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