Divisibility Revised

Every natural number (greater than 1) is divisible by two natural numbers, i.e. 1 and itself. Therefore, every natural number greater than 1 has at least two factors. Prime numbers have only two factors.

Note: 1 is neither a prime nor a composite number.

Divisibility hints:

2- Any natural number is divisible by 2 when the the digit at its unit place is any one among 0,2,4,6,8. Example: 22,4,36.

3- A natural number is divisible by 3 when the sum of its digits is divisible by 3.
For example, To check if 35892 is divisible by 3, we find the sum of digits.
3+5+8+9+2=27
27 is divisible by 3. Therefore, 35892 is divisible by 3.

4- A natural number is divisible by 4 when the number formed by digits at ten's and unit's place respectively is divisible by 4.
For example, To check if 35892 is divisible by 4:
The number formed by ten's and unit's place is 92
92 is divisible by 4. Therefore, 35892 is divisible by 4.

5- Any natural number is divisible by 5 when the the digit at its unit place is any one among 0 or 5. Example: 25, 93670.

7- A natural number is divisible by 7 when the difference between twice the digit at unit's place and the number formed by rest of the digits is 0 or a multiple of 7.
For example, Consider 196.
Twice the digit at unit's place=2*6=12
Number formed by rest of the digits=19
19-12=7
Therefore, 196 is divisible by 7.

11- Take the sum S1 of the digits at odd places and the sum S2 of the digits at even places. If the difference between S1 and S2 is 0 or any multiple of 11, the number is divisible by 11.
For example, To check if 3852101 is divisible by 11:
S1= 1+1+5+3=10, S2= 0+2+8=10, S1-S2=0
Therefore, the given number 3852101 is divisible by 11.

2^n- If the number formed by last n digits of the natural number is divisible by 2^n, the number is divisible by 2^n.
For example,
If the number formed by last 4 digits of a natural number is divisible by 16, the natural number is divisible by 16. Consider 7856981600.
Similarly, If the number formed by last 5 digits of a natural number is divisible by 32, the natural number is divisible by 32.

5^n- If the number formed by last n digits of the natural number is divisible by 5^n, the number is divisible by 5^n.
For example,
If the number formed by last 2 digits of a natural number is divisible by 25, the natural number is divisible by 25. Consider 625.

3^n- If the sum of the digits of a natural number is divisible by 3^n, the natural number is divisible by 3^n.
For example,
If the sum of the digits of a natural number is divisible by 9, the natural number is divisible by 9. Consider 728109.

6^n- If a natural number is divisible by 2^n and 3^n, it is divisible by 6^n.
For example,
If a natural number is divisible by 4 and 9, it is divisible by 36.

10^n- If a natural number has n consecutive zeroes beginning from the unit's place of the number, it is divisible by 10^n.
For example,
If a number has 3 consecutive zeroes beginning from the unit's place, it is divisible by 1000. Consider 378000.

Note: If a natural number is divisible by two co-prime numbers, it is also divisible by the product of the co-prime numbers.
Thus, if a natural number is divisible by 3 and 4, it is also divisible by 12. Consider 576.
The vice versa of the statement is also true.
Thus to check if a number is divisible by 42, it suffices to check if it is divisible by 6 and 7.

- 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