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Given $p, q, r$ are prime numbers, and $n = pqr = 398585546147$, how do I obtain $\phi(pqr) \bmod 32$?

I tried equating it to $(p - 1)(q - 1)(r - 1) \bmod 32$ but I do not know how to proceed from there. Calculators are not allowed for this question as well.

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    $\Phi(pqr)$ is either 8, 16, 24 or 0, because $(p-1)$ -and similarly the others- are even, hence $8$ divides their product. I don't know more at the moment, but I'm keeping on thinking.. – 2017-02-27
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    @poopapole: Are you sure the problem doesn't provide more information? – 2017-02-27
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    @poopapole: Also, given that you wrote: "Calculators are not allowed for this question as well", it suggests that the question is one problem from a set of problems. Is the problem from a book? If so, which one? If not, what is the source of the problem? – 2017-02-27
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    @quasi Hi, sorry I did not fully explain the context of the question. It was an MCQ question in an exam practice paper. Out of all the answers, only 1 of it was even and that was supposed to be the answer. After consulting my prof, what he wanted was not for us to attempt to solve and obtain the values per se, but deduce the answer through knowing the properties of primes numbers, like what you have listed below. Thank you so much though for your answer, I really learnt a lot from it. – 2017-02-28

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Partial answer . . .

Although it was not explicitly specified, I suspect it was intended that the primes $p,q,r$ are distinct. Thus, in what follows, I'll assume $p,q,r$ are distinct.

As you noted, $n = pqr \implies \phi(n) = (p-1)(q-1)(r-1)$.

\begin{align*} n \text{ odd}&\implies p,q,r \text{ are odd }\\[6pt] &\implies p-1,q-1,r-1 \text{ are even}\\[6pt] &\implies 8\,|\,(p-1)(q-1)(r-1)\\[6pt] &\implies \phi(n) \equiv 0 \pmod 8\\[6pt] \end{align*}

Since $n \equiv 3 \pmod 4$, either

  • Case $(1)\;\;$Exactly one of $p,q,r$ is congruent to $3$ mod $4$, and the remaining two are congruent to $1$ mod $4$.
  • Case $(2)\;\;p,q,r$ are all congruent to $3$ mod $4$.

For case $(1)$, two of $p-1,q-1,r-1$ are multiples of $4$, and the remaining one is even, hence $\phi(n)=0 \pmod{32}$.

For case $(2)$, each of $p-1,q-1,r-1$ is even, but not a multiple of $4$. It follows that $\phi(n)$ is a multiple of $8$, but not a multiple of $16$, hence $\phi(n) = 8 \text{ or }24 \pmod{32}$.

Thus, $\phi(n)\text{ mod }32$ is one of $0,8,24$.

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    I guess even this requires $p$, $q$, and $r$ to be distinct, which isn't obvious to me. But maybe we're allowed to assume that. – 2017-02-28
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    I was assuming that problem intended 3 distinct primes, but yes, if that was intended, it should have been explicitly stated. I'll edit my answer to indicate that my reasoning assumes 3 distinct primes. – 2017-02-28