Magic squares-4/4 cells

I have so far dealt with 3/3 cells.
Let me now deal with 4/4 cells formed with numbers 1 to 16-
I will first show how the magic square can be formed by way of steps-
Step 1-You write down the 16 numbers in 4 rows/4 columns,starting with 16 first,and going down upto 1.
Step 2-The numbers in the 2nd and 3rd columns are now changed to read from bottom to top-viz the 2nd column will now read as 3,7,11,15 instead of 15,11,7,3.same step in 3rd column.
Step3-The numbers in the 2nd and 3rd columns are now interchanged.
Step 4-The numbers in the 2nd and 3rd rows are now altered by reading from right to left,viz for example the 2nd row will now read as 9,7,6,12 instead of 12,6,7,9 .same process 3rd row.
Step 5-The numbers in the 2nd and 3rd rows are now interchanged.
You will now find the magic square  as follows-shown in groups of each row-
(16,2,3,13)...(5,11,10,8)...(9,7,6,12)..(4,14,15,1)....The totals of each row,column and both diagonals will be 34..This is a total which is double the sum of the first and last number viz twice (1+16).
Same procedure can be followed for forming squares with any consecutive 16 numbers.
Also by taking the first 16 even numbers  or 16 odd numbers,or starting with  any even or odd number. 

More friendly numbers

In an earlier blog I had shown how the numbers 13 and 16 can be treated as friends.These 2 numbers are in fact a particular version of a general case of friendship.The friendship I suggest is between the digits 4 and 7.Details are as follows. You can take any number like 25 or 34 where the sum of the digits is 7.Work out the squares-giving us 625 and 1156.Add the digits of these two you will get 4....6+2+5=13and 1+3=4.Likewise 1+1+5+6=13 and 1+3=4.Similarly take any  number like 31 and 22 where the sum of the digits is 4.Square and add the digits.You will have 7 as the result.....,31*31=961....9+6+1=16  and 1+6=7....22*22=484  4+8+4=16 and 1+6=7.
You can take any number with as many digits as you like where the sum of the digits is 4 or 7.The friendship will be revealed in every case.
Some more cases of friendship will follow

Perfect numbers

Let me introduce a new subject
Every number can be broken up into factors.The factors can then be added up.In many cases the total will be more than the number under consideration and  similarly in many other cases it will be less.But only in a very very few cases the total will be the same as the number itself.Such numbers are called Perfect numbers.The few examples are 6......,28     ,496      8128   and believe me the next number is 32949336.There is a formula to work out such numbers but I do not wish to burden readers with those calculations.Readers may themselves try and find the same.

Numbers can be friends

Two numbers are treated as friends,when certain operations performed on one number result in the other and viceversa.
Example 16 and 13.
When the number 16 is squared and the digits in the result are added up,you get 13.viz 16*16=256.and
2+5+6=13
Same procedure with 13...13*13=169...1+6+9=16 as mentioned by me
There are several other cases of such friendships which I will deal with later on

Ramanujan number

Let me tell a story about the number 1729,which is known by the name Ramanujan number.
When Ramanujan ,the famous Indian mathematician was travelling in a car with Prof Hardy,the English mathematician,they witnessed a vehicle with the number 1729 on it.When prof Hardy mentioned about the same being a silly number,Ramanujan said that it was not so.He mentioned that it was the smallest number which could be expressed as a sum of 2 cube numbers in 2 different ways,and also that in one case the cubes were of consecutive numbers.Prof Hardy was surprised when Ramanujan gave the details.1729 is the sum of 729(cube of 9) and 1000(cube of 10).also the sum of 1(cube of 1) and 1728(cube of 12)
Let me now pose a question .Considering what Ramanujan said we can find that the difference between the cube of 12(1728) and cube of.10(1000) which is 728 will also be the difference between the cube of 9(729) and cube of1(1)Can we deduce that 728 is the smallest such number? 

General Formula for Divisibility tests

I had given in an earlier blog a procedure for divisibility of any number by the prime 41.I had mentioned about a sequence of numbers to be remembered for this procedure viz 1,10,18,16,37...to be repeated as required.
These numbers are arrived at by finding the remainders when 1,10,100,1000,10000,100000,...etc are divided by 41.For 1 and 10 as they are less than 41,they should be considered as remainders.18 is the remainder when 41 divides 100,...16 the remainder when it divides 1000 etc etc.After 5 steps the same remainders get repeated
The sequence of numbers in some cases go a long way.For instance for 17,you will have 16 numbers before repetition starts,for 19
you will have 18 numbers before repetition etc
I am giving below the sequences for some other primes.Where the sequence contains more than 10 numbers,I am giving only the first 10 numbers of the sequence.
Prime 17- 1,10,15,14,4,6,9,5,/16,7....etc  The mark (/)  shown after the first 8 numbers means that the remaining 8 numbers can be obtained by deducting the first 8 numbers successively from 17.
Prime 19-1,10,5,12,6,3,1115,17,/18......etc....
Prime 73-1,10,27,51,/72,63,46,27 repeated later
Prime 23-1,10,8,11,18,19,6,14,2,20......etc
Prime 29-1,10,13,16,15,5,21,7,12,6......etc
Prime 31-1,10,7,8,18,25,2,20,14,16.....etc
Prime 43-1,10,14,11,24,25,35,6,17,41...etc
Prime 47-1,10,6,13,36,31,28,45,27,35...etc
Prime 79-1,10,21,52,46,65,18,22,62,67,....etc
You could find that all the numbers in each sequence have a connection to the cyclic number generated by the corresponding prime...Viz 17 has 16 numbers in its sequence,19 has 18 numbers,41 has 5 numbers etc
  .  

Divisibility test for 41-also general formula for such tests

I am now presenting a test for divisibility of numbers by the prime number 41.This is based on a general formula useful for testing divisibility by any other prime number say 17,19,23,29,31,43 ...etc
In the case of each prime,you will have to remember certain numbers for testing the divisibility.
For 41, you will have to remember the numbers...1,10,18,16 and 37
Example of number to be tested-1174814....
Step 1-Read the digits from right to left...4,1,8,4,7,1,1
Step2-Multiply these digits successively by the numbers mentioned above viz 1,10,18,16,37,1,10
Step3-Add up the products-(4*1)+(1*10)+(8*18)+(4*16)+(7*37)+(1*1)+(1*10)=4+10+144+64+259+1+10=492.
Step 4-Test the result for  divisibility by 41..If the result is divisible by 41,the number under test will be divisible by 41.In our example 492 is divisible by 41 and so 1174814 will be divisible.
Try testing more numbers say 4560102...1330573..
I will explain in a future blog how the numbers 1,10,18,16 and 37 were arrived at.If you know the method,there will be no necessity to remember them.
Depending on the number of digits in the number under test,you can use the numbers 1,10,18,16,37 given by me successively-like me using 1,and 10 for the 6th and 7th  digits.