Frankenodd Theory Of Odd Numbers

Correction: In the third sentence "He took 7 consecutive positive integers and he added them. He noticed that all the results were divisible by 5."
$5$ must be replaced with $7$

Let's say that the first term is $m$ and last is $n$ then

$m+(m+1)+...+n=$

$(1+2+...+n) - (1+2+...+m-1)=$

$\large\frac{n(n + 1)}{2}$ - $\large\frac{(m-1)m}{2}=$

$\large\frac{n^2 + n -m^2 + m}{2}=$

$\large\frac{(n + m)(n - m) + n + m}{2}=$

$\large\frac{(n + m)(n - m + 1) }{2}$

So this sum is always divisible by $n-m+1$ which is the number of terms

i.e. $22+23+24+25+26+27+28+29+30$ has $9=30-22+1$ terms

First let's generalize. Start with the sum of $n$ (where $n$ is odd) consecutive positive integers, starting with $x$ : $x+(x+1)+(x+2)+\cdots+(x+(n-1))$ $\implies n\times x + \frac{(n-1)((n-1)+1)}{2}=nx+\frac{n(n-1)}{2}$ Now, we need to show that, when divided by $n$ , the expression is a whole number: $\frac{\left(nx+\frac{n(n-1)}{2}\right)}{n}$ $\implies \frac{nx}{n}+\frac{n(n-1)}{2n}$ $\implies x+\frac{n-1}{2}$ We know that $x$ is a positive integer, so that will be a whole number. Therefore, we can exclude it from the expression: $\frac{n-1}{2}$ . Since we know $n$ is odd, it can be written in the form $2i+1$ , where $i$ is a non-negative integer. Let's replace it: $\frac{(2i+1)-1}{2}=\frac{2i}{2}=i$ Therefore, knowing that $i$ is an integer, we know that Dr. Frankenodd's theory MUST be true.

If the order of the integers is always consecutive, the sum of the integers will always be divisible by the median number to equal the number of integers used.

Here is a solution using modulo arithmetic. So say you are summing up $k$ consecutive integers then modulo $k$ their residue classes will be $0,1,2,\ldots, k-1$ (perhaps in a different order: it could look like $5,6, \ldots , k-1,0,1,2,3,4$ ). Now you can also write $k-1 \equiv -1 \text{ mod } k$ , $k-2 \equiv -2 \text{ mod } k$ , ... So basically you are summing up: $0+1+2+\ldots + k-2 + k-1 \equiv 0+1+2+ \ldots + -2 + -1 \text{ mod } k$ Now a simple counting argument will show you that when $k$ is odd you will have: 0, every integer up to $k/2$ and their negative counterparts: $0+1+2+\ldots + \lfloor \tfrac{k}{2} \rfloor+ \lceil \tfrac{k}{2} \rceil + \ldots + k-2 + k-1 \equiv 0+1+2+ \ldots \lfloor \tfrac{k}{2} \rfloor -\lfloor \tfrac{k}{2} \rfloor+ \ldots+ -2 + -1 \text{ mod } k$ So it sums to 0.

Note: when $k$ is even this does not work because the integer $k/2$ will be left unpaired with a negative counterpart. So more generally when you sum up $k$ consecutive integers with $k$ an even number, then it is divisible by $k/2$ but not by $k$ .

Alternative: you can also sum up $0+1+2+\ldots + k-2 + k-1 = \frac{k(k-1)}{2}$ . Since $k$ is odd, 2 divides $k-1$ and so this sum is also divisible by $k$ .