help

We can use Legendre's formula: https://en.wikipedia.org/wiki/Legendre%27s_formula

By Legendre's formula, the number of factors of 3 is

\(\left\lfloor \frac{99}{3} \right\rfloor + \left\lfloor \frac{99}{3^2} \right\rfloor + \left\lfloor \frac{99}{3^3} \right\rfloor + \left\lfloor \frac{99}{3^4} \right\rfloor = 33 + 11 + 3 + 1 = 48\)

I get the following:

Edit:  That last line should say 3ⁿ not 3n.

As a check, here is WolframAlpha's factorisation:

Thanks Alan, I had not seen this of course or I would not have answered the new thread.

Also looked at here.

https://web2.0calc.com/questions/help_59968