2023

National Mathematics Day

Srinivasa Ramanujan, often referred to as Ramanujan, indeed displayed extraordinary mathematical talent from a very young age, earning him the label of a child prodigy. He was born on December 22, 1887, in Erode, a town in Tamil Nadu, India. Ramanujan’s mathematical abilities were evident early in his life, and he had an innate talent for discovering and formulating mathematical theorems.

Here are some notable aspects of Ramanujan’s early mathematical prowess:

  1. Self-Taught Genius: Ramanujan was largely self-taught in mathematics. His formal education in the subject was limited, and he had minimal exposure to advanced mathematical literature. Despite this, he independently developed numerous theorems and results.
  2. Mathematical Notations: Ramanujan often created his own unique notations and symbols to represent mathematical concepts. His notebooks were filled with formulas and theorems that he had derived on his own.
  3. College Years: Ramanujan entered the Government Arts College in Kumbakonam but faced challenges in completing formal education due to financial difficulties and his focus on independent mathematical exploration. His lack of interest in subjects other than mathematics affected his academic progress.
  4. Letter to G.H. Hardy: One of the turning points in Ramanujan’s life was when he wrote a letter to the British mathematician G.H. Hardy in 1913. In this letter, Ramanujan included a list of mathematical results he had discovered, many of which were new and profound. Recognizing the brilliance of Ramanujan’s work, Hardy invited him to England.
  5. Collaboration with G.H. Hardy: Ramanujan’s collaboration with G.H. Hardy at Cambridge University led to numerous groundbreaking contributions to mathematics. Hardy later described Ramanujan as one of the most original mathematicians of his time.

Despite facing challenges and health issues, Ramanujan’s contributions to mathematics had a lasting impact. Mathematicians all over the world continue to study and admire his work in number theory, modular forms, and other areas. Ramanujan’s story remains an inspiring example of raw mathematical talent and intuition.

The Indian government officially observed Srinivasa Ramanujan’s birthday, December 22, 2012, as National Mathematics Day in 2012. National Mathematics Day has been observed annually ever since. Additionally, 2012 was designated and observed as the National Mathematics Year.

Is a Soccer ball just a ball?

Even though this question might seem simple, a soccer ball isn’t a “ball”, as many people think. 

As crazy as it may sound, “A soccer is a Polyhedron”.

Polyhedron

A polyhedron is a three-dimensional shape with flat faces, straight edges, and sharp corners (vertices). The term “polyhedron” derives from the Greek words poly, meaning “many,” and hedron, meaning “surface.” When many flat surfaces are joined together, a polyhedron is formed. The names of these shapes are based on their faces, which are typically polygons.

It’s a truncated, 32-sided icosahedron with both pentagonal and hexagonal surfaces. It has 60 vertices and is one of the least known and most widely used shapes in the entire world. This is the geometry of soccer.

It has 20 hexagons and 12 pentagons on it.

A short video summarises this:

Pi (⌅) Day!!!

Earnmath wishes you all a very happy International pi day!!!

Pi day!!!

Math titbits

For example:

If we consider the three consecutive numbers 7,8 and 9

82 = 64

7 X 9 = 63.

Five Postulates of Euclidean Geometry

The elements of geometry, an exquisite and influential geometry treatise written around 300 B.C by the Greek mathematician Euclid, explains Euclidean geometry.

The five postulates form the foundation of euclidean geometry.

Are there only 64 squares present on the chess board?

This is a trick question because it is possible for someone to get confused and immediately calculate the total number of squares in a chessboard with eight rows and eight columns by using the formula (number of rows X number of columns)

= 8 X 8 = 64.

Let’s start by reducing the complexity of the situation.

The total number of 1X1 squares is presented in eight rows by eight columns on the chess board.

= 8 X 8 =64.

However, if you start thinking about 2X2 squares, 3X3 squares, and so on up to 8X8 squares, you can figure out how to answer the question.

The total number of 2X2 squares is presented in seven rows by seven columns on the chess board.

= 7 X 7 =49.

The total number of 3X3 squares is presented in six rows by six columns on the chess board.

= 6 X 6 =36.

The total number of 4X4 squares is presented in five rows by five columns on the chess board.

= 5 X 5 =25.

The total number of 5X5 squares is presented in four rows by four columns on the chess board.

= 4 X 4 = 16.

The total number of 6X6 squares is presented in three rows by three columns on the chess board.

= 3 X 3 = 9.

The total number of 7X7 squares is presented in two rows by two columns on the chess board.

= 2 X 2 = 4.

The total number of 8X8 squares is presented in one row by one column on the chess board.

= 1 X 1 = 1.

Therefore, the total number of squares presented on the chess board is found by summing up all the values obtained by 1X1, 2X2, 3X3, 4X4, 5X5,6X6,7X7 and 8X8 squares.

i.e., 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 204.

Therefore, the total number of squares on the chess board is 204.

Suitcase lock problem

You have recently purchased a suitcase that features a combination number lock. Each slider provides a selection of ten numbers, ranging from 0 to 9. In order to prevent anyone from opening your suitcase, you will need to come up with a secret number combination. Do you know how many different possible choices there are for secret codes?

Solution:

You have an option of ten different digits, ranging from zero to nine, for each of the digits in the slider.

Therefore, the number of different combinations that can be made are

10 X 10 X 10 = 1000 options.

If there is a condition that the digit in the slider should not be repeated, then the following applies:

You will have a choice of 10 numbers for the first slider, and 9 for the second slider and 8 for the third slider.

Which means, we have

10 X 9 X 8 = 720 options.

The following video will explain the given problem: