How many trailing zeroes will be present in the base 12 representation of 55!?
Explanation
The question can be restated as follows - "What is the highest power of 12 that divides 55!"
Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.
So, in order to find the highest power of 12 that divides 55!, we need to look at the highest powers of 2 and 3 that divide 55!.
Successive division by 2 will give us the highest power of 2 that divides 55!:
55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.
Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 = 50
Successive division by 3 will give us the highest power of 3 that divides 55!:
55/3 = 18, 18/3 = 6, 6/3 = 2.
Highest power of 3 that divides 55! = 18 + 6 + 2 = 26.
So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide 55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!
We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's.
Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.