Let's Compute this Manually!

I have only recently gained level 4 in Number Theory and I am still a learner, so I am not able to provide a rigorous mathematical solution.

For $10 | \phi(n)$ , either the prime factorisation of $n$ should contain 5 and 2 or 5 and 2 should be introduced while calculating $\phi(n)$ from $n$ . I gave more importance to 5 than 2 because it occurs less frequently than 2 in numbers from 1 to 50. I know this is not a satisfactory reason, which is why I said I do not have a rigorous solution.

Let us take the first case. In case there is only one 5 in the prime factorisation of $n$ , it will get cancelled out when we calculate $\phi(n)$ . So there must be two 5's in the prime factorisation of $n$ . Only 25 and 50 satisfy this condition. On calculating $\phi(50) \text{and} \phi(25)$ , we do find that they are divisible by 10.

The second case will only come when the prime factorisation of $n$ contains a prime $m$ such that $m -1$ is a multiple of 5. The only possible values of $m$ are 11, 31 and 41. So, the possible values for $n$ in this case are 11, 22, 31, 33, 41 and 44. Again, checking $\phi(n)$ for these values of $n$ confirms the initial condition.

I don't have any argument to show that these are the only numbers satisfying this condition. I would appreciate it if someone would help me in this regard.

So, the answer is $11 + 22 + 25 + 31 + 33 + 41 + 44 + 50 = \boxed{257}$

Case 1: $\phi(p) = p - 1 = 10k$ for any prime p in the form of 10k+1 for some positive integers k

As we can see that p = 11, 31, 41 satisfy the condition.

Case 2: $\phi(11k) = \phi(11)\phi(k) = 10\phi(k)$ for some k such that $gcd(11,k) = 1$

We don't need to consider the values k because it already restricts that $n \leq 50$ .

So n = 11, 22, 33, 44 satisfy the condition.

Case 3: Some specific numbers

$\phi(25) = \phi(5^{2}) = 5^{2} - 5^{1} = 20$

$\phi(50) = \phi(2 \times 25) = \phi(2)\phi(25) = 20$

So the possible values are 11, 22, 33, 44, 31, 41, 25, 50

The sum $= \boxed{257}$