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I have two questions.

  1. If $f$(n) is a multiplicative function defined on the positive integers, is

$g(n)=$$\frac{f(n)}{n}$ multiplicative as well?

I think the answer is yes, but I don't know how to prove it.

  1. Evaluate $$\sum_{d|2016} \frac{\phi(d)}{d}$$

where $\phi(n)$ is the totient function.

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  • $\begingroup$ The product of two multiplicative functions is multiplicative. $\endgroup$ – Lord Shark the Unknown May 16 '17 at 6:03
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As regards the second question, note that $$g(n):=\sum_{d|n} \frac{\phi(d)}{d}$$ is multiplicative (it is the Dirichlet convolution of two multiplicative functions).

Moreover, $n=2016=2^5\cdot3^2\cdot 7$, and for any prime $p$, $$g(p^k)=\sum_{d|p^k} \frac{\phi(d)}{d}=1+\sum_{j=1}^k\frac{p^j-p^{j-1}}{p^j}=1+k\left(1-\frac{1}{p}\right).$$ Hence $$g(2016)=g(2^5)\cdot g(3^2)\cdot g(7)=\frac{7}{2}\cdot\frac{7}{3}\cdot \frac{13}{7}=\frac{91}{6}.$$

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  • $\begingroup$ That helps a lot. I'm getting 323/42 with that formula. $\endgroup$ – shrindle May 16 '17 at 6:14
  • $\begingroup$ It should be 91/6. Check your computations. $\endgroup$ – Robert Z May 16 '17 at 6:16
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1) Show that for positive integers $m,n$, one has $g(mn) =g (n)g(m) $. Note this is precisely the definition of a function being 'multiplicative.'

2) We have $d$ as any divisor of 2016. Now observe that $2016=2^5×3^2×7$.

Use the fact that the Euler Totient function is multiplicative for coprime numbers to simplify your counting process.

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  • $\begingroup$ That's the part I'm confused about. How do I separate this sum in such a way that I can "cleanly" solve it utilizing the multiplicative functions you discuss. I do realize that Euler's totient function is multiplicative when the numbers are relatively prime, but I'm still confused. $\endgroup$ – shrindle May 16 '17 at 6:04
  • $\begingroup$ You should have g(mn) = f(mn)/mn. Then because f is multiplicative by assumption, write f (mn) =f(n)f(m). Can you see it now? $\endgroup$ – thedilated May 16 '17 at 6:07
  • $\begingroup$ I understand the first question. What I'm confused by is the second one with the sum. $\endgroup$ – shrindle May 16 '17 at 6:09
  • $\begingroup$ Let us see an e.g. Note that 63 is a divisor of 2016. So one of the terms in the summand is phi(63)/63. But 63 = 7×9 and gcd(7,9) =1. So we can write phi(63) = phi(7)phi (9) which is v easy to compute. $\endgroup$ – thedilated May 16 '17 at 6:15

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