Adding a list of integers will result in an odd answer when there are an odd number of odd numbers in the list.

• If the list consists of 8 consecutive numbers there will always be 4 odd numbers -> result will be even.
• If the list consists of 7 consecutive numbers, there can be 3 or 4 odd numbers -> result can be even or odd.
• If the list consists of 5 consecutive numbers, there can be 2 or 3 odd numbers -> result can be even or odd.

With 6 consecutive numbers there will always be 3 odd numbers. Then the result will be odd.

In general the result will always be odd with 2 modulo 4 consecutive numbers.

Sum of any 5 consecutive positive integers = $\frac{n(n+1)}{2}$ - $\frac{ (n - 5)(n - 4)}{2}$ = $\frac{(n^2 + n) - (n^2 - 9n + 20)}{2}$ = $5n - 10$ (n here is positive integer greater than or equal to 5 )

Sum of any 6 consecutive positive integers = $\frac{n(n+1)}{2}$ - $\frac{(n - 6)(n - 5)}{2}$ = $\frac{(n^2 + n) - (n^2 - 11n +30)}{2}$ = $6n - 15$ (n here is positive integer greater than or equal to 6 )

Sum of any 7 consecutive positive integers = $\frac{n(n+1)}{2}$ - $\frac{ (n - 7)(n - 6)}{2}$ = $\frac{(n^2 + n) - (n^2 - 13n + 42)}{2}$ = $7n - 21$ (n here is positive integer greater than or equal to 7 )

Sum of any 8 consecutive positive integers = $\frac{n(n+1)}{2}$ - $\frac{ (n - 8)(n - 7)}{2}$ = $\frac{(n^2 + n) - (n^2 - 15n + 56)}{2}$ = $8n - 28$ (n here is positive integer greater than or equal to 8 )

$5n - 10$ , $7n - 21$ , $8n - 28$ could be even as well as odd whereas $6n - 15$ is always a odd number because $6n$ is even and $even - odd = odd$

In order for the sum of a sequence of consecutive numbers to be odd, the sequence must contain an odd number of odd numbers.

x=5: Can have either 2 odds and 3 evens or 3 odds and 2 evens.

x=6: Must have 3 odds and 3 evens.

x=7: Can have either 3 odds and 4 evens or 4 odds and 3 evens.

x=8: Must have 4 odds and 4 evens.

Of the choices, only n=6 is guaranteed to have 3 odd numbers.

WE KNOW THAT,

$\text{Even+Odd=Odd; Odd+Odd=Even; Even+Even=Even}$

$\text{so, for every consecutive system, the pattern will be Even+Odd+Even+... or Odd+Even+Odd+...}$

$\text{now,for 5,it will be=Even+Odd+Even+Odd+Even=Odd+Odd+Even=Even}$

$\text{or,it will be=Odd+Even+Odd+Even+Odd=Odd+Odd+Odd=Odd}$ ....................[different result]

$\text{now, for 6,it will be =Even+Odd+Even+Odd+Even+Odd=Odd+Odd+Odd=Odd}$

$\text{or, it will be=Odd+Even+Odd+Even+Odd+Even=Odd+Odd+Odd=Odd}$ ......[same result]

$\text{now,for 7 it will be=3 times Odd+Odd=Even or,it will be=3 times Odd+Even=Odd}$ .............[different result]

$\text{ now for 8 it will be=4 times odd=Even or, it will be =4 times Even=Even}$ ..............................[same result but Even})

$\color{#3D99F6}\text{so, the answer has to be=6}$

We will assume the consecutive numbers are n ,n+1 ,n+2 ...

● So the sum of 5 consecutive numbers is : n+(n+1)+(n+2)+(n+3)+(n+4) = 5n+(1+2+3+4)
ie. 5n+10 .. which is odd or even depending on 'n'

●Now , the sum of 7 consecutive numbers is 7n+(1+2+3+4+5+6) =6n+21 . Which is odd or even depending on 'n'

●Similarly, the sum of 8 consecutive numbers is 8n+(1+2+3+4+5+6+7) = 8n+28 Which is a even number irrespective of 'n'

●Coming to 6: Sum of 6 consecutive numbers is: 6n+15 Observe it carefully.. It is odd whatever the value of n may be.. Since even+odd=odd... Hence the answer is 6

This is very simple to understand.

We know that, Even (E) + Odd (O) = Odd (O)

Consecutive integers would generate a pattern that is : E O E O E O... or O E O E O E... Eg: 1 2 3... or 2 3 4...

