Math

In these type of question. Proceed in the following way..
Now n= 30!
and 18 = \(3^2 * 2\)

Find the maximum power in the numerator(30!) of all the prime no. available in the denominator.

The maximum power of 2 can be determined in the following way..

30!/2 =15
30!/4 = 7
30!/8 = 3
30!/16 = 1
30!/32 = 0

so the maximum power of 2 available is 15+7+3+1= 26

The maximum power of 3 can be determined in the following way..

30!/3 = 10
30!/9 = 3
30!/27 = 1
30!/81 = 0

so the maximum power of 3 available is 10+3+1= 14

Now we have lot of 2's and we have few 3's.. and if we need to make maximum 18's, we need at least two 3's and and one 2.

so even if we have 14 threes, we can make only seven 18's.

So the maximum power (k) is 7...!!