Number Theory and Logic!

The number of trailing zeros $z(n)$ of $n!$ is given by:

$\begin{aligned} z(n) & = \sum_{k=1}^\infty \left \lfloor \frac n{5^k} \right \rfloor \\ \implies z(123456789) & = \sum_{k=1}^\infty \left \lfloor \frac {123456789}{5^k} \right \rfloor \\ & = \left \lfloor \frac {123456789}{5} \right \rfloor + \left \lfloor \frac {123456789}{5^2} \right \rfloor \left \lfloor \frac {123456789}{5^3} \right \rfloor + ... + \left \lfloor \frac {123456789}{5^{11}} \right \rfloor & \small \color{#3D99F6}{\text{Note that for }k \ge 12, \ \left \lfloor \frac {123456789}{5^k} \right \rfloor = 0} \\ & = 24691357 + 4938271 + 987654 + 197530 + 39506 \\ & \quad + 7901 + 1580 + 316 + 63 + 12 + 2 \\ & = \boxed{30864192} \end{aligned}$ .

A number gets a zero at the end of it if the number has $10$ as a factor. For instance, $10$ is a factor of $50$ , $120$ , and $1234567890$ etc.. So I need to find out how many times $10$ is a factor in the expansion of $123456789!$ .

But since $5×2 = 10$ , All the products of $5$ and $2$ needs to be taken account of. Looking at the factors in the expansion, there are many more numbers that are multiples of $2$ ,i.e., $(2, 4, 6, 8, 10, 12, 14,\dots)$ than are multiples of $5$ ,i.e., $(5, 10, 15,\dots)$ . That is, if all the numbers with $5$ as a factor is taken, there will be way more than enough even numbers (multiples of $2$ ) to pair with them to get factors of $10$ (and another trailing zero on the factorial). So to find the number of times $10$ is a factor, all I really need to worry about is how many times $5$ is a factor in all of the numbers between $1$ and $123456789$ .

The largest multiple less than $123456789$ is $123456785$ . So the number of multiples of $5$ between $1$ and $123456789$ is $123456785\div5\,=\,24691351$ .

But, $25\,=\,5\,\times\,5$ , so each multiple of $25$ has an extra factor of $5$ , which also needs to be taken account of. $123456789\,\div\,25\,=\,4938271.56$ . So there are $4938271$ factors of $25$ between $1$ and $123456789$ .

In this way, we arrive at a general rule:

• Take the number that you've been given the factorial of.
• Divide by $5$ ; if a decimal is obtained, truncate to a whole number.
• Divide by $5^{2} = 25$ ; if a decimal is obtained, truncate to a whole number.
• Divide by $5^{3} = 125$ ; if a decimal is obtained, truncate to a whole number.
• Continue with higher powers of $5$ , until the division results in a number less than 1. Once the division is less than 1, stop.
• Sum all the whole numbers obtained from the divisions. This is the number of trailing zeroes.

Using the above procedure , we have:

$\begin{aligned} 123456789\, \div \, 5^{ 1 }\, \, & =24691357.8 \\ 123456789\, \div \, 5^{ 2 }\, \, & =\, \, \, 4938271.56 \\ 123456789\, \div \, 5^{ 3 }\, \, & =\, \, \, \, \, \, 987654.312 \\ 123456789\, \div \, 5^{ 4 }\, \, & =\, \, \, \, \, 197530.8624 \\ 123456789\, \div \, 5^{ 5 }\, \, & =\, \, \, \, \, \, \, \, 39506.17248 \\ 123456789\, \div \, 5^{ 6 }\, \, & =\, \, \, \, \, \, \, \, \, \, \, 7901.234496 \\ 123456789\, \div \, 5^{ 7 }\, \, & =\, \, \, \, \, \, \, \, \, \, 1580.264899 \\ 123456789\, \div \, 5^{ 7 }\, \, & =\, \, \, \, \, \, \, \, \, \, \, \, \, 316.0493798 \\ 123456789\, \div \, 5^{ 8 }\, \, & =\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 63.20987597 \\ 123456789\, \div \, 5^{ 9 }\, \, & =\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 12.64197519 \\ 123456789\, \div \, 5^{ 10 }\, \, & =\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, 2.528395039 \\ & \end{aligned}$

The number of trailing zeros is the sum of the the above divisions after their answer is truncated .

So if $k$ is the number of trailing zeroes,

$\large{\begin{aligned} k\quad & =24691357+4938271+987654+197530+39506+7901+1580+316+63+12+2 \\ & =\boxed { 30864192 } \end{aligned}}$