# Question #45f47

##### 1 Answer
Feb 28, 2017

It is impossible. See explanation.

#### Explanation:

The formal proof could look like:

Let the odd numbers be:

${n}_{1} = 2 {k}_{1} + 1$,
${n}_{2} = 2 {k}_{2} + 1$,
${n}_{3} = 2 {k}_{3} + 1$,
${n}_{4} = 2 {k}_{4} + 1$, and
${n}_{5} = 2 {k}_{5} + 1$,

The sum would be:

$S = 2 \cdot \left({k}_{1} + {k}_{2} + {k}_{3} + {k}_{4} + {k}_{5}\right) + 5 =$

$2 \cdot \left({k}_{1} + {k}_{2} + {k}_{3} + {k}_{4} + {k}_{5} + 2\right) + 1$

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