# Find the maximum number of right claims

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?

• can't they all be $8!$?
– JMP
Oct 7, 2017 at 23:09
• @JonMarkPerry it cant, then for example all of them can be divided by 2 :)
– Oray
Oct 7, 2017 at 23:10
• Are what Alice written are 8 distinct integers ? Oct 8, 2017 at 1:19
• If you can divide by 4 without remainders then you can divide by 2 without remainders Oct 8, 2017 at 1:24

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}

• A number is expected as a final answer to this question, I suppose ! Oct 8, 2017 at 1:15
• @MeaCulpaNay if you look carefully, you will find a number in 'So:'. Also, I have noticed that you write terminating punctuation as its own separate word in your comments - usually it is appended to the final word instead (e.g. 'I suppose!'). [Note: that is not criticism, just an observation. If you are learning English, I am happy to help] Oct 8, 2017 at 1:24