Question

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!.