# Highest Power in the Expansion of n! Highest Power Formula

How many zeroes at the end $25!$?If $30!$ is expressed in its prime factorization form, what is the power of $10$? What is the highest power of 15 in the expansion in $30!$? These are just some examples and different problem but the method used to solve these problems is the same. The number of zeroes at the end $n!$ is commonly called trailing zeroes.

For trailing zeroes, $n!=n(n-1)(n-2)\ldots(2)(1)$. The number of zeroes here is dependent on the number of $10s$. We observe that $10=5\cdot 2$. Surely the number of 2’s will always be more than the number of 5’s.

Let’s take for example in the expansion of $10!$. $10=5\cdot 2$ $9=3^2$ $8=2^3$

If we continue this and we count the number of 5’s. There are only two of them. 1 from the factor of 10 and the other is the 5 itself. Two is also the highest power of $5$.

The number of $2s$ is 8. 1 from 10, 3 from 8, 1 from 6, 2 from 4, 1 from 2. Meaning the number of 10 that is dependent in number of 5’s. Since there are only two 5’s. The number of zeroes in the expansion of 10! is also 2.

Generally, the highest power of $lates r$ such that $r=a\cdot b\cdot c\ldots$  where $a>b>c\ldots$  in the expansion of $n!$ can be found using the following formula. $Highest Power=\lfloor\displaystyle\frac{n}{a}\rfloor+\lfloor\displaystyle\frac{n}{a^2}\rfloor+\lfloor\displaystyle\frac{n}{a^3}\rfloor+\ldots$

Where $\lfloor x\rfloor$ is the greatest integer function of x. $\lfloor 2.5\rfloor=2$, $\lfloor 3.9\rfloor=3$

Worked Problem 1:

How many zeroes at the end $25!$?

Solution:

This problem is the same in our example above. $Trailing Zeroes=\lfloor\displaystyle\frac{25}{5}\rfloor+\lfloor\displaystyle\frac{25}{5^2}\rfloor+\lfloor\displaystyle\frac{25}{5^3}\rfloor+\ldots$ $Trailing Zeroes=\boxed{6}$

Worked Problem 2:

What is the highest power of 15 in the expansion in $30!$?

Solution: $15=5\cdot 3$. Which means that the highest power of 15 is dependent on the highest power of 5. $Highest Power=\lfloor\displaystyle\frac{30}{5}\rfloor+\lfloor\displaystyle\frac{30}{5^2}\rfloor+\lfloor\displaystyle\frac{30}{5^3}\rfloor+\ldots$ $Highest Power=6+1+0=\boxed{7}$

Worked Problem 3:

What is the highest power of $105$ in $2014!$?

Solution: $105=7\cdot 5 \cdot 3$. Since 7 is highest prime factor of 35, the highest power of 105 is dependent on the highest power of 7. $Highest Power=\lfloor\displaystyle\frac{2014}{7}\rfloor+\lfloor\displaystyle\frac{2014}{7^2}\rfloor+\lfloor\displaystyle\frac{2014}{7^3}\rfloor+\lfloor\displaystyle\frac{2014}{7^4}\rfloor+\ldots$ $Highest Power=287+41+5+0=\boxed{333}$