Answer:

• subtract eqn(b) from eqn(a);

• find x :

Answer:
rdh
Step-by-step explanation:
The prisoners could do a pattern that could signal who has already been in the room. So for example, one light off, both lights off, one light on, both lights off. But with left and right lights to make the sequence last as long as possible. Then after that sequence to have the last one be the signal that everyone has been in there. I dunno this is tricky lol
We have been given that Judy’s brother Sam has a collection of 96 comic books. We have been asked to find the 10 ways Sam could divide his comic books into equal groups.
This can be done as follows:
96 can be written as
. Thus, we can either have 2 groups of 48 comics or 48 groups of 2 comics. As we can see we have 2 ways to divide the comic books in equal groups.
Likewise, 96 can also be written as
. Here, we can similarly see that we can either have 3 groups of 32 comics or 32 groups of 3 comics. As we can again see we have 2 more ways to divide the comic books in equal groups, taking the total number of ways to 4.
Continuing in this manner we will see that we can have 4 groups of 24 or 24 groups of 4, 6 groups of 16 or 16 groups of 6 and lastly 12 groups of 8 or 8 groups of 12 comic books, thus taking the count to a <u>total of 10 ways</u> in which Sam can divide his comic books into equal groups.
Our question has been answered.