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This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with prime numbers, but other than that, the textbook gave no hints really and I’m really not sure about how to approach it.

So anyway, here the problem goes:

- Is there a better upper bound for the primorial $x\#$ than $4^x$
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The $\text{Factof}$ (Integer on number of factors) of an integer is the integer divided by the number of factors it has. For example, $18$ has $6$ factors so $\text{Factof}(18) = \frac{18}{6} = 3$, and $27$ has $4$ factors so $\text{Factof}(27) = \frac{27}{4} = 6.75$.

Show that the square of any prime number is the $\text{Factof}$ of some integer.

**Edit: I have reworded the question so that it is more specific and can be answered coherently in the context of the rules.**

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Given the prime $p\neq 3$, the integer number you are looking for is $n(p) = 3^2p^2$. It has $3^2 = 9$ factors ($1$, $3$, $p$, $3p$, $3^2p$, $3p^2$, $3^2$, $p^2$, $3^2p^2$) and

$$

\mbox{Factof}(n(p)) = \frac{3^2p^2}{3^2} = p^2

$$

If $p=3$ then the number $n(p)=2^2 3^3 = 108$. In this case

$$

\mbox{Factof}(108) = \frac{2^23^3}{2^2 3} = 3^2

$$

*‘A prime no. always has two factors.’*

So the product of two primes $p$ and $q$ will have $4$ factors:

$1$, $p$, $q$, $pq$

Now if the primes are distinct then the product $pq$ is not a multiple of $4$ since $2$ is the only even prime (and hence the other prime will be odd).

So, $\text{Factof}(pq)$ is not an integer.

*For (b) part,*

Number of factors of ${(pq)}^4$ = $25$

So, for the $\text{Factof}{(pq)}^4$ to be an integer, it needs to be divisible by $25$ or in other words,

${(pq)}^4$ = $25k$ where $k$ is some integer

*For (c) part,*

I’m still working on the (c) part but maybe these inputs might help:

Notice that the $\text{Factof}$ of that integer has to be an integer itself in the first place.

It also needs to be the square of a prime no.

So the integer must be equal to the product of its number of factors and the square of a prime number.

Moreover, every integer $I$ can be expressed as:

$I$ = ${p_1}^{k_1}\cdot{p_2}^{k_2}\cdot{p_3}^{k_3}…$

where ${p_1}, {p_2}, {p_3}$ are distinct prime nos. and ${k_1}, {k_2}, {k_3}$ are integers.

same question here what a coincidence !!1!1!

a) 4

b)all primes have 2 factors and by multiplying 2 primes you get 4 factors,

the product of these factors must be divisible by 4 and 2 is the only even prime number however 2 x all the other primes NEVER equal a multiple of 4 therefore the answer is never an integer (basically, you can reword and explain more)

c) so far all I’ve got is 2 as P, then 5 as q and then reverse it so 5 as p and 2 as q (I’m sure you can figure out why) bu basically

5 x 2^4 or 16 = 80

80 has 10 factors

80 divided by 10 = 8 which is an integer!

IF anyone can help with D then plz HMU coz i don’t think anyone in Australia has figured this IONOF question

xox and doing my part for all those struggling maths students

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