Answer:
There are 64,000 possible codes. Jillian is incorrect, since he added the possible options in each trial, instead of multiplied them(he did 40+40+40 instead of 40*40*40).
Step-by-step explanation:
From 0 to 39, there are 40 numbers.
The code has three values:
V1 - V2 - V3
When we have n trials with m possible options, the total number of opitons is:

In this question:
3 trials(values), with 40 options. So

There are 64,000 possible codes. Jillian is incorrect, since he added the possible options in each trial, instead of multiplied them(he did 40+40+40 instead of 40*40*40).
1 plus to the negative Peter plus the relative integers
It would be a decimal which is 3.363636363636364 and if you convert it into a fraction it is 9090909091/25000000000 so in real terms it is 3 students per computer.