Question #45f47

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Question #45f47

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Answer 1 · 2017-02-28T12:00:48

It is impossible. See explanation.

Explanation:

The formal proof could look like:

Let the odd numbers be:

$n_{1} = 2 k_{1} + 1$ $n_1=2k_1+1$ ,
$n_{2} = 2 k_{2} + 1$ $n_2=2k_2+1$ ,
$n_{3} = 2 k_{3} + 1$ $n_3=2k_3+1$ ,
$n_{4} = 2 k_{4} + 1$ $n_4=2k_4+1$ , and
$n_{5} = 2 k_{5} + 1$ $n_5=2k_5+1$ ,

The sum would be:

$S = 2 \cdot (k_{1} + k_{2} + k_{3} + k_{4} + k_{5}) + 5 =$ $S=2*(k_1+k_2+k_3+k_4+k_5)+5=$

$2 \cdot (k_{1} + k_{2} + k_{3} + k_{4} + k_{5} + 2) + 1$ $2*(k_1+k_2+k_3+k_4+k_5+2)+1$

The las experssion is an even number plus 1, so it is an odd number.

Informally you could say that:

The sum of an even number of odd numbers is always even, but the sum of odd number of odd numbers is always odd

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Question #45f47

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