Telephone number puzzle

A person described the telephone number (of his friend) consisting of  7 digits as follows-
The initial 3 digit part is 447 more than the last 3 digits of the four digit second part,but less than the first 3 digits of the second part by 447.Find the telephone number

Harshad numbers-Rejoinder.-

I had mentioned about a Multiple Harshad Number in my previous blog .Same can be called MHN-12 as the division by the sums of the digits can be repeated 12 times.
The same is 20165 02858 57988 44661 76-The sum of the digits being 108,you get the result of division
18671 32276 46285 59872-sum now is 99 which can be used for the next division.
Proceeding repeatedly you will get the final two numbers 392 and 28.
In general  numbers of the form-10080,100800,1008000,10080000.. etc are all such multiple numberss

Diagonals-rectangular cube

Imagine any rectangular cube say with side lengths x,y and z
There will be 3 diagonals -we can call them as a,b and c
It will be very difficult to find values of x,y and z so that the diagonal lengths
a,b and c have integral values.
There is a complicated formula to find such values-
Here are some examples of such values-
Example 1) side lengths 44,117 and 240 .The diagonals have values  125,244 and 267
Example 2) side lengths 352,  936 and 1920.  The diagonals have values 1000,1952 and 2136.
Example 3) side lengths 828,2035 and 3120.The diagonals have values 2197,3228 and 3725.  

Perfect triangles

Triangles for which the perimeter and the area have the same integral values are called
perfect triangles-Some examples below-
1)Sides 6,8,10,-perimeter and area=24
2)Sides 5,12,13-perimeter and area=30
3)Sides 9,10,17-perimeter and area=36
4)Sides 7,15,20-perimeter and area=42
5)Sides 6,25,29-perimeter and area=60
You can see all the integral values are multiples of 6.
For your information the areas can be calculated using the following formula-
Find the perimeter 'p' and divide it by 2 to get 's' the semi-perimeter.
Assuming the sides are designated by a, b and c find the product of s,s-a,s-b and s-c
The square root of the product is the area
Example (4) above......a=7,b=15,c=20.....p=42....s=21
s-a=14,s-b=6,s-c=1.Hence the product is 21*14*6*1=1764...Sq root=42=area

Connection between numbers 39 and 1200

You can split up the number 39 into 3 parts in 3different ways,and in each case the products of the 3 parts will be the same viz 1200 as shown below-
25+8+6=39...25*8*6=1200
24+10+5=39....24*10*5=1200
20+15+4=39....20*15*4=1200

Using odd numbers for squares/cubes

Here are ways for obtaining squares/cubes using odd numbers-
Consider the following set of odd numbers-
1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37etc
For getting squares you simply add them consecutively-
1=1*1
1+3=4=2*2
1+3+5=9=3*3
1+3+5+7=16=4*4etc
1+3+5+.......37=361=19*19 and further squares.
For getting cubes you add as follows-
1=cube of 1
3+5=8=cube of 2
7+9+11=27=cube of 3
13+15+17+19=64=cube of 4
21+23+25+27+29=125=cube of 5 etc
Considering  the above groups as the 1st,2nd,3rd,4th &5th, the 'n'th group will be as follows-
First term n^2-n+1
Last term n^2+n-1
No of terms-n
Sum of all the terms=n*average of the first &last term=n*n^2=n^3=cube of n
For example the 30th group will have 871 as the first term,929 as the last term and the sum of
all the terms will thus be 30*(871+929)/2=30*900=27000=cube of 30

bbbirthday trick

Let me introduce a trick for getting some calculations made from the data relating to one's birthdate and announcing the birthdate after the results of the calculation are given to you.
The following are the calculations-
Multiply the year(all 4 digits) by  671
Multiply the month figure (viz 1 upto 12) by 66462
Multiply the date figure (viz 1 upto 31) by 26182
Add up the three results and announce the total.
You will then be able to find the birthdate from this announced result

Multiplication trick

Let me introduce you all to a different way of multiplication.
Suppose you want to multiply 401 with 25
Enter them in 2 columns -the smaller number 25 preferably in the first column
In the first column write down one by one values after dividing by 2(If there is a remainder ignore it ).
In the second column write down one by one values after multiplication by 2 ,corresponding to each entry in the first column.
Select values in first column which are odd numbers and write down the corresponding entries in the second column.Add these entries of the second column.The total  will be the required product.-See below the working-
25-------401
12...........802
6............1604
3............3208
1............6416
The odd numbers in the first column are 25,3and 1.The corresponding entries in the second column are 401,3208 and 6416.Add these you get 10025 which is the product of 25 and 401.
Another Example-27 to multiply with 48
27...........48
13...........96
6............192
3............384
1............768..The required product   48+96+384+768=1296.
You can also enter the bigger number in the first column but this will only result in more entries in each column..

