How many zeroes end 2718#? [Primorial Prime Topic]

The primorial of n, written n#, is the product of the prime numbers less than or equal to n.

I cannot find anywhere how to determine the answer to finding the number of trailing zeroes for 2718#. I did plug this number in Wolfram Alpha which shows only 1 zero at the end[which I am not sure that is correct].

Can you provide a simple example that I can follow? Previously to this question, I found the factorial and double factorial of 2718.

• Of course there is only one $0$. That number is divisible by $2\times 5=10$ but clearly not by $10^2$. – lulu Mar 4 '18 at 21:43
• Each primorial ends with at most one 0 since it's never divisible by 100. – krirkrirk Mar 4 '18 at 21:45
• Notation: usually the primorial $n\#$ is defined to be the product of the first $n$ primes, not the product of the primes $≤n$. See this for example. Of course, this does not effect the number of zeroes at the end of your expression. – lulu Mar 4 '18 at 21:48
• @lulu Wikipedia, MathWorld, Prime Glossary disagree – Wojowu Mar 4 '18 at 21:49
• @Wojowu My error. I stand corrected! Thanks. – lulu Mar 4 '18 at 21:52

The number of trailing zeros of any number can be computed by finding the highest (integer) power of $2$ which is a divisor of the number, the highest power of $5$ which is a divisor of the number, and taking the smaller of the two exponents. This is because $10 = 2 \cdot 5$, and the highest power of $10$ dividing a number is equal to the number of trailing zeros.
In the case of $n \#$ for any $n \ge 5$, there is exactly one power of $5$ and one power of $2$ dividing $n \#$, so there is exactly one trailing zero.
Notice that this is vastly easier than computing the number of trailing zeros in $n!$.
$2718\#$ is divisible by $2\times5 = 10$ but is not divisible by $2^2\times5^2 = 100$, because it has only one of each prime on its factorization. So there is only one $0$ at the end of $2718\#$.