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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Divisibility Rules

Today we will look at how to quickly tell if a number is divisible by

$2$ - A number is divisible by $2$ if it's last digit is even $(0, 2, 4, 6,$ or $8)$.
$3$ - A number is divisible by $3$ if the sum of it's digits is divisible by $3$.
$4$ - A number is divisible by $4$ if it's last $2$ digits are divisible by $4$.
$5$ - A number is divisible by $5$ if it's last digit is either a $0$ or $5$.
$6$ - A number is divisible by $6$ if it is divisible by both $2$ and $3$.
$7$ - A number is divisible by $7$ if you get a multiple of $7$ when you subtract twice the last digit of the number from the remaining number.
(e.g. $665$ is divisible by $7$ because $66 - 2 \cdot 5 = 56$, which is a multiple of $7$.)
$8$ - A number is divisible by $8$ if the last $3$ digits are divisible by $8$.
$9$ - A number is divisible by $9$ if the sum of it's digits adds up to a multiple of $9$.
$10$ - A number is divisible by $10$ if the last digit of the number is $0$.
$11$ - A number is divisible by $11$ if the alternating sum of the digits is divisible by $11$.
(e.g. $2871$ is divisible by $11$ because $2 - 8 + 7 - 1 = 0$, and $0$ is a multiple of $11$.)

 Example
Find all possible values of $a$ such that the number $759a2$ is divisible by $3$.

Solution
From the rules of divisibility, the number $759a2$ is a multiple of $3$ only if the sum of it's digits $7+5+9+a+2=23+a$ is a multiple of $3$. Since $0 \le a \le 9$, this implies that $\boxed{a = 1, 4, 7}$ are all the possible values.



Comments

  1. UnknownSeptember 28, 2019 at 8:31 AM

    Could you explain how 0 is a multiple of 11?

    ReplyDelete
    Replies
    1. akaliaSeptember 30, 2019 at 5:03 PM

      Thanks for your question.

      0 is a multiple of every number because anything multiplied by 0 equals 0.

      Another way to think about it is to let $x$ and $y$ be integers. Then, $x$ is a multiple of $y$ if there is some other integer $z$ such that $x=y \cdot z$.

      Now, let $x=0$ and $y$ be an arbitrary integer. We then need to find $z$ such that $0=y*z$, and we can see that choosing $z=0$ will do the trick every time. Thus, zero is a multiple of every integer.
      Hopefully this makes sense. Let me know if you have further questions.

      Delete

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Equilateral and Equiangular Polygons

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