Another general method-divisibility tests

I am now introducing details another general method for divisibility tests.
Let me show the way for conducting a test for divisibility by 19.
Number to be tested-4783915
Break this into 2 parts-478391 and 5
Multiply the second part by 2 and add to first part-478391+10=478401
Repeat the procedure-47840+2=47842
Repeat again-4784+4=4788
Repeat 478+16=494-
Test the result for divisibility.--494 is divisible by 19 and so is therefore the original start number 4783915
Another Example-Number to be tested-15857799
1585779+18=1585797....158579+14=158593...15859+6=15865..1586+10=1596
159+12=171...171 is divisible by 19 and so is original start number 15857799
Similar  method can be adopted for conducting test for divisibility by other numbers like 17,23,29 ,79,89 etc
However the number to be used for multiplying the second part in each case is different-they are shown below-
For 17 it is 12
For 23 it is 7
For 29 it is 3
For 79 it is 8
For 89 it is 9
You will notice that these are the same numbers which can be worked out based on my blog Multipliers-Rejoinder posted a couple of days back.-though with reference to cyclic numbers.
Try numbers 159327944.....,65482725,....45327029331.-for testing divisibility by 17 

Multiplers-Rejoinder

This has reference to my previous blog posted some hours ago today.
I was worried the process mentioned would be difficult to understand.
There are two ways of looking at the issue of movement of digits in the cyclic number 142857.
Let us consider the movement of the first digit to the right now(instead of the last digit to the front as considered earlier).Thus we now have 42857l,285714,857142,571428 and 714285.The multipliers required now are 3,2,6,4 and 5(These you can see are the same multipliers considered in the reverse order)
This new set of multipliers are more easier to find.
They are all the successive remainders when you take the reciprocal of 7
While taking the reciprocal of 7 we first add zero to 1 getting us 10 and divide the same by 7.The remainder is 3 which is the first multiplier now.Next we add zero to 3 getting us 30 and divide the same by 7.The remainder is 2 which is the second multiplier now.Next we add zero to 2 getting us 20 and divide the same by 7.The remainder  is 6 which is  the third multiplier now.The process can be continued to get the next multipliers viz 4 and 5.
You have to understand the necessity of the 2 processes viz.the one in the previous blog and the one now shown.
This will be clear when you consider a cyclic number of too many digits
For instance the cyclic number for the prime 97 has 96 digits.The first 10 of them are 0103092783 and the last 10 of them are 0206185567.If you want to know what multiplier will make the last digit 7 go to the front you have to perform the difficult process of finding the reciprocal right till the end. as per this blog.But if you adopt the process in the previous blog you will see that the value  of 'f' is 68 and this multiplier will make the last digit 7 to go to the front.
You may choose in every case whichever process you consider easier and best. 

