Multiplicative Perfect

Number Theory Level 2

Let $\kappa(n)$ denote the product of all the divisors of positive integer $n$ (inclusive of 1 and itself).

Let $n$ be called a multiplicative perfect when $\kappa(n)=n^2.$

Find the sum of the first three multiplicative perfects greater than $1$ .

The answer is 24.

3 solutions

Kalpok Guha
Mar 18, 2015

$\kappa(n)=n^{\frac{\tau(n)}{2}}$ [ $\tau(n)$ denotes the number of divisors of $n$ ]

Thus if $\kappa(n)=n^2$

Then $\tau(n)=4$

The first three integers with $\tau(n)=4$ are $6,8,10$

The sum $6+8+10=24$

This is not good solution as I did not prove the first assumption.Friends and foes if you have a good solution please post it.

Lol, don't worry, there are no foes here. We're just all friends :)

Trevor Arashiro - 6 years, 2 months ago

I know you are my friend but anyone may be my foe

Kalpok Guha - 6 years, 2 months ago

$\kappa(16) = 1*2*4*8*16 = 1024 = {32}^{2}$ and $\tau(16) = 5$ . I want to see a better (more general) solution. PS : I'm not your foe.

Bhaskar Sukulbrahman - 6 years, 2 months ago

I don't see the problem here. The formula works even if $n$ is a perfect square. Demonstrating it for the example you provided:

$\large \kappa(16)=(16)^{\frac{\tau(16)}{2}}=16^{5/2}=4^5=1024=32^2$

It agrees with the result you found manually. If you want, you can even check out the Math SE discussion on this. Here 's the link to it.

Prasun Biswas - 6 years, 2 months ago

Oh yes thanks Prasun da .I did not notice that

Kalpok Guha - 6 years, 2 months ago

But $\kappa(16) = {32}^{2} \neq {16}^{2}$ and $\tau(16) \neq 4$ as it is described in the solution in line 2 and line 3.

Further more, $\kappa(64) = {64}^{7/2} \neq {64}^{2}$ is not a perfect square. $\tau(64) = 7 \neq 4$ .

Bhaskar Sukulbrahman - 6 years, 2 months ago

@Bhaskar Sukulbrahman – Well, who ever said that $16$ and $64$ are multiplicative perfects? Note that the terms "multiplicative perfect", "perfect number" and "perfect square" have different meanings and are in no way synonymous.

I'm elaborating his solution here. See if you can understand. If any part confuses you, notify me.

Solution: For any positive integer $n\gt 1$ , we have $\large \kappa(n)=n^{\frac{\tau(n)}{2}}$ .

Also, if $n$ is a multiplicative perfect, then, by the definition given in the problem, we must have $\kappa(n)=n^2$ . Since we are looking for multiplicative perfects $\gt 1$ , both the results we have are applicable for those numbers. So, we proceed to solve for multiplicative perfects $n\gt 1$ as follows:

$\large n^{\frac{\tau(n)}{2}}=n^2\implies \frac{\tau(n)}{2}=2\implies \tau(n)=4=1\times 4=2\times 2$

By the theorem of divisors and the result we obtained just now, the multiplicative perfects $(n)$ will be of the form $p_1p_2$ or $(p_3)^3$ where all $p_i$ 's are primes.

Now, starting with the smallest primes, we can easily get $2\times 3=6$ , $2^3=8$ and $2\times 5=10$ as the required values. Any other value that can be formed will definitely be bigger than these three values.

P.s- You can easily see that $16$ and $64$ do not fit any one of the two general forms and hence will not be a multiplicative perfect.

Prasun Biswas - 6 years, 2 months ago

@Prasun Biswas – Thanks for the explanation... I get it now.

Bhaskar Sukulbrahman - 6 years, 2 months ago

@Prasun Biswas – Thank you for explaining the solution.Really Thanks

Kalpok Guha - 6 years, 2 months ago

The formula $\ \kappa (n) = n^{\frac{\tau (n)}{2}}$ only works for non-square n because you can pair up factors a,b of n (ab=n) with $\ a < \sqrt{n} < b$ from which you have $\frac{\tau (n)}{2}$ such pairs each contributing a single multiple of n. However, for $\ n= k^2$ we only have one factor of k in $\ \kappa (n)$ so we must divide $\ n^{\frac{\tau (n)}{2}}$ by k so that $\ \kappa (n) = n^{\frac{\tau (n) -1}{2}}$ in this case so we require $\ \tau (k^2) = 5$ . This is nice as 5 is prime so you find that in fact $\ \tau (n) = 5$ if and only if $\ n= p^4$ for some prime p (Can you show this? If not I can elaborate).

Curtis Clement - 1 year, 10 months ago

I am completely confused here. I thought the question was asking for the first three numbers from which the product of their divisors was equal to their own square, in which case the first three prime numbers would suffice. Where is this about needing to have four divisors coming from?

Tristan Goodman - 3 months, 3 weeks ago

Ankit Kumar Jain
Mar 19, 2015

You just have to find the numbers having 4 divisors, and that's it.

Plz explain your solution

Kalpok Guha - 6 years, 2 months ago

See the product of factors = $n^{\frac{N}{2}}$ , where n is the number and N is the number of factors. So if N is 4 , then the result is $n^{2}$ .

Ankit Kumar Jain - 6 years, 2 months ago

Naman Kapoor
Mar 19, 2015

My sol is bit different we should the nos that have two different divisors other than n and 1 which r 6,8,10 and u should not consider anyone as ur foe @kalpok guha

Multiplicative Perfect

The answer is 24.

3 solutions

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