Where is/are the problem(s)?

Given this problem(link here):

d29233ϕ(ϕ(d))ϕ(d)\large\sum_{d\mid 29233}\frac{\phi(\phi(d))}{\phi(d)}

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


  • The sum is taken over all positive integer divisors dd of 29233=23×31×4129233 = 23 \times31\times41.

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

  • aba\mid b denotes "aa divides bb".

For prime numbers:

If d1=p1a×p2b×p3c×d-1=p_1 ^a\times p_2^b\times p_3^c\times\cdots, where pnp_n are prime numbers.

ϕ(ϕ(d))ϕ(d)=ϕ(ϕ(d)d)=ϕ(d(11d)d)=ϕ(11d)=ϕ(d1d)=ϕ(d1)ϕ(d)=(d1)(p11p1)(p21p2)(p31p3)d×d1d=(p11p1)(p21p2)(p31p3)\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 Γ(d)\Gamma(d) denote ϕ(ϕ(d))ϕ(d)\cfrac{\phi(\phi(d))}{\phi(d)}

If d=p1a×p2b×p3c×d=p_1 ^a\times p_2^b\times p_3^c\times\cdots:

Γ(d)=Γ(p1)a×Γ(p2)b×Γ(p3)c×\large\Gamma(d)=\Gamma(p_1)^a\times \Gamma(p_2)^b\times \Gamma(p_3)^c\times\cdots


The divisors of 2923329233 are: 123314123×31×4123×3123×4131×41\begin{array}{cccc} 1&23&31&41\\ 23\times31\times41&23\times31&23\times41&31\times41 \end{array}

















d29233ϕ(ϕ(d))ϕ(d)=1+511+415+25+433+211+875+8165=5×(11×3×5+5×3×5+4×11+2×3×11+4×5+2×3×5+8)+8×1111×3×5×5=5×(165+75+44+66+20+30+8)+88825=2128825\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=56732128+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

No vote yet
1 vote

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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

Mahdi Raza - 1 month, 3 weeks ago

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

Mahdi Raza - 1 month, 3 weeks ago

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  • Yeah. This function is multiplicative.

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

  • I will use composite.

  • Thank you!

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

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Hi Páll, the summation is not equal to 2128825\frac{2128}{825} .

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

Brilliant Mathematics Staff - 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?

Percy Jackson - 1 month ago

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