Question

This question is so fun :DDDDD
What is the maximum \(k\) such that \(\large12^k|49!\)?

Answer 1

\(12^k | 49!\)

\(\implies 49! = 12^k*m = (2^2\times 3)^k*m\)

Answer 2

I have to have dinner now, so I'll check you guys' answers when I finish my dinner :D

Answer 3

max k = least (#of 2^2 in 49!, #of 3 in 49!)

Answer 4

nice one :) i saw u working on similar question few days back... did u make this question ha ?

Answer 5

nope, I found it in http://brilliant.org/
but you didn't calculate the answer @ganeshie8

Answer 6

Okay, rest is trivial right

#of 2^2 in 49! = (49/2 + 49/2^2 ... 59/2^5)/2 = (24 + 12 + 6 + 3 + 1)/2 = 23
#of 3 in 49! = (49/3 + 49/3^2 ... 59/2^3)/2 = (16 + 5+ 1)/2 = 22


max k = min(23, 22) = 22

Answer 7

u have a shortcut ? using wilson theorem or something ?

Answer 8

corrected typoes :)

#of 2^2 in 49! = (49/2 + 49/2^2 ... 59/2^5)/2 = (24 + 12 + 6 + 3 + 1)/2 = 23
#of 3 in 49! = (49/3 + 49/3^2 ... 49/3^3) = (16 + 5+ 1) = 22


max k = min(23, 22) = 22

Answer 9

@ganeshie8 I'm back :D

you forgot all the floor signs but other than that you're all correct :D

Answer 10

floor signs are implicit lol, dont be picky :P

Answer 11

lolz xD

Answer 12

and the plural of typo is typos

Answer 13

just fyi :D

Answer 14

c'mon

Answer 15

lolz sry :p

Answer 16

lol xD

Answer 17

wondering there must be a shorter way to knw the powers of a prime number in n!

Answer 18

don't think there's any :P

Answer 19

there must be, i feel lazy to do n/2, n/2^2.... humm

Answer 20

\[\Large\sum_{n=1}^{\left\lfloor\ln_2x\right\rfloor}\left\lfloor\frac{x}{2^n}\right\rfloor\]

Answer 21

this is the "shortcut" @ganeshie8

Answer 22

clever :|

Answer 23

lolz

Answer 24

or you can do so: