Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Number of Trailing 0's

Today, we will be looking at problems that ask you to find the number of trailing $0$'s in any factorial. Say for example, we are asked to find the number of trailing zeroes of $101!$
Reminder: $n! = n \cdot (n-1) \cdot (n-2) \cdot ... \cdot 3 \cdot 2 \cdot 1$

Simply put, trailing zeroes are zeroes at the end of the number without any non zero digits to the right of them. Ex. - there are $4$ trailing zeroes in the number $20340000$.

Trailing $0$'s are formed when a multiple of $2$ is multiplied by a multiple of $5$. So, all we have to do is count the number of $5$'s and $2$'s in $101!$.

Let's start with counting the number of $5$'s.
The numbers $5, 10, 15, 20, 25, ... 95, 100$ all contribute one factor of $5$ to the factorial, so we have $20$ factors of $5$. However, some numbers have more than one multiple of $5$. For example, the numbers
$25 = 5 \cdot 5$
$50 = 2 \cdot 5 \cdot 5$ 
$75 = 3 \cdot 5 \cdot 5$
$100 = 4 \cdot 5 \cdot 5$ 
all have an extra factor of $5$, so we have to add $4$ to our count. This results in a total of $24$ $5$'s in $101!$.

We can choose to count the number of $2$'s in a similar fashion, but if we think a little harder, we would see that there will be many more multiples of $2$ than there would for $5$ because of the fact that $2<5$. Thus, we can be assured that for every $5$, we will have a $2$ to pair it up with. This means that the number of trailing zeroes in $101!$ is the same as the number of $5$'s, so our final answer would be $\boxed{24}$

Let's look at another example.
Problem: Find the number of trailing zeroes in $731!$

Solution: As we saw above, we only need to focus on the number of $5$'s. Let's first look at the numbers that have at least $1$ factor of $5$. These are all multiples of $5$ less than $731$. To count these, we just find $731\div 5$, and keep the quotient. This gives us $146$ multiples of $5$ under $731$.

Next, we need to account for numbers that have $2$ factors of $5$. In other words, we need to find the number of multiples of $5^2 = 25$ under $731$. We do the same thing as before and find the quotient of $731 \div 25$, which equals $29$. We can keep going to the next power of $5$, which is $5^3$, or $125$. Dividing $731 \div 125$ gives us a quotient of $5$. We repeat this process one last time with $5^4 = 625$, and $731 \div 625$ gives us a $1$ for our quotient. We can't go any higher for powers of $5$, so we just have to add all our values to finish. Thus, our final answer is
$$146 + 29 + 5 + 1 = \boxed{181} \text{ trailing zeroes in 731!}$$
Feel free to comment if you have any questions!

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