A simple riddle related to addition of odd numbers I'm not sure if this type of question can be asked here, but if it can then here goes:
Is it possible to get to 50 by adding 9 positive odd numbers? The odd numbers can be repeated, but they should all be positive numbers and all 9 numbers should be used.
PS : The inception of this question is a result of a random discussion that I was having during the break hour :)
 A: A direct approach: 
Any given integer is either odd or even. If $n$ is even, then it is equal to $2m$ for some integer $m$; and if $n$ is odd, then it is equal to $2m+1$ for some integer $m$. Thus, adding up nine odd integers looks like
$${(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)\atop +(2f+1)+(2g+1)+(2h+1)+(2i+1)}$$
(the integers $a,b,\ldots,i$ may or may not be the same).
Grouping things together, this is equal to
$$2(a+b+c+d+e+f+g+h+i+4)+1.$$
Thus, the result is odd.

An simpler approach would be to prove these three simple facts:
$$\begin{align*}
\mathsf{odd}+\mathsf{odd}&=\mathsf{even}\\
\mathsf{odd}+\mathsf{even}&=\mathsf{odd}\\
\mathsf{even}+\mathsf{even}&=\mathsf{even}
\end{align*}$$
Thus, starting from
$$\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}+\mathsf{odd}$$
and grouping into pairs,
$$\mathsf{odd}+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})+(\mathsf{odd}+\mathsf{odd})$$
we use our facts to see that this is
$$\mathsf{odd}+\mathsf{even}+\mathsf{even}+\mathsf{even}+\mathsf{even}.$$
Grouping again,
$$\mathsf{odd}+(\mathsf{even}+\mathsf{even})+(\mathsf{even}+\mathsf{even})$$
becomes
$$\mathsf{odd}+\mathsf{even}+\mathsf{even}$$
becomes
$$\mathsf{odd}+(\mathsf{even}+\mathsf{even})=\mathsf{odd}+\mathsf{even}=
\mathsf{odd}$$
A: $$\sum_{k=1}^{9}(2n_k+1)= 2\sum_{k=1}^9n_k+9$$
$$41=2\sum_{k=1}^9n_k$$
There is no integer $\sum_{k=1}^9n_k$ that satisfies this.
