How many integers between 2 and 2015 inclusive cannot be expressed as a sum of at least two consecutive positive integers?

The answer is 10.

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True that, will amend accordingly. Thanks! :)

Wee Xian Bin
- 6 years, 5 months ago

$\text{To add two or more consecutive numbers at least one has to be odd. Any}\\ \text{ power of 2 can not be built by summation that include an odd number.} \\~~\\ Any~\color{#3D99F6}{ odd~ number} > 1, \text{can always be obtained by adding the following }\\\text{two consecutive numbers.,}\\\dfrac{The~ odd ~number- 1} 2 +\dfrac{The~ odd ~number+ 1} 2.\\Any~\color{#D61F06}{ even~ number} \text{ may be obtain by adding only odd number of }\\ \text{consecutive numbers, if the middle term is even. This middle term is also} \\ \text{the average. So the sum would be } \\ \text{number of terms (which is odd) * (middle term)} \neq 2^{an~ integer}$

Niranjan Khanderia
- 5 years, 11 months ago

2 (power) number's are only not consecutive : 2

4

8

16

32

64

128

256

512

1024

2048

4096 etc.

I don,t know why 2 (power) number's are not consecutive. But this is true.

From 2 to 2015 range our answer is 10.

answer = 10.

1 Helpful
0 Interesting
0 Brilliant
0 Confused

This solution has been marked incomplete. You did not explain your reasoning.

Why is it that only numbers of form $2^n$ are not expressive as sum of consecutive integers ?

Venkata Karthik Bandaru
- 6 years, 3 months ago

In response with respect to Challenge Master : Have you any explain that this solution is incorrect ?

Shohag Hossen
- 6 years ago

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Shohag....look at Niranjan's solution... he was successful in explaining why only 2^n is a number that cant be represented as the sum of consecutive integers.

Anurag Singh
- 5 years, 6 months ago

But 6 also cannot be expressed in that way

Rishik Jain
- 6 years, 1 month ago

**
[2]
**

4

8

16

32

64

128

256

512

1024

**
[2015]
**

2048

Based on the intervals from 2 to 2015, or [2,2015], there are only
**
10
**
numbers.

0 Helpful
0 Interesting
0 Brilliant
0 Confused

This solution has been marked incomplete. You did not explain your reasoning.

In response with respect to Challenge Master : Have you any explain that this solution is incorrect ?

Shohag Hossen
- 6 years ago

I have also same concept

Shohag Hossen
- 5 years, 6 months ago

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The problem should really say "consecutive $\text{positive}$ integers". Otherwise the answer would be 0, since $\sum_{k=-n}^{n+1}k=n+1$ for any natural number $n$ .