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I need help finding all natural numbers n such that $\phi(n)$=12.

My professor told us to use $\phi$(mn)=$\phi$(m)$\phi$(n).

I know that $12$ can be factored into $3 \times 4$, $12 \times 1$ and $6 \times 2$.

Jyrki Lahtonen
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ematth7
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  • Hint: no primes greater than $13$ can be factors of the number $n$, unless $n=13$, because $13$ has $12$ coprime numbers less than it. You can improve greatly on this. – Colm Bhandal Dec 08 '15 at 11:17

1 Answers1

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HINT:

As $\phi(\prod p_r^{x_r})=\prod\phi(p_r^{x_r})$ where $p_r$s are distinct primes

So, if $p^a|n, \phi(p^a)|\phi(n)=12$

Clearly $p\le13$

So, we need to check for $p=2,3,5,7,11,13$

Now, $\phi(p^a)=p^{a-1}(p-1)$ for integer $a\ge1$

lab bhattacharjee
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