A simple riddle related to addition of odd numbers

Question

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 :)

Answer 1

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.

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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 $$\mathsf{odd}+(\mathsf{even}+\mathsf{even})=\mathsf{odd}+\mathsf{even}= \mathsf{odd}$$

Answer 2

$$\sum_{k=1}^{9}(2n_k+1)= 2\sum_{k=1}^9n_k+9$$

$$41=2\sum_{k=1}^9n_k$$

There is no integer $\sum_{k=1}^9n_k$ that satisfies this.