Poisoned wine

I give the answer for 7 bottles, it’s the same for 100. Let’s number the bottles in binary from 0 to 6 :

0<->000

1<->001

2<->010

3<->011

4<->100

5<->101

6<->110

We then need $\boxed{3}$ rats Why? Each rats tests the bottles where there is a « 1 » in the k position for k in {1,2,3}. Then if the first rat dies the poisoned bottles is necessarily the number 4 (always starting from 0). You have then 7 different cases whether it’s only the rat 1,2 or 3 who dies or it’s both rats 1 and 3 or 1 and 2 or 2 and 3 or 1 and 2 and 3 who die. Same logic for 100 which is 1100100 in binary

Number the 100 bottles of wine from 0 to 99 and then convert them to base 2 that is from $0_2$ to $1100011_2$ . There is a total of 7 binary digits. Now use 7 rats to represent the 7 $2^n$ -digit $d_n$ , where $n=0,1,2,3,4,5,6$ . The $d_n$ rat is to drink from the bottle with $d_n = 1$ . Since each of the 100 bottles has a unique 7-digit binary code, The poisoned bottle will kill the specific digital rats. For example, if none of the 7 rats dies, then bottle 0 is poisonous. If $d_0$ , $d_1$ and $d_2$ rats die, then bottle 7 ( $111_2$ ) is poisonous. Therefore, we need only $\boxed{7}$ rats of which at most 6 will be killed.

$\begin{array} {crrrrrrr} \text{Decimal} & d_6 & d_5 & d_4 & d_3 & d_2 & d_1 & d_0 \\ \hline 0 & 0 & 0 &0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 &0 & 0 & 0 & 0 & 1 \\ 2 & 0 & 0 &0 & 0 & 0 & 1 & 0 \\ \cdots & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\ 97 & 1 & 1 &0 & 0 & 0 & 0 & 1 \\ 98 & 1 & 1 &0 & 0 & 0 & 1 & 0 \\ 99 & 1 & 1 &0 & 0 & 0 & 1 & 1 \end{array}$