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

Checking if a number is prime - A faster way

In order to check if a number(say n) is prime or not, you take the following steps:

1. If n is even and n!=2, it is not a prime.
2. Check for the square root of n, say x. 
3. If x is not divisible by any prime number less than x, it is a prime.
4. Again while checking for all prime numbers less than x, you only need to look for the odd numbers. 

Example, 

1. n=9923. We check if 9923 is prime or not.
2. Since 9923 is odd, 2 is discarded.
3. Square root of 9923 is roughly 99.614, we round it off to 100.
4. Now we check if any prime less than 100 divides 9923.
5. Since we have to check for primes, for convenience, we opt for a larger subset, all odd numbers.
6. Set A={3,5,7,9,11,...,91,93,95,97,99}.
7. We check if any of these numbers divides n.
8. More efficiency is possible if we apply Sieve of Eratosthenes and strike out the numbers divisible by 3,5, and 7. We only will need to check the divisibility from the remaining numbers.

Since, we didn't check for all the numbers from 1 to 100 and only the odd ones, it made the program 200% efficient. Forget checking all numbers from 1 to 9922.

© Rishabh Bidya