even; if "a" is odd, the other way is true)

We want to know if our formula [1] is able to match all positive integer p (except the powers of 2). 

To show this perhaps is better to separate the values of p depending on parity: 

a) p= odd positive integer
All odd numbers can be expressed as sum of consecutive numbers. To prove it just assign a=1 and the expression become p=2n+1 which cover the entire field of odd positive integers, when n assumes positive integer values.

b) p=even positive integer.
All even positive integers (except the powers of 2) will have one or more odd divisors. Then they could be express as the product of an odd factor for an even factor. If the odd factor is bigger than the even one, then we express that odd factor in terms of (2n+a) and the even factor in terms of a+1. 

This procedure is because 2n+a is always bigger than a+1, and always will be possible for 2n+a to reach a number superior to a+1 and with a different parity of a+1. In this way the formula [1] always can match the number p.

If it is the other way (the odd factor lower than the even) we assign the odd factor to a+1 and the even one to 2n+a. 

Ex 1: 68=17*4=17*8/2 
a+1=8 a=7 2n+a=17 n=5
So a=7 n=5 68=5+6+7+8+9+10+11+12

Ex 2: 286=2*11*13=11*52/2 
a+1=11 a=10 2n+a=52 n=21
So a=10 n=21 286=21+22+23+24+25+26+27+28+29+30+31

See that this also implies that, if p is even, for any different odd divisor of p (except 1) there will be one different sum
(if p is odd for each pair of odds divisors there will be one sum).

F. ex 286 will have another 2 different sums:

a+1=13 a=12 2n+a=44 n=16


a+1=4 a=3 2n+a=143 n=70

But if there is none odd divisor in the number p (only the powers of 2), then the expression [1] would not be able to match p.

Edited on July 13, 2016, 12:11 pm
  Posted by armando on 2016-07-13 04:04:48

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