5 Answers
the hint is that think of each prime factor of 30!
Then see what you can think..
Then post what you can and what you cannot.. then you get the remaining hint ;)
30!=226.314.57.74.112.132.17.19.23.29
by same i mean only the power of given prime nos.
now find same for 3000!.
Similarly 3000!=22993.31498........29106........2999.
Therefore for 3000! to be completely divisible by 30! all the prime factors till 29 in 3000! have to cancel out...
as power of 29 in 3000! is 106...the power of 29 in 30! in the denominator has to be 106.
Therefore the reqd power of 30! is 106
aniket.. I am not very sure about ur argument..
you need to justify that no other prime will create a trouble..
Finally you may or may not be correct..