Pattern in cubes of even numbers : Last two digits of cubes of even numbers at a difference of 10

The procedure:
Take any even number (Say 2x).
Form a set of even numbers by adding 10 (Your set={2x,2x+10,2x+20,2x+30,...}).
Cube every member of your obtained set.
Observe the last two digits (taken from right to left) of the obtained cubes.
The last two digits of the cubes repeat after every five members of the set.

The Proof:
Any 6 terms of the series will be of the format 2x,2x+10,2x+20,2x+30,2x+40,2x+50.
We need to prove that the last two digits of the cube of 2x and the cube of 2x+50 are equal.

Similarly, last two digits of cubes of 2x+10 and 2x+60 will be equal.
Similarly, last two digits of cubes of 2x+20 and 2x+70 will be equal.
And so on ...
Thus, the pattern continues to repeat after every five terms.

Note:
In the first image above, we didn't consider numbers ending with zero, because the cubes of such numbers anyways end with 00. So, such numbers, too can be terms of such a set of numbers.

Example:
We choose 4 as our even number.
The set of even numbers obtained is {4,14,24,34,44,54,64,74,84,94,104,114,...}.
We cube the members of the obtained set {64,2744,13824,39304,85184,157464,262144,405224,592704,830584,1124864,1481544,...}.
We observe the last two digits of the obtained cubes {64,44,24,04,84,64,44,24,04,84,64,44,...}.
The pattern {64,44,24,04,84} repeats itself.

Derived Corollary:
The last two digits of an even number's cube and another number's cube are the same when another number is formed by adding a multiple of 50 to the first number.

Example of the corollary:
We choose 6 as our first number.
We add multiples of 50 to 6. {6,56,156,306}
We cube these numbers. {216,175616,3796416,28652616}
The last two digits of the numbers obtained are the same, i.e. 1 and 6.

Q.E.D

© 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

Surprise your buddies: A wonderful number trick

The trick:

We tell our friend to think of any random number. Then perform some random operations on it and .. Surprise! We tell him the final answer.

Note: Before we start the trick, we must think of two numbers, a and m. a is the number that we want as our answer and m is an arbitrary multiplier. Preferably a and m should be small for our convenience during arithmetic operations. Of course, we never disclose a and m to our friend before or during the trick !!

Let us begin !!

i) Think of a and m.
ii) Multiply a and m. P=a*m.

iii) Tell your friend to think of any number, say x.
iv) Tell him to multiply x with m.
v) Tell him to subtract P from the obtained product of x and m.
vi) Now, tell your buddy to divide the obtained difference by m.
vii) Now, you tell your friend to subtract the obtained quotient from x.

TADA !! Tell your friend - The answer is a.

Let us try the trick (An example) :

i) I think of two numbers:
   a=3. (3 should be my final answer.)
   m=2.
ii) P=6.

iii) You think of a number, say 7.         (x=7)
iv) Multiply 7 with 2. Result is 14.
v) Subtract 6 from 14. Result is 8.       (P=6)
vi) Divide obtained difference by 2, i.e. 8 by 2. Result is 4      (Here our m=2. Therefore, we divide difference by 2).
vii) Subtract obtained result from the number you had thought first, i.e. Subtract 4 from 7.

The answer is 3 !!!



© Rishabh Bidya

Total number of triangles in an inverted triangle system

Question:

To find total number of triangles in an inverted triangular system which has n triangles, each of which is inverted and contained inside its previous triangle.

Example: 

Solution : 

If the total number of triangles is n, 

Except for the innermost triangle, every triangle made results in three smaller triangles, so the number of such smaller triangles is 3*(n-1)

Apart from the smaller formed triangles, there are n triangles that form the figure.

Thus, the total number of such formed triangles will be the sum of both these numbers, i.e. 3*(n-1)+n

Therefore, we obtain the total number of triangles in an inverted triangle system is 4n-3

© Rishabh Bidya