131) No. of zeroes at d end of n! = min.(p,k) where p - power of 5 when n! is written as a product of prime no.s and k - power of 2 when n! is written as a product of prime no.s ...
2) power of a prime no. p in n! = [n/p] + [n/p2] + [n/p3] + ...
from d above two key points... v get d answer as 74
[1]
13try using those two points... if u dont get d answer... den i'll post d complete solution...
62Good work MAQ :)
To add to what MAQ said..
this is the same sas highest power of 5
the reason behind that is that there will be some power of 5 and some power of 2... Out of these, the power of 5 will be lower.. so the number of zeroes will be dicided by the power of 5
which is given by the expression given by MAQ
1hey dude we can do that using simple logic ...
no need of these complex formula's
62well MAQ i thought u still remember rule 1 and 2!!!
And that this equation you wrote is a derivation!
We had done this question very recently.. Did u not see how and why? or did u miss that thread?
13yeah i do remember it bro... i myself gave d derivation of dis equation dat day rite...!!!
1@MAQ
can u plz provide the derivation of the formula u gave?...
62Hint:
what no of numbers are divisible by p
how many by p^2
how many by p^3
and so on
13@surbhi... Have a luk at d pinked post in dis thread...
http://targetiit.com/iit_jee_forum/posts/9th_january_2009_1549.html
even then if u dont understand... i'll explain it in more detail... but first have a luk at it...
[1]
1I have a shortcut for these questions
300/5 -60
60/5 -12
12/5-2
60+12+2
samajh aaya toh thik hai nahi toh rehne do
lol
full fargi hai
13ur method is same as dat of d formula i gave akshay...[3] [6]
ur method fails when we get decimal values upon division... [2] ..... then v must take GIF of dat value which is nothing but d given equation... [4]
62yes akshay.. this is absolutely what maq has done.. :)
Only that you din follow the two rules of permutation combination ;)