Then the sum in both cases would be: [ a x(E+E+..)] + [ b x(O+O+...)] = X, where a and b denote the number of even and odd numbers, respectively.

For X to be odd, we need to focus on b . b should be odd.

This is only possible, if we have a total set of consecutive integers to be even. Given the options, 6 is the minimum acceptable answer.

The sum of $n$ consecutive even numbers is always even, therefore for the sum of $X$ consecutive integers to be odd it must mean that $X= odd + even$ . Since the sum of $X$ consecutive integers is always odd independent of whether we start from an odd or even number, we must always have a sum of an odd amount of odd numbers. It cant therefore be $X = 5$ as if we start from an even number we will be adding an even amount of odd numbers. Similarly it can't be $X = 7$ if we start from an odd number as we would be adding an even number of odd numbers and for the same reason it can't be $X = 8$ if we start from an even number.

Therefore the answer is $X = \boxed{6}$ .

2 consecutive positive integers will always add up to an odd number. Therefore for total to be odd it needs to have odd number of pairs. 2 | 2 | 2 ( 6 numbers). Hence answer is 6. 5 can be eliminated as last digit may align with single even number.

All series of X consecutive integers where X is even are $X/2$ pairs of odd/even integers each pair of which must add up to an odd.

We need X to be an even number that is odd when divided by 2.

Although there are infinity many such - ( $2(2n+1)$ ) where n is a positive integer - , 6 is the only one offered in the solution set.

(n)+(n+1)+(n+2)+(n+3)+(n+4)+(n+5)=6n+15

6n is even, 15 is odd. Even + Odd = Always Odd

the sum of consecutive integers is x + (x+1)+...+(x+n-1) where n is the amount of terms added. the amount of terms must be even to insure we have an even bases of ax + b where a is the amount of terms added. Looking at 6 we have 6x + (5+4+...+1) = 6x +15. 6 is even so any number times 6 is even. 15 is odd. even + odd = odd so out of the answer choices given 6 is the answer.

My solution was a bit more general, without really taking into account the options (at least at first). The sum of $X$ consecutive integers is $S=\frac{(n+X)(n+X+1)}{2} - \frac{n(n+1)}{2}$ This can be written as: $S = \frac{X(2n+1)+X^2}{2}$ For $S$ to be odd, and since $(2n+1)$ is always odd, $X$ must be an odd multiple of 2, say $2m$ with odd $m$ , which gives: $S = m(2n+1)+2m$ This is where I looked at the options The only option where $X$ is an odd multiple of 2 is 6, the solution.

Notice that:

even+even=even

odd+even=odd

odd+odd=even

Addition has commutative property(odd1+even2+odd3 = odd1+odd3+even2)

When adding X amount of even numbers the result will always be even.

So you can simplify X amount of even number within the addition into one even number.(even2+even4+even6 = stillEven#)

Since odd+even=odd, you can ignore any even numbers and count just the odd numbers.

Having odd amounts of odd numbers in an addition will always result in odd.

When:

X=5 there can be 2 or 3 amount of odd numbers so result will be even or odd

X=6 there can be only 3 amount of odd numbers so result will be odd only

X=7 there can be 3 or 4 amount of odd numbers so result will be odd or even

X=8 there can be only 4 amount of odd numbers so result will be even only

Let $j$ be a positive integer.

$\sum_{k = 0}^{X - 1} (j + k) = \dfrac{X(X + 2j - 1)}{2}$

For $X = 6 \implies \sum_{k = 0}^{5} (j + k) = 3(2j + 5) = 3m$ , where $m$ is odd.

$\therefore X = 6$ .

Note: I fail to understand why the question is, $''$ Which of the following could be $X?"$ , since $X = 6$ guarantees that the sum above is odd.

For sum of x consecutive numbers x=b-a $\sum_{i=a}^bn = \sum_{i=1}^bn - \sum_{i=1}^an$ $= \frac{b(b+1)}{2} - \frac{a(a+1)}{2}$ $= \frac{b^2-a^2 + b-a }{2}$ $= \frac{(b-a)(b+a+1)}{2}$ $= \frac {(x)(b+a+1)}{2}$ Assume $x= k2^p$ Where k is coprime with 2 $\sum_{i=a}^bn = 2^{p-1}k(a+b+1)$ If sum is odd so p=1 Thus only 6 in the options satisfies the condition hence it is the answer.

