Question

Set of eleven integers, there are two integers whose difference in divisible by 10

Answer 1

A set of 7 distinct positive integers will either contain

n∗10+xn∗10+xand m∗10+xm∗10+x, so their difference is (n−m)∗10(n−m)∗10 or

n∗10+xn∗10+xand m∗10+(10−x)m∗10+(10−x), so their sum is (n+m)∗10+x+(10−x)=(n+m+1)∗10(n+m)∗10+x+(10−x)=(n+m+1)∗10.

where x,n,m∈Ix,n,m∈I and 0≤x<100≤x<10.

This is because, when you pick any set of 6 or more integers modulo 10, it will have the property that there will either be two xxs, or an xx and a (10−x)(10−x).

Let’s try to construct as large a set that doesn’t have the above property:

00, can't have 1010, 2020, etc, as 0+10=0−10=0mod100+10=0−10=0mod10

11, can’t have 9,19,etc9,19,etc, as 1+9=0mod101+9=0mod10

22, can’t have 88, as 2+8=0mod102+8=0mod10

33, can’t have 77, as 1+9=0mod101+9=0mod10

44, can’t have 66, as 1+9=0mod101+9=0mod10

Can’t have 55, as 5+5=5−5=0mod105+5=5−5=0mod10, can't have 1111 as 11−1=0mod1011−1=0mod10, can't have 12, as 12−2=0mod1012−2=0mod10, etc. As you can see, we can have at most 5 elements that don’t have the property. Any more and the property must be satisfied.