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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 :)

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Hint: adding an odd number of odd numbers will give ... (is this a real question ?) –  Raymond Manzoni May 25 '13 at 8:27
Hmm..let me ponder over that –  403 Forbidden May 25 '13 at 8:29
This says sum of 9 odd numbers can't be even… True? –  403 Forbidden May 25 '13 at 8:32
Yes: consider the last digit in binary if it helps. –  Raymond Manzoni May 25 '13 at 8:36

2 Answers 2

up vote 6 down vote accepted

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

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I'm sorry. I don't think I understood your query. –  403 Forbidden May 25 '13 at 8:34
Shouldn't that be +9? –  Ataraxia May 25 '13 at 8:35
Or just $$\mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd+odd+odd}\\= \mathsf{\underbrace{even+odd}+odd+odd}\\= \mathsf{\underbrace{odd+odd}+odd}\\= \mathsf{\underbrace{even+odd}}\\= \mathsf{odd}$$ –  Rahul May 25 '13 at 8:43
@Rahul: Very clever user of MathJax! I approve :) –  Zev Chonoles May 25 '13 at 8:44
@403Forbidden you can always select it (the checkbox under the voting arrows) if you found it most helpful. –  Ataraxia May 25 '13 at 8:51

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


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

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But my colleague says otherwise. And he refuses to let me in on the solution until the end of day..which is what is killing me. Argh! These math riddles are so addictive –  403 Forbidden May 25 '13 at 8:33
@403Forbidden he would have to demonstrate that it is possible for the sum of a set of integers to be a non-integer, and if he could do that he's worthy of an award. –  Ataraxia May 25 '13 at 8:34
As it turns out, it was just a trick question. He admitted to this (getting a result of 50) being not possible :) –  403 Forbidden May 25 '13 at 9:09

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