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. 

Magic Square with numbers from 6 to 14/Strange property.

In an earlier blog I had asked readers to verify how the strange property found in the square formed with numbers from 1 to 9 could be found in other squares like that formed from numbers 6 to 14.
The rows in the square with numbers from 1 to 9 will have digits...4,9,2...3,5,7...8,1,6.The number in the first row 492 is actually  4*100+9*10+2.So the numbers in the first row of the square from digits 6 to 14 namely 9,14 and 7 can be joined using a similar method to get 9*100+14*10+7=1047.Just like 294 formed using 4,9 and 2 in the reverse direction,we can get the second  number 7*100+14*10+9=849...Similarly the numbers in the second and third rows will give us 8*100+10*10+12=912,     12*100+10*10+8=1308.....13*100+6*10+11=1371 and 11*100+6*10+13=1173.
The squares of 1047,912 and 1371 will add up to the total 3807594 ,the same relating to the squares of 849,1308 and1173.
The numbers similarly formed using the columns are 993,1506 and 831  giving the total of squares 3944646 the same like the second set of numbers formed in the reverse direction viz 1389,714 and 1227.
Only a little imagination was required to find how the same property could be found in other magic squares.

Generator 41 for cyclic number.

The number 41 generates the cyclic number 02439.You will have surprises by multiplying this cyclic number with  numbers from 1 to 40.You will find 7 extra cyclic numbers,all of them multiples of 02439-so that all the 40 numbers from 1 to 40 are covered to show the cyclic nature.Details are as follows-
Multiplying the cyclic number with 1,10,18,16,and 37 {shown by me with the notation S(1),S(10),S(18),S(16) and S(37)}will give you 02439,24390,43902,39024 and 90243.
S(2) gives you 04878 and this is covered by S(2),S(20),S(36),S(32) and S(33) viz 04878,48780,87804,78048 and 80487.
Likewise-S(3)-07317,S(30)-73170,S(13)-31707,S(7)-17073 and S(29)-70731
S(4)-09756,S(40)-97560,S(31)-75609,S(23)-56097, and S(25)-60975.
S(5)-12195,S(9)-21951,S(8)-19512,S(39)-95121 and S(21)-51219
S(6)-14634,S(19)-46341,S(26)-63414,S(14)-34146 and S(17)-41463.
S(11)-26829,S(28)-68292,S(34)-82926,S(12)-29268 and S(38)-92682 and finally
S(15)-36585-S(27)-65853,S(24)-58536,S(35)-85365 and S(22)-53658.
Does this not appear amazing?

Cyclic numbers-General information

I have done extensive research on cyclic numbers.I wish to present some facts/data.
1)    Each cyclic number comes from taking the reciprocal of a prime like 7,13,17,31,41 ...etc or a  reciprocal of a composite of  primes like 91,49,133,343,1729...We can call these numbers as generators.Thus 7 is a generator of 142857,13 is a generator of 076923..etc
2)    Generally for a generator like n=7,the cyclic number will have digits totalling to( n-1)=6.Thus for generator 17 you will have 16 digits,for 19 you will have 18 digits,for 97 you will have 96 digits,for 601 you will have 600 digits etc.
3)    Alternately,and in much more number of cases,the cyclic number for 'n' will have digits totalling to a factor of
(n-1).Thus for 13,you will have 6 digits,for 31 you will have 15 digits,for 41 you will have 5 digits,for 73 you will have 8 digits etc
4)     For a composite number like 287 which is a product of 7 and 41,you will have digits totalling to an LCM of 6 (relating to 7)and 5(relating to 41) or 30....
5)    There are small numbers like 601 which have cyclic numbers of a large number (600) of digits,but there are also large numbers like 60249167 which have quite a lesser number of digits -to be exact 22 digits.-another example -1900381976777332243781 which has just 52 digits.
6) For some numbers like 127,you will have inter-relationship between the digits in the cyclic number.This has in fact 42 digits but you can split it up into 21  digits each-with the digits in the first group and those in the second group adding successively upto 9...Actually the cyclic number is 00787/40157/48031/49606/2 followed by 99212/59842/51968/50393/7....I have shown the digits seperated in groups of 5 to have a meaningful reading.
I will present some more interesting data in future blogs..

Magic squares-strange property

In a magic square formed with the digits 1 to9,I had shown in an earlier blog that the squares of the numbers formed with digits read from left to right will add up to the same total even when the numbers are formed with the digits read from right to left-as shown below-
Sq 492+Sq 357+Sq 816=1035369=Sq 294+Sq 753+Sq 618
Also considering the columns
Sq 438+Sq 951+Sq.276=1172421=Sq 834+Sq 159+Sq 672
Can you expect whether a similar property can be found  from the squares formed with numbers having more than one digit  as per example below?
Rows---9,14,7......8,10,12....13,6,11
Columns....9,8,13...14,10,6...7,12,11...totals 42 for each row,each column and both diagonals
The property can  be  found to be fulfilled in such cases also
Readers can try and find how the property is fulfilled.

Cyclic number 076923

Under my blog title Fun with numbers I had introduced information about 2 cyclic numbers viz 142857 and 0588235294117647.The first one is found by taking the reciprocal of 7 and the second one by taking the reciprocal of 17.
Now I am introducing another cyclic number 076923.This is obtained by taking the reciprocal of 13.
The number 142857 has 6 digits-one less than 7 used for taking the reciprocal-and shows cyclic nature by multiplying with numbers from from 1 to  6.The second 16 digit number shows cyclic nature by multiplying with numbers from 1 to 16.
You will ask a question about the cyclic number now introduced by me viz 076423-which has only 6 digits-what happens when the same is multiplied by the 12 numbers from 1 to 12.Actually the number 076923 and the number 153846 which is its double act as cyclic numbers.These 2 account for showing the cyclic nature by multiplying 076923 by the digits from 1 to 12.as shown below-
Multiplying with 1(M-1) results in 076923
M-3 results in 230769
M-4 results in 307692
M-9 results in 692307
M-10 results in 769230
M-12 results in 923076
M-2 results in 153846
M-5 results in 384615
M-6 results in 461538
M-7 results in 538461
M-8 results in 615384
M-11 results in 846153
Like in case of the other cyclic numbers multiplication by 13 will result in all 9's viz  999999. 

Response to my blog-Digits being the same.

Regarding my blog mentioning the digits 8,6,4,5 and1,one of the readers has given 3 examples viz 46851 ,14685 and 48651.-These will give rise to the numbers 93702,29370 and 97302.
I wish to indicate that there are some more cases-at least 3 more.They could also be located easily.. 

Response to question N-1-Black hole number 123

It has been suggested in one of the comments that zero should not be counted either as odd or even.It is not correct to leave out zero without considering it as odd or even.Taking the analogy of 1,3,5,7 and 9 counted as odd and 2,4,6 and 8 counted as even,it would be reasonable to count zero under the 'evens' category.In that case the number mentioned 303 will lead to the same blackhole number 123,like 323