Multifactorial and divisors

Number Theory Level 3

Fill in the blank.

$N$ has $3$ prime factors.

$N^2$ has $7!!$ positive divisors.

$N^3$ has $10!!!$ positive divisors.

$\text{\_\_\_\_}$ has $13!!!!$ positive divisors.

Note: You might want to read up the definition of multifactorials first.

N^6

N^4

N^5

N^7

1 solution

A Former Brilliant Member
May 22, 2017

Let $N=p^aq^br^c$ , where $p,q,r$ are primes ans $a,b,c$ are positive integers.

Thus $N^2 =p^{2a}q^{2b}r^{2c}$

The no. of divisors of $N^2$ are $(2a+1)\times(2b+1)\times(2c+1)$

$(2a+1)\times(2b+1)\times(2c+1)=7!!=7\times5\times3$

Without any loss of generality $a=1,b=2,c=3$

Thus $N^k =p^{ka}q^{kb}r^{kc}$

Therefore no. of divisors of $N^k$ are $(ka+1)\times(kb+1)\times(kc+1)$

$(ka+1)\times(kb+1)\times(kc+1)=13!!!!=13\times9\times5$

$ka+1=k(1)+1=5 \Rightarrow k=4$

Thus ans is $\boxed{4}$

I don't think I like multifactorials, they make the factorial symbol so confusing. You need parenthesis now to clarify, and honestly, if the whole point is to say "multiply every n numbers", it should really not be something that will take forever to write for large n. (ex. 800!!!!!!!!!! = 800 790 780...).

I couldn't figure this problem out because I wasn't sure what 10!!! or 13!!!! meant (the wiki didn't help)... I really don't like the notation at all. I hope it is changed to something more reasonable, like $64(!^3) = 64!!!$

Alex Li - 4 years ago

I agree that the notation is a bit confusing.

And we can see that $n! > n!! > n!!!$ then using a notation like this $n(!^3) = n!!!$ will be a bit miss leading as that makes the number look much bigger than it is.

(I know that there is no reason for it to be that way but a good notation is one which is intuitive...)

A Former Brilliant Member - 4 years ago

Multifactorial and divisors

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