Pattern in sixth powers of numbers : Last two digits of sixth powers of numbers at a difference of 10

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

(Note the numbers highlighted in yellow repeating in the same column)

The Proof:
Any 6 terms of the series will be of the format x,x+10,x+20,x+30,x+40,x+50.
We need to prove that the pattern of the sixth powers repeats after every 5 terms.
Therefore, it suffices to prove that the digits at the unit place and tens place of sixth power of a number are equal when the number after five terms is raised to the sixth power.
Therefore, we need to prove that the last two digits of the sixth power of x and the sixth power of x+50 are equal. (Since the numbers are taken at a difference of 10 and there are 5 such terms.)

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

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

Important Observation:
If any number has 5 at its unit place, its sixth power ends with 2 and 5 at ten's and unit's place respectively. (This, too can be proved easily).
A similar pattern is observed in the squares, cubes, fourth powers and fifth powers of numbers.

Q.E.D

© Rishabh Bidya

Pattern in fifth powers of numbers : Last two digits of fifth powers of numbers at a difference of 20

The procedure:

Take any number (Say x).

Form a set of numbers by adding 20 (Your set={x,x+20,x+40,x+60,...}).

Raise every member of the obtained set to its fifth power.

Observe the last two digits (taken from right to left) of the obtained fifth powers.

The last two digits of the fifth powers of every member of the set are the same.

The Proof:

Any term of the series will be of the format x+20k(where k must be a non negative integer).

We need to prove that the digits at the unit's and tens's place of the fifth powers of x and x+20k are same.

Example:

We choose 3 as our x.

The set of numbers obtained is {3,23,43,64,...}.

We raise the terms of this set to their fifth powers {243,6436343,147008443,992436543,...}.

We observe the last two digits of the obtained fifth powers.

The number at unit's place is 3 whereas the number at the tens' place is 4.

Derived Corollary:

The last two digits of an odd number's fifth power and another number's fifth power are the same when another number is formed by adding a multiple of 20 to the first number.

Q.E.D

© Rishabh Bidya

Pattern in fourth powers of numbers : Last two digits of fourth powers of numbers at a difference of 10

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

The Proof:
Any 6 terms of the series will be of the format x,x+10,x+20,x+30,x+40,x+50.
We need to prove that the pattern of the fourth powers repeats after every 5 terms.
Therefore, it suffices to prove that the digits at the unit place and tens place of fourth power of a number are equal when the number after five terms is raised to the fourth power.
Therefore, we need to prove that the last two digits of the fourth power of x and the fourth power of x+50 are equal. (Since the numbers are taken at a difference of 10 and there are 5 such terms.)

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

Example:
We choose 4 as our number.
The set of numbers obtained is {4,14,24,34,44,54,64,74,84,94,104,114,...}.
We raise the terms of this set to their fourth powers {256,38416,331776,1336336,3748096,8503056,16777216,29986576,49787136,78074896,116985856,168896016...}.
We observe the last two digits of the obtained fourth powers {56,16,76,36,96,56,16,76,36,96,56,16...}.
The pattern {56,16,76,36,96} repeats itself.

Derived Corollary:
The last two digits of a number's fourth power and another number's fourth power 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}
We raise these numbers to their fourth powers. {1296,9834496,592240896}
The last two digits of the numbers obtained are the same, i.e. 9 and 6.

Important Observation:
If any number has 5 at its unit place, its fourth power ends with 2 and 5 at ten's and unit's place respectively. (This, too can be proved easily)
A similar pattern is observed in the cubes of even numbers and in the squares of numbers  .

Q.E.D

© Rishabh Bidya

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

The procedure:

Take any number (Say x).

Form a set of numbers by adding 10 (Your set={x,x+10,x+20,x+30,...}).

Square every member of your obtained set.

Observe the last two digits (taken from right to left) of the obtained squares.

The last two digits of the squares repeat after every five members of the set.

The Proof: 

Any 6 terms of the series will be of the format x,x+10,x+20,x+30,x+40,x+50.

We need to prove that the last two digits of the square of x and the square of x+50 are equal.

Similarly, last two digits of squares of x+10 and x+60 will be equal.

Similarly, last two digits of squares of x+20 and x+70 will be equal.

And so on ...

Thus, the pattern continues to repeat after every five terms.

Example:

We choose 4 as our number.

The set of numbers obtained is {4,14,24,34,44,54,64,74,84,94,104,114,...}.

We square the members of the obtained set {16,196,576,1156,1936,2916,4096,5476,7056,8836,10816,12996,...}.

We observe the last two digits of the obtained squares {16,96,76,56,36,16,96,76,56,36,16,96,...}.

The pattern {16,96,76,56,36} repeats itself.

Derived Corollary:

The last two digits of a number's square and another number's square 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 square these numbers. {36,3136,24336,93636}

The last two digits of the numbers obtained are the same, i.e. 3 and 6.

Important Observation: 

If any number has 5 at its unit place, it ends with 2 and 5 at ten's and unit's place respectively. (This, too can be proved easily)

A similar pattern is observed in the cubes of even numbers. Pattern in cubes of even numbers at a difference of 10

Q.E.D

© Rishabh Bidya

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