Random Integers Sum or Difference

We will try to find a largest set of integers, such that no pair has either a sum or a difference that is a multiple of 10. To do this, consider $\mathbb{Z}/10\mathbb{Z}$ , the integers modulo 10; rather than the usual coset representation, we will use $\mathbb{Z}/10\mathbb{Z} = \left\{ \overline{\text{-}4}, \overline{\text{-}3}, \overline{\text{-}2}, \overline{\text{-}1}, \overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5} \right\}$ .

[In simple terms, we are partitioning the set of integers into ten sets, based on their residues or remainders when they are divided by 10. Any two integers $a$ and $b$ are members of the same coset iff $a - b$ is an integer multiple of 10. For example, 2, 52, -8 and -28 are all members of $\overline{2}$ , because the difference of any two of those numbers is a multiple of 10; similarly, -3, -43, 7 and 67 are all members of $\overline{\text{-}3}$ .]

We group the cosets as follows: $A = \left\{ \overline{0} \right\}, B = \left\{ \overline{5} \right\}, C = \left\{ \overline{1}, \overline{\text{-}1} \right\}, D = \left\{ \overline{2}, \overline{\text{-}2} \right\}, E = \left\{ \overline{3}, \overline{\text{-}3} \right\}$ and $F = \left\{ \overline{4}, \overline{\text{-}4} \right\}$

First, by the very definition of the cosets, two integers will only have a difference of a multiple of 10 if they are members of the same coset; so if we choose at most one integer from each coset we will avoid any difference of a multiple of 10.

Next, it is easy to see that one integer in $\overline{0}$ will only form a sum of a multiple of 10 with another member of $\overline{0}$ ; the same can be said for members of $\overline{5}$ . Also, an integer in $\overline{1}$ will only form a sum of a multiple of 10 with a member of $\overline{\text{-}1}$ , an integer in $\overline{2}$ will only form a sum of a multiple of 10 with a member of $\overline{\text{-}2}$ , etc.

Thus as long as we only take one integer from each of the coset groups $A, B, C, D, E$ and $F$ , we can avoid any sums or differences that are multiples of 10, but if we take more than one integer from any one of the groups we will guarantee a sum or difference of a multiple of 10.

Therefore a set of integers can have at most six elements in order to avoid a sum or difference of a multiple of 10, which means that to guarantee either a sum or difference of a multiple of 10, a set would have to contain a minimum of $\boxed{\text{seven}}$ integers.

P.S. I am very unused to expressing myself in the language of quotient groups, cosets, etc, so any suggestions on how I could improve any of the phrasing used here would be welcome and appreciated.