Question

the sum of r consecutive positive integers will always be divisible by 2 if r is a multiple of
1. 6
2. 5
3. 4
4. 3
5. 2

Answer 1

i dont know how to explain this one very well >.<

Answer 2

Xand, do you know about looking at integers mod 2?

Answer 3

If you dont, I guess one way to do this to try to think of counter examples for each answer.
For example, if i thought 3 was the answer, i would pick:
2, 3, 4 as my counter example. That sum isnt divisible by 2, so it cant be the right answer.
Just repeat this process until you have one answer left.

Answer 4

can't we just look at

\[(k)+(k+1)+(k+2)+\cdots (k+(r-1))\]
\[=rk+\sum_{i=0}^{r-1}i\]
\[=rk+\frac{(r-1)r}{2}\]

then look at what happens when \[r=2a,r=3a,\ldots,r=6a\]
we will be able to see that if \[r=4a\] then

\[=rk+\frac{(r-1)r}{2}\]

\[=4ak+\frac{(4a-1)4a}{2}=2\left(2ak+\frac{(4a-1)2a}{2}\right)\]

\[=2\left(2ak+(4a-1)a\right)\]

Answer 5

A real simple way of doing this:

even + even is even
odd +odd is even

odd+even is odd

so we want a sum of and even number of odd term

clearly the only one that will guarantee this is a multiple of 4

since

2) o+e=o
3) e+o+e=o
4) e+o+e+o=e and o+e+o+e=e (both ways even)
5) o+e+o+e+o=o
6) o+e+o+e+o+e=o

Answer 6

yep, that last post of yours is what i wanted to do with the mod 2 business. nice nice :)