Let us take the first no. as (n ) then it's consecutive numbers will be (n+1) ,(n+2), (n+3) ..... And so on... Now for the case of 5 consecutive numbers add (n) + (n+1) + (n+2) + (n+3) +(n+4) The sum is 5(n+2) for 5 consecutive numbers. 3(2n+3) for 6 no. 7(n+3) for 7 no. 4(2n+7) for 8 no. Now you will find that by substituting random starting numbers . The 6 no. Sum always gets an odd no. !

solved

Write the sum of $x$ consecutive integers as $a + (a+1) + ... + (a+ x-1)=ax+\frac{x (x-1)}{2}$ . The residue of this quantity mod 2 must be 1:

Let $x=2k$ . Then, $ax+\frac{x (x-1)}{2} \equiv 2ka+2k^2-k \equiv k \mod 2$

So $k\equiv 1\mod 2$ . Write $k$ as $2z+1$ , then $x = 4z+2$ . Therefore, $x\equiv 2\mod 4$ . The answer is 6.

Note that if our initial assumption that $x=2k$ is wrong and $x=2k+1$ , then the residue of the sum depends on $a$ .

6.

X must be an odd multiple of 2. 6 is the only odd multiple of 2 among the choices.

Only an odd multiple of 2 assures that the sequence will have an odd sum, as follows:

Every pair of consecutive integers sums to an odd number. 2n + (2n +1) or (2n +1) + 2n = 4n + 1

Thus any odd number of pairs will sum to an odd number.

A sequence of an even number of pairs will always sum to an even number.

A odd number of consecutive integers will sum to even or odd depending on the number of pairs within it.

If the odd number is of the form 4n + 1 with n= 0,1, ... (an even number of pairs), the sequence of 4n numbers will be have an even sum, so if the last number is also even, the entire sequence sum will be even.

if the odd number is of the form 2(2n+1) + 1 with n = 0,1,2,... (an odd number of pairs) the pair sequence will sum to an odd number, so if the last number is odd, the entire sequence will sum will be even.

Thus only a sequence of an odd number of pairs will always sum to an odd number.

5, 7 and 8 are not odd multiples of 2, and cannot work.

$\begin{aligned}&\text{The sum of x consecutive positive integers is given by}\\ &\left(\sum_{i=1}^{x} i \right) +x(n-1) \hspace{4mm}\color{#3D99F6}\text{Where n is the first term}\\ &=\dfrac{x(x-1)}{2}+x(n-1)\\ &=\dfrac{x\left((x-1)+2(n-1)\right)}{2}\\ &=\dfrac{x(x+2n-3)}{2}\\\\ &\text{Note that } x \text{ and } (x+2n-3) \text{ are of opposite parity. }\\\\\ &\underline{Case :1\hspace{4mm} \color{#3D99F6}x=1,3 \pmod{4}}\\ &(x+2n-1) \text{ will be } 0 \text{ or } 2 \pmod 4 \text{ depending on n }\\ &\implies \dfrac{x(x+2n-3)}{2} \text{ can be } \equiv 0,1\pmod{2}\\\\ &\underline{Case :2\hspace{4mm} \color{#3D99F6}x=0 \pmod{4}}\\ &\dfrac{x}{2} \equiv0 \pmod{2}\\ &\implies \dfrac{x(x+2n-3)}{2} \equiv 0\pmod{2}\\\\ &\underline{Case :3\hspace{4mm} \color{#3D99F6}x=2 \pmod{4}}\\ &\dfrac{x}{2} \equiv1 \pmod{2}\\ &(x+2n-3) \equiv 1 \pmod{2}\\ &\implies\dfrac{x(x+2n-3)}{2} \equiv 1\pmod{2}\\ &\text{Thus for sum of x consecutive integers to be always odd, } x\equiv 2\pmod{4}\\ &\text {which makes the answer } \color{#EC7300}\boxed{\color{#333333}6}\end{aligned}$

Out of the four options, only 6 consecutive positive integers has to always have an odd number off odd numbers. The sum of odd number of odd numbers is always odd. The number of even numbers does not have an effect in the parity of the sum.

Relevant wiki: Parity of Integers

Let the first of the consecutive integers be $x$ . Then the six consecutive integers will be $x, x+1, x+2, x+3, x+4, x+5$ . Taking the sum of these gives us $6x+15 = 3(2x+5)$ . We know that $2x + 5$ $LaTeX$ must always be odd since $2x$ will be even, and even plus odd gives us an odd number. Furthermore, multiplying this odd number by $3$ will still return an odd number. Thus, the sum of any six consecutive integers will always be odd.