 Φ(11) = 10, because again, 11 is a prime number and, so, cannot be coprime with any of the numbers smaller than it. This Φ(11) = 111 =10
 Φ(24) = 8. This is because the prime factors of 24 are 2 and 3, and there are 8 numbers i.e. 1, 5, 7, 11, 13, 17, 19 and 23, which are not divisible by 2 or 3, so all of them are coprime with 24.
 Φ(18) = {1, 5, 7, 11, 13, 17} = 6
 Φ(41) = 40, because it is a prime number, so it has no factors other than one or itself and cannot have common factors.
 Φ(7) = 6, again because it is a prime number and cannot share factors with any numbers smaller than it.
 Φ(6) = {1, 5} = 2
 Φ(25) = {1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24} = 20
Part 2
 Φ(7 x 4) = Φ(28) = 12
Φ(7) x Φ(4) = 6 x 2 = 12
Thus, Φ(7 x 4) = Φ(7) x Φ(4)
 Φ(6 x 4) = Φ(24) = 8
Φ(6) x Φ(4) = 2 x 2 = 4
Therefore, Φ(6 x 4) ≠ Φ(6) x Φ(4)
 Check whether or not Φ(mn) = Φ(m) x Φ(n) for at least two separate choices of n and m
 Φ(5 x 10) = Φ(50) = 20
Φ(5) x Φ(10) = 4 x 4 = 16
Thus, Φ(5 x 10) ≠ Φ(5) x Φ(10)
 Φ(7 x 6) = Φ(42) = 12
Φ(5) x Φ(10) = 4 x 4 = 16
Therefore, Φ(5 x 10) ≠ Φ(5) x Φ(10)

In some cases, Φ(mn) = Φ(m) x Φ(n), whilst in other cases this is not so. Investigate this situation.
There are endless values of m and n which I can try, but it is not possible to investigate all of them, so instead, I will start with the smallest positive integers, and go up steadily to see if I can find a pattern or a link between m and n when Φ(mn) = Φ(m) x Φ(n).
Table showing values of Φ(mn) when m=1
Table showing values of Φ(mn) when m=2
Table showing values of Φ(mn) when m=2
Since, the tables above do not show any clear link or pattern between m and n, I have now made two tables comparing the values of m and n that do work for the equation Φ(mn) = Φ(m) x Φ(n) on the right, and the values that don’t on the left. Maybe this will help me spot a relationship between them.
At a first look, it seems that for all the values of m and n that do work, one number is even and the other odd. However, there are some exceptions, which show that this cannot be the case. For example, 1 and 3, 1 and 5, and 3 and 5 are all combinations of m and n which are both odd and yet they fulfil the equation Φ(mn) = Φ(m) x Φ(n).
Another possible relationship between m and n is that one cannot be a multiple of another. This is because all the values of n in the table on the right are divisible by m, while none of the values of n in the table on the left are divisible by m. However, when I carried out some further investigation to prove this theory, I found various larger combinations where n is not divisible by m, so according to my theory, they should fulfil the equation Φ(mn) = Φ(m) x Φ(n),but they do not. Some of them are listed below:
This means, that there is an even more general rule which defines the relationship between m and n. The two tables comparing values of m and n that fulfil or do not fulfil the equation have been placed below again, this time with some of the ‘anomalous’ pairs of m and n added.
Now, the only remaining connection between all the pairs that do not work is that m and n have common factors, or are not coprime. All the pairs listed on the left that do work however, are ones where m is coprime with n.
Thus I can conclude that Φ(mn) = Φ(m) x Φ(n), only when m and n are coprime. The numbers must be coprime to ensure that there are no repeated factors.

If p and q are prime numbers investigate Φ(pnqm)
Before I investigate this situation, there are a number of rules I have to establish. First of all, for any prime number p, Φ(p)= p1, because it is only divisible by 1 and itself, and so, must be coprime with all the numbers smaller than it. For example, Φ(3) = {1,2} = 2 (which is 31) and Φ(7) = {1,2,3,4,5,6} = 6 (which is 7 1).
Next, if p is a prime number and pk is the power of that prime number, then Φ(pk) tells us the number of positive integers smaller than pk that are coprime with it.
The prime factor of pk is p, so all the numbers which share common factors with pk must be multiples of p e.g. p, 2p, 3p ... and so on. Thus all the numbers which are not multiples of p, must be coprime with pk. This means that, if the number of multiples of p smaller than then pk is m, then Φ(pk) = pk – m.
The value of m for any value of pk can be determined by. This can also be written as
pk1, as is shown here:
In other words, m=, pk1where m represents the number of multiples of p less than pk.
We know that pk1 gives us m, which is the number of multiples of p smaller than then pk, because it represents the number of lots of p in pk. For example, 33 can also be written 3x3x3 which is 32 (9) lots of 3, so in other words, in 33 there are 331 or 32 multiples of 3.
Another thing we need to know, is the rule we learnt from Part 3, that Φ(xy) = Φ(x) x Φ(y), when x and y are coprime. Applying this rule to Φ(pnqm) through substitution, if we say that x = pn and y = qm, then Φ(pnqm) = Φ(pn) x Φ(qm). The rule works because, as I explained earlier, the prime factors of pn and qm will be p and q respectively because they are both prime numbers, and so pn and qm will always be coprime too, because they cannot have any common factors.
So we now know how to find Φ(pn) for any prime value of p by using the formula Φ(pn) = pn  pn1 or Φ(qm) for any value of q using Φ(qm) = qm – qm1. In this case, we must understand that n and m are used as constants instead of k, and that the meaning is one and the same. Using this information and the knowledge that Φ(pnqm) = Φ(pn) x Φ(qm), we can then find out Φ(pnqm) for any prime values of p and q.