Help with a simple number theory proof Prove, that no matter how we give $8$ three-digit numbers, we can always choose $2$ of them, which we write next to each other, that six-digit number will be divisible with $7$.
Example: I have $123$, and $456$ and I can make $123456$ or $456123$ as a $6$-digit number.
Edit: I think it has to do something with 8. From 8 numbers you can always choose 2 which are congruent to each other(mod7), and these 2 numbers, wrote down as 6 digit number are good.

Hints:


*

*$1000\equiv -1\mod 7$

*$8>7$

*Pigeon Hole Principle

