Number of trailing zeros

In base 10, we have to find the highest power of 10 in n! to find the number of zeroes trailing it. In base, 7, we have to find the highest power of 7 in n! to find the number of zeroes trailing it.

Highest power of 7 in 134! is [134/7]+[134/49]= 19+2 = 21.

Highest power of 10 in 134! is [134/5]+[134/25]+[134/125] = 26+5+1 = 32.

Therefore, the difference in number of zeroes is 32-21 = 11.

When a factorial is expressed in base 10, the number of zeroes at its end is the highest power of 10 in the factorial. When the factorial is expressed in base 12, the number of zeroes at the end is the highest power of 12 in the factorial. Since $$12 = 2^2*3$$, we have to find the highest powers of 3 and $$2^2$$ in 25! and take the minimum of the two.

Highest power of $$3 = [25/3]+[25/9] = 8+2 = 10.$$

Highest power of $$2^2 = [{[25/2]+[25/4]+[25/8]+[25/16]}/2] = [(12+6+3+1)/2] = 11.$$

Therefore the highest power of 12 in 25! = min(10,11) = 10.

So, when expressed in base 12, 25! has 10 trailing zeroes.

For every power of 5 in the number, there is one trailing zero and for every power of 25, there are two trailing zeroes and so on. (Since the powers of 2 and 4 is much larger than 5 or 25, we need to consider only the power of 5 and 25).

Power of 5: [1/5]+[2/5]+[3/5]+...+[20/5] = 4*0+5*1+5*2+5*3+1*4 = 34.

Power of 25: [1/25]+[2/25]+...+[20/25] = 0.

Therefore, the number of zeroes trailing the number is 34.