Can 5 odd numbers be added to get 30?

Sep 10, 2015

No.

Explanation:

The sum of an odd number of odd numbers is odd.

Every odd number can be written as 2i+1 for an integrer, $i$, so
For this question in particular, if we add:

$2 i + 1$
$2 j + 1$
$2 k + 1$
$2 l + 1$
$2 m + 1$

We get:

$2 \left(i + j + k + l + m\right) + 5$

 = 2((i+j+k+l+m+2) +1#

which is of the form $2 n + 1$, so it is an odd number and cannot be equal to $30$.

Sep 10, 2015

In normal integer arithmetic - no.

In modular arithmetic - yes with any odd modulo > 30.

Explanation:

Normal arithmetic

Suppose your $5$ odd numbers are:

${a}_{1} = 2 {k}_{1} + 1$
${a}_{2} = 2 {k}_{2} + 1$
${a}_{3} = 2 {k}_{3} + 1$
${a}_{4} = 2 {k}_{4} + 1$
${a}_{5} = 2 {k}_{5} + 1$

where ${k}_{1} , {k}_{2} , {k}_{3} , {k}_{4} , {k}_{5} \in \mathbb{Z}$

Then:

${a}_{1} + {a}_{2} + {a}_{3} + {a}_{4} + {a}_{5}$

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

which is odd.

$30$ is even - not odd - so ${a}_{1} + {a}_{2} + {a}_{3} + {a}_{4} + {a}_{5} \ne 30$

Arithmetic modulo $31$

Let ${a}_{1} = {a}_{2} = {a}_{3} = {a}_{4} = 13$ and ${a}_{5} = 9$

Then ${a}_{1} + {a}_{2} + {a}_{3} + {a}_{4} + {a}_{5} = 61 = 30$ modulo $31$

In fact in modular arithmetic modulo an odd base all numbers are both odd and even.