Find the maximum number of right claims

Question

Alice wrote $8$ positive numbers on the board and Bob claims that:

• Exactly two of them can be divided by $2$ without remainders.
• Exactly three of them can be divided by $3$ without remainders.
• Exactly four of them can be divided by $4$ without remainders.
• Exactly five of them can be divided by $5$ without remainders.
• Exactly six of them can be divided by $6$ without remainders.
• Exactly seven of them can be divided by $7$ without remainders.
• Exactly eight of them can be divided by $8$ without remainders.

But as you suspected some of these claims were not right.

What is the maximum number of claims that could be right?

Answer 1

I'm going to number the claims from 2 to 8, then it is quite easy to see that:

A claim eliminates its factors and multiples

I.e.:

8 eliminates 2 and 4
4 eliminates 2 and 8
2 eliminates 4, 6 and 8
6 eliminates 2 and 3
3 eliminates 6

So:

We can only have one of the claims from the set $\{2,4,8\}$, and one from $\{3,6\}$. So we can have at most 4 true claims.

Example of maximal configuration:

$\begin{align}2\times3\times5\times7&=210\\2\times3\times5\times7&=210\\3\times5\times7&=105\\5\times7&=35\\5\times7&=35\\&7\\&7\\&1\end{align}$