Question

HELP with part2

http://www.artofproblemsolving.com/Wiki/index.php/Legendre's_Formula

Answer 1

\[\large \sum \limits_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor = \dfrac{n-S(n)}{p-1}\]

Answer 2

so what is difficile in part 2 ?

Answer 3

i think this is more easy understandably wrote there

Answer 4

@dan815 your opinion ?

Answer 5

thanks for replying :)
it seems i have difficulty in working in bases other than 10

im not getting how to interpret below form :
\[\large \left\lfloor \dfrac{n}{p^i} \right\rfloor=e_xe_{x-1}\ldots e_{x-i}\]

Answer 6

Let the base p representation of n be e_xe_{x-1}e_{x-2}\dots e_0 where the e_i are digits in base p. Then, the base p representation of \lfloor \frac{n}{p^i}\rfloor is e_xe_{x-1}\dots e_{x-i}. Note that the infinite sum of these numbers (which is e_p(n!)) is

Answer 7

so e_x,e_x-1 ... these are the base p representation of n

exactly how have wrote above