Using 2,0,1,6 and arithmetic operations to get every digit from 0 to 9

2x0x1x6=0

(2^0)^16=1

2+0x1x6=2

-(2+0+1-6)=3

20-16=4

-(2x0+1-6)=5

2x0x1+6=6

2x0+1+6=7

2+0x1+6=8

2+0+1+6=9

Thus, 2,0,1,6 (or 2016 ;) ) gives us 0 to 9 (Everything in numbers). I wish 2016  to provide you 'EVERYTHING' in terms of joy, desires, goals and love (and all else that you seek). Wishing you a 'HAPPY NEW YEAR' :D Cheers!! Enjoy

© Rishabh Bidya

Magic Squares 6 : Creating a 5x5 magic square using 25 consecutive integers given the sum of a row or a column

Problem Background : A 5x5 magic square is an arrangement of 25 squares arranged in 5 rows and 5 columns. We are given a sum S which is the sum of any row or column. We need to arrange 25 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

General solution of a 5x5 magic square when sum of a row or column is given.

Derived corollary : It can be noted that for the numbers in individual squares to be positive integers, S must be a multiple of 5 and S must be greater than or equal to 60.

Note that this solution is however only a special case of the general solution:

General solution of a 5x5 magic square when sum of a row or column is given.

Note that in the image above, substitute d as the difference between the terms.

Say if the terms are 1,3,5,7,.. d=2.

Observations:

The numbers in individual squares are positive integers when:

S>60d and S is a multiple of 5.

The above result can be derived from Magic Squares 5 .

Thus, S=5a+60d. After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 5 : Creating a 5x5 magic square using 25 terms in Arithmetic Progression

Problem Background : A 5x5 magic square is an arrangement of 25 squares arranged in 5 rows and 5 columns. We are given 25 terms in an arithmetic progression(AP) to place one term in each of the squares such that the sum of each of the rows and columns is equal.

Solution : We assume the terms of the AP are a,a+d,a+2d,...,a+23d,a+24d where a is the first term and d is the common difference.

General Solution of a 5x5 magic square

Derived corollary : It can be noted that we get a magic square of any 25 consecutive integers when d=1 and a is the first term among the 25 integers. 

Magic square of any 25 consecutive integers when a is the first integer

Example: If a=1 and d=1, we get a magic square of the integers from 1 to 25.

Magic square of first 25 natural numbers

Q.E.D

© Rishabh Bidya

Magic Squares 4 : Creating a 4x4 magic square using 16 consecutive integers given the sum of a row or a column

Problem Background : A 4x4 magic square is an arrangement of 16 squares arranged in 4 rows and 4 columns. We are given a sum S which is the sum of any row or column. We need to arrange 16 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

Derived corollary : It can be noted that for the numbers in individual squares to be positive integers, S must be of the form 2(mod4) and S must be greater than or equal to 34.

Note that this solution is however only a special case of the general solution:

Observations:
The numbers in individual squares are positive integers when:

1. S>30d
2.  If d is even, S is divisible by 4. If d is odd, S is of the form 2(mod4)

The above result can be derived from  Magic Squares 3 :

Thus, S=4a+30d. After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 3 : Creating a 4x4 magic square using 16 terms in Arithmetic Progression

Problem Background : A 4x4 magic square is an arrangement of 16 squares arranged in 4 rows and 4 columns. We are given 16 terms in an arithmetic progression(AP) to place in each of the squares such that the sum of each of the rows and columns is equal.

Solution : We assume the terms of the AP are a,a+d,a+2d,...,a+14d,a+15d where a is the first term and d is the common difference.

General Solution of a 4x4 magic square

Derived corollary : It can be noted that we get a magic square of any 16 consecutive integers when d=1 and a is the first term among the 16 integers.

Magic square of any 16 consecutive integers where a is the first integer

Example: If a=1 and d=1, we get a magic square of the integers from 1 to 16.

Magic square of first 16 natural numbers

Q.E.D

© Rishabh Bidya

Magic Squares 2 : Creating a 3x3 magic square using 9 consecutive integers given the sum of a row or a column

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given a sum S which is the sum of any row or column. We need to arrange 9 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

Derived corollary : It can be noted that for the numbers in individual squares to be integers, S must be divisible by 3, i.e. S must be 0(mod3) and S must be greater than 12.

The above result can be derived from  Magic Squares 1 :

After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 1 : To create a 3x3 magic square given 9 consecutive integers

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given 9 consecutive integers to place in each of the squares such that the sum of each of the rows and columns is equal.

Solution : If the given integers begin from a, rest if the integers are a+1, a+2, a+3, ..., a+7, a+8.

In the above figure, Sum of every row as also sum of every column is 3a+12.

Q.E.D

© Rishabh Bidya

Total number of triangles in a figure with h horizontal non intersecting lines and v vertical lines dropped from a vertex to its opposite side(cevians)

Question:

To find total number of triangles in a triangular body which has h horizontal lines, two of which do not intersect and v vertical lines which are all dropped from one vertex to its opposite side.

Solution:

The total number of triangles in the figure is:

(h+1)*(h+2)*(v+1)/2

Example :

If h=3 and v=2,

substituting the values of h and v in the formula given above, we get,

Total number of triangles = 30


Note : If a line is dropped from a vertex to its opposite side, it is called as a cevian.

© Rishabh Bidya