# Where is/are the problem(s)?

$\large\sum_{d\mid 29233}\frac{\phi(\phi(d))}{\phi(d)}$

Given that the sum above is equal to $\dfrac{p}{q}$, where $p$ and $q$ are coprime positive integers, find $p+q+2720$

Clarification

• The sum is taken over all positive integer divisors $d$ of $29233 = 23 \times31\times41$.

• $\phi(\cdot)$ denotes the Euler's totient function.

• $a\mid b$ denotes "$a$ divides $b$".

## For prime numbers:

If $d-1=p_1 ^a\times p_2^b\times p_3^c\times\cdots$, where $p_n$ are prime numbers.

\large\begin{aligned} \cfrac{\phi(\phi(d))}{\phi(d)} &=\phi\left (\cfrac{\phi(d)}{d} \right )\\ &=\phi\left (\cfrac{d\left ( 1-\cfrac{1}{d} \right )}{d} \right )\\ &=\phi\left ( 1-\cfrac{1}{d} \right )\\ &=\phi\left ( \cfrac{d-1}{d} \right )\\ &=\cfrac{\phi\left ( d-1 \right )}{\phi\left ( d \right )}\\ &=\cfrac{(d-1)\left ( \cfrac{p_1-1}{p_1} \right )\left ( \cfrac{p_2-1}{p_2} \right )\left ( \cfrac{p_3-1}{p_3} \right )\cdots}{d\times\cfrac{d-1}{d}}\\ &=\left ( \cfrac{p_1-1}{p_1} \right )\left ( \cfrac{p_2-1}{p_2} \right )\left ( \cfrac{p_3-1}{p_3} \right ) \end{aligned}

## For composite numbers:

Let's $\Gamma(d)$ denote $\cfrac{\phi(\phi(d))}{\phi(d)}$

If $d=p_1 ^a\times p_2^b\times p_3^c\times\cdots$:

$\large\Gamma(d)=\Gamma(p_1)^a\times \Gamma(p_2)^b\times \Gamma(p_3)^c\times\cdots$

## Solution:

The divisors of $29233$ are: $\begin{array}{cccc} 1&23&31&41\\ 23\times31\times41&23\times31&23\times41&31\times41 \end{array}$

#### $1$:

$1$

#### $23$:

$\cfrac{1}{2}\times\cfrac{10}{11}=\cfrac{5}{11}$

#### $31$:

$\cfrac{1}{2}\times\cfrac{2}{3}\times\cfrac{4}{5}=\cfrac{4}{15}$

#### $41$:

$\cfrac{1}{2}\times\cfrac{4}{5}=\cfrac{2}{5}$

#### $23\times31$:

$\Gamma(23)\times\Gamma(31)=\cfrac{4}{33}$

#### $23\times41$:

$\Gamma(23)\times\Gamma(41)=\cfrac{2}{11}$

#### $31\times41$:

$\Gamma(31)\times\Gamma(41)=\cfrac{8}{75}$

#### $23\times31\times41$:

$\Gamma(23)\times\Gamma(31)\times\Gamma(41)=\cfrac{8}{165}$

\large \begin{aligned} \displaystyle\sum_{d\mid 29233}\cfrac{\phi(\phi(d))}{\phi(d)} &=1+\cfrac{5}{11}+\cfrac{4}{15}+\cfrac{2}{5}+\cfrac{4}{33}+\cfrac{2}{11}+\cfrac{8}{75}+\cfrac{8}{165}\\ &=\cfrac{5\times(11\times3\times5+5\times3\times5+4\times11+2\times3\times11+4\times5+2\times3\times5+8)+8\times11}{11\times3\times5\times5}\\ &=\cfrac{5\times(165+75+44+66+20+30+8)+88}{825}\\ &=\cfrac{2128}{825} \end{aligned}

So the solution should be: $2128+825+2720=5673$

# Where is the problem?

"And, if you observe closely, the right answer is 2953 = 2128 + 825. So maybe you need not add 2720"

This means the problem was wrong.

Note by Páll Márton (no activity)
1 month, 3 weeks ago

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@Páll Márton (no activity) - What's with the (no activity)??? Does it mean you're offline?

- 1 month ago

I think your answer is correct, I just want to ask whether this function is multiplicative or not?

And, if you observe closely, the right answer is 2953 = 2128 + 825. So maybe you need not add 2720

- 1 month, 3 weeks ago

• Yeah. This function is multiplicative.

• 2953 is working(lol). I don't understand it.

• I will use composite.

• Thank you!

- 1 month, 3 weeks ago

Also, it should be 'composite' instead of 'complex', unless there are different ways of saying 'not-a-prime'

- 1 month, 3 weeks ago

- 1 month, 3 weeks ago

Hi Páll, the summation is not equal to $\frac{2128}{825}$.

In your solution, the "Gamma function" is wrong.

Staff - 1 month, 3 weeks ago