Help please

1) How many zeroes does 66! end in when written in base 12?

I think that this is an easier way to express the question.

66!=k*12^n       k is a positive integer and so is n.   Find the largest value of n

12=2*2*3

We have the factors 1 to 66

half of those are divisable by 2  that is 33 of them

So now I will divide those ones  by 2 and I get the numbers 1 to 33

ok , 16 of those are divisable by 2

So I will divide those by 2 again and I get the numbers 1 to 16

8 of these will be divisable by 2

Divide these by 2 again and I get the numbers 1 to 8

4 of these will be divisable by 2

Divide these by 2 again and I get the numbers 1 to 4

2 of these will be divisable by 2

Divide these by 2 again and I get the numbers 1 to 2

1 of these will be divisable by 2

33+16+8+4+2+1 = 64       

So the highest power of 2 that goes into 66! is    2^64     

so

mod(66!,2^64) = 0

mod(66!,4^32) = 0

Now I need to look for the highest power of 3 that will go into 66!

We have the factors 1 to 66

One third of those are divisable by 3  that is 22 of them

So now I will divide those ones  by 3 and I get the numbers 1 to 22

ok , 7 of those are divisable by 3

So now I will divide those ones  by 3 and I get the numbers 1 to 7

ok , 2 of those are divisable by 3

So I will divide those by 3 again and I get the numbers 1 to 2

0 of these will be divisable by 2

22+7+2= 31         

So the highest power of 3 that goes into 66! is      3^31 

mod(66!,3^31) = 0

mod(66!,12^31) = 0

So it looks to me like there will be 31 zeros.

1 - 66!_(12) =a162b4b289973a1a4b2336638b21ba13a710b618a5bb5a83343a2840,000,000,000,000,000,000,000,000,000,000_12

It has 31 zeros