I've been asked the following question:

What five odd integers from the set $\{1, 3, 5, 7, 9, 11, 13, 15\}$ that when summed together equals to $30$? Note that any integer can be used more than once.

If my limited knowledge of maths is correct, there should be no answer, as no odd number of odd integers summed together can give an even number.

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    $\begingroup$ Your answer is correct $\endgroup$ – ajotatxe May 18 '14 at 15:57
  • $\begingroup$ Is that the exact wording of the question you have been asked? $\endgroup$ – Mark Bennet May 18 '14 at 15:57
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    $\begingroup$ A trick question!!! $\endgroup$ – PA6OTA May 18 '14 at 15:59
  • $\begingroup$ It's actually a riddle kind of question i got over mobile messaging. $\endgroup$ – Malfunction May 18 '14 at 16:01
  • $\begingroup$ @PA6OTA seems to have the same idea as I do, but the wording matters as I can see two possible tricks. $\endgroup$ – Mark Bennet May 18 '14 at 16:04

As straightforward mathematics there is no answer.

As anyone who has ever placed hymn numbers in a hymn board will know, it is possible to turn $9$ upside down to get $6$, and if this is allowed by the wording you can get a sum of $30$.

Likewise if it is odd numbers which are chosen, but the digits rather than the numbers which are added, the set $3,5,7,9,15$ gives $3+5+7+9+1+5=30$ - again this depends on precisely how the question is worded.

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    $\begingroup$ The wording doesn't seem to imply either of these cases. I guess it's just somebody screwing around after all. $\endgroup$ – Malfunction May 18 '14 at 16:27

By definition, odd integers are of the form $2n+1$ where $n\in \mathbb{Z}$. Since we want the sum of $5$ odd integers to be equal to $30$ this would imply that for $a,b,c,d,e \in \mathbb{Z}$ $$(2a+1)+(2b+1)+(2c+1)+(2d+1)+(2e+1)=2(a+b+c+d+e+2)+1=30$$ which is impossible since $(a+b+c+d+e+2)\in \mathbb{Z}$ and $2(a+b+c+d+e+2)+1$ is an odd integer by definition.

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  • $\begingroup$ plus one because this looks like a mathematical proof to me, and that's what I was googling for. $\endgroup$ – matiu Apr 7 '15 at 11:59

I think this is a valid question. Normal addition base 10 you can't achieve this since

(odd+odd)+(odd+odd)+odd=even+even+odd=even+odd= always odd or (2x+1)+(2y+1)+(2z+1)+(2u+1)+(2v+1)=2(x+y+z+u+v)+5 is odd

A good answer seems to be an addition with Base 5 arithmetic


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