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Mathematics
OpenStudy (anonymous):

Prime factorization of 20!

OpenStudy (anonymous):

really?

OpenStudy (sriram):

2*2*5*19*1*2*3*3*17*1*2*2*2*2*3*5*7*2*13*2*3*2*11*5*2*3*3*2*2*2*7*3*2*5*2*2*3*2*1=20! lol!!!!!

OpenStudy (anonymous):

some fancy method? all the primes from 2 to 20 will be there for sure answer is \[2^{18}×3^8×5^4×7^2×11×13×17×19\]

OpenStudy (sriram):

never tried this before but is lovely

OpenStudy (sriram):

someone plz check it wid my ans

OpenStudy (anonymous):

ok i will rewrite what you wrote. easier to see without the ones \[2*2*5*19*2*3*3*17*2*2*2*2*3*5*7*2*13*2*3*2*11\] \[*5*2*3*3*2*2*2*7*3*2*5*2*2*3*2=20! \]

OpenStudy (sriram):

my god it matches how did u get it satellite

OpenStudy (anonymous):

Just making sure u were all awake....:-)

OpenStudy (anonymous):

grind it till you find it. you know all the primes from 2 to 19 will be there. then just a matter of how many times right? basically what you did. there may be some snap way to do it but not so clear from the pattern

OpenStudy (anonymous):

U know 11 to 19 are there once, 7 twice (7 and 14), 5 in 5,10,15,20 etc..

OpenStudy (sriram):

got it had read it somewhere long ago no of twos=[20/2]+[20/4]+[20/8]+[20/16]+[20/32] where [] stand for greatest integer

OpenStudy (anonymous):

or maybe you can do this. how many 5s? 5,10,15,20 4 of them how many 7's? 7,17 2 a little harder for the 3's right?

OpenStudy (anonymous):

because you can't just count. you have to count 9 and 18 twice

OpenStudy (anonymous):

i meant 7, 14 2 of them doh

OpenStudy (sriram):

no of 3's=[20/3]+[20/9]+[20/27]=8 my method works!!

OpenStudy (anonymous):

cool!

OpenStudy (sriram):

dont remember the derivatn though

OpenStudy (anonymous):

ok cool but i am not sure i understand

OpenStudy (sriram):

look [ ] stands for greates integer like [20/3]=6 we ignre things after decimal

OpenStudy (anonymous):

why \[[\frac{20}{3}]+[\frac{20}{3^2}]+[\frac{20}{3^3}]\]

OpenStudy (anonymous):

no i get what it means, but how do you know to use the denominators 3,3^2,3^3 how do you know when to stop?

OpenStudy (anonymous):

oh never mind dumb question

OpenStudy (anonymous):

since that is 0

OpenStudy (sriram):

as u go further [20/27] =0 so u keep adding 00

OpenStudy (sriram):

so u stop at 9

OpenStudy (anonymous):

yeah <dopeslap>

OpenStudy (sriram):

hahaha

OpenStudy (anonymous):

so lets try how many 3's in 30!

OpenStudy (sriram):

10+3+1=14

OpenStudy (anonymous):

[30/3]+[30/9]+[30/27]=10+3+1

OpenStudy (anonymous):

cool. now why?

OpenStudy (anonymous):

Multiplicative functions

OpenStudy (sriram):

i think it is like this 20/3 gives how many mutiples of 3 before20 so as we keep adding 20/9 it 3 repeats in a no. divisible by 9 so we add it again and so on

OpenStudy (anonymous):

god i love the internet. if they had this when i was in school i would have had several degrees by now! http://2000clicks.com/mathhelp/NumberTh10FactorialDivisors.aspx

OpenStudy (anonymous):

yeah you have the reasoning. thanks!

OpenStudy (anonymous):

@estudier, did you know this question was so interesting when you posted?

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