If 1000! is divisible by 13^n than find the largest value of n.

Question

praxer · Answer

@ganeshie8

praxer · Answer

@ikram002p  bachao bachao.. :)

praxer · Answer

I wasted last one hour in posting the questions. OS Seems damn sloppy today..

praxer · Answer

my queries never bump.. :(

ikram002p · Answer

well (100!) the idea is to get all numbers that devieds 13 which are less than 1000 like :- 13,2.13,3.13..... n.13 (PS dot means x ) 13^2,2.13^2 , .., n.13^2 13^3,2.13^3,.....n.13^3 ...... 13^n,2.13^n,.....n13^n and count them . do you know legendre theorem ? (its already been given in this formula ) http://www.cut-the-knot.org/blue/LegendresTheorem.shtml

ikram002p · Answer

OS is lagging :)

ikram002p · Answer

so according to the link i gave
greatest n = 1000-(sum of digits of (1000)base 13) /13-1

zarkon · Answer

$$\left\lfloor{\frac{1000}{13}}\right\rfloor+\left\lfloor{\frac{1000}{13^2}}\right\rfloor$$

ikram002p · Answer

Zacron , but following this is much easier greatest n = 1000-(sum of digits of (1000)base 13) /13-1 according to this rule http://prntscr.com/4s4x97

ikram002p · Answer

thus
greatest n = (1000-(sum digits of (5bc)))/(13-1)
=1000-28/12=81
thus
13^81 | 1000!

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=prime+factor+1000%21

ikram002p · Answer

yeahhh so i was correct ,first time i used legendre lol