How to find $N-60$ when it is based on the sum of its prime divisors?

Question

The problem is as follows:

A certain number coincides with the sum of the prime positive divisors of the number of zeros in which the product of the first $784$ integers ends positive, when expressed in base $15$ (pentadecimal). Find the number minus $60$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\textrm{33}\\ 2.&\textrm{39}\\ 3.&\textrm{30}\\ 4.&\textrm{28}\\ \end{array}$

The official method of solution according to my book is as follows:

$784!=a(15)^n$

Then it goes to say that

$n=194=2\times 97$

to finally say that:

Then it declares that $2+97=99$ and finally saying this:

$99-60=39$

declaring that the official answer is $39$.

But that's my problem with this explanation. How does it get the number from that factorial?

The number it seems way to big to be computed by hand needless to say that from the sole equation of

$784!=a(15)^n$

it doesn't say what is $a$ or what is $n$? What does this has to do with pentadecimal base?. I don't get it.

Can someone explain me what is intended in this question and a better argument of justification in other words a better solution method which implies not fancy calculations or too encyclopedical in nature?

This request is based that my knowledge on the subject is somewhat limited so I need all the steps necessary so I can understand.

How can this problematic number be found?. This problem belongs to a section in my book about the fundamental theorem of arithmetic although I don't see it in any way how could this be applied in this peculiar scenario.

Can someone help me here with an easy to follow and understand method of solution?

Answer 1

To get the number of zeros $n$ with which the product of the first 784 integers ends,

when expressed in base $15$ (pentadecimal), we need to find the highest power of $15$ dividing $784!.$

$15=3\times5$, so $n$ is also the highest power of $5$ dividing $784!$.

That can be obtained using Legendre's formula:

$n=\left\lfloor\dfrac{784}{5}\right\rfloor+\left\lfloor\dfrac{784}{25}\right\rfloor+\left\lfloor\dfrac{784}{125}\right\rfloor+\left\lfloor\dfrac{784}{625}\right\rfloor=156+31+6+1=194.$

Answer 2

$784!$ has $1931$ decimal digits. The last $194$ are zero because of the $194$ factors $5$ in its factorization.

$194$ factors $=5$ are given by all multiples of $5$ which are $156=784$ divided by $5$.

Plus $5^2$ and its $31$ multiples $$25,50,75,100,125,150,175,200,225,250,275,300,325,350,375,400,425,\\450,475,500,525,550,575,600,625,650,675,700,725,750,775$$

$5^3$ and its $6$ multiples $$125,250,375,500,625,750$$

and $5^4=625$. Finally $156+31+6+1=194$

Therefore the factorization of $784!$ goes as: $$784!=2^{781}\cdot 3^{390}\cdot 5^{194}\cdot\ldots$$ Prime factors of $194$ are $194=2\cdot 97$ and $2+97=99$.

So the answer is $99-60=39$