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I have a question about (k+l=n)

Possible Duplicate:
Why is the sum of all integers that are multiples of $11$ equal to $22$?

I have a question about the following equation:

(k+l=n)

I’ve tried to approach it in a lot of ways, but all seem to lead to the same answer, which is $11n$.
Would someone mind explaining this equation to me?

A:

The easiest way to prove this is to rewrite the equation so that you’re starting with $n$ and counting:
$$(n-k)\cdot 11 + k \cdot 1 = n.$$
The left side counts how many multiples there are for $n$ of $11$, so it should be $11n$.

A:

HINT:
$k+l=n\implies k=n-l\;,\;\;\;k,l,n\in\Bbb Z$
Now, as explained by @Ivan, $\;\;11\mid k\;\;$ iff $\;\;1\le k\le 11$
Now, $11n=11\cdot n\iff 1\cdot n\le 11\cdot n\iff 1\cdot n\le 11n$
and hence $\;\;k=n-l\le 11\iff 1\cdot n\le 11n\iff 1\cdot n-1\le 11n$
Finally, using that $\;\;n\cdot 1\le n\;\;$ iff $\;\;n\le 1$

A:

$\;k+l=n \iff n=k+l-k=l+k-k=l+(-k)=l+(-1)k$ and by the fact that $\;\mathbb Z
i k\mapsto k+(-1)k\in\math
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