Multipliers for cyclic numbers

In a previous blog I had mentioned that by using the multipliers 5,4,6,2,and 3 on the cyclic number 142857,one can see the last digit at right moving successively to the front viz producing the results 714285,571428,857142,285714 and 428571,the other digits remaining in the same order in each case.
The multipliers 5,4,6,2 and 3 can be found as follows-
I had mentioned that the cyclic number 142857 was found by taking the reciprocal of prime number 7.We can call 7 as the 'generator'(g) of the cyclic number.
The first multiplier (f) was shown by me as 5
The relation between 'f' and 'g' is that (10f-1) should be a multiple of 'g'
In the case of g=7,we have f=5 as 50-1=49 is a multiple of 7.
For the subsequent multipliers you have to repeatedly multiply 5 by itself and  find modulus 7 of the result.
(I now have to explain the term modulus.The modulus of any number say 45 with  reference to another number say 7 is the remainder when 45 is divided by 7.In this case the remainder is 3 and so modulus 7 of 45 is 3)
For our case we have 5 multiplied by 5 gives us 25 and modulus 7 of 25 is 4 which is the second multiplier.
Next 4 multiplied by 5 gives us 20 and modulus 7 of 20 is 6,which is the third multiplier.
Next 6 multiplied by 5 gives us 30 and modulus 7 of 30 is 2 which is the fourth multiplier.
Next 2 multiplied by 5 gives us 10 and modulus 7 of 10 is 3 which is the fifth multiplier.
The process stops here as 3 multiplied by 5 gives us 15 and the modulus 7of 15 is 1.
You  may find the processes a bit difficult to understand but read again you will understand the same .
Now let me give you  the problem -whether you will be able to work out how the multipliers 4,3,12,9 and 10 can be found for the cyclic number 076923  relating to g=13(The cyclic number 076923 can be found by taking the reciprocal of 13). 

Harshad numbers

A number which is divisible by the sum of its digits has been given the name as  a Harshad number.Examples are given below-
153 divisible by 1+5+3=9 giving the result 17
133 divisible by 1+3+3=7 giving the result 19.
There are 23 numbers below 100 and 26 numbers between 100 and 199
Can  you locate them?
There are also multiple Harshad numbers.An example is 6804.It is called a multiple Harshad no for the reason given below-
6804 divided by the sum of its digits 18 gives the result 378.
378 divided by the sum of its digits 18 gives the result 21.
21 divided by the sum of its digits 3 gives the result 7.
As process is repeated 3 times for 6804 it is called Grade 3 Multiple Harshad number.
There is a Grade 12 Multiple Harshad number which I will reveal if anyone is interested.

One more cyclic number

In the cyclic number 142857  the digits at right gradually shift to the left on multiplication by 5,4,6,2 and 3.viz producing 714285,571428,857142,285714 and 428571.This cyclic number is actually found by taking the reciprocal of the prime number 7.viz in the reciprocal the group of digits 1,4,2,8,5,7 repeat continuously..
You can find more cyclic numbers by finding similarly the reciprocals of more prime numbers or multiples of prime numbers.  
If you take the reciprocals of numbers having 2 digits or more you will face with one or more zeros at the beginning,like 076923 for 13.......,012345679 for 81.....00729927 for 137 etc and you should allow the zeros to remain while making the multiplications.
Now I am presenting a cyclic number with six zeros in the beginning.
The same is 00000048269001213 and this can be found by taking the reciprocal of the prime 2071723.
If you multiply this with the number 621517,you will find the last digit at right viz 3 moving to the front viz producing 30000004826900121.You can see that except for 3 all the other digits appear in the product in the same order.
I shall inform you later how this number 621517 used for multiplication has been found.I am in the process of finding more numbers which when used for multiplication allow the other digits at right to gradually move to the left.There is a method for finding them but because the numbers are large,the process is not easy and will take time.

Countdown

This is based on a TV programme shown by BBC and French TV.
You are given a set of numbers and you are required to arrive at a particular result using those numbers and the the usual mathematical signs relating to addition,subtraction,multiplication and division.Although the TV programmes had provided a small time period for working out the solution,I am not suggesting any period.Two examples below for you to try.
1) Numbers 6,1,25,8,5,8---result required 640
2) Numbers 4,7,9,25,6,4....result required 758
Except where there is already a repetition like 8 in example 1 and 4 in example 2,no repetition of numbers are allowed.  

Arrange digits

Here is a small puzzle to solve-
Arrange the digits 1,2,3,4,5,6,7,8,9 to get a fraction with equivalent value 1/3

Get a total 900

You have the nine distinct digits 1,2,3,4,5,6,7,8 and 9.You are allowed to form 3 numbers each of 3 digits from these,without using any digit more than once..The requirement is that the total of these 3 numbers should be 900.Try forming the numbers.

Two sets of five digits each

I had dealt with earlier 2 sets of five distinct digits each,where the numbers formed from the first,could be double the numbers you can form from the second-examples below-
76902(38451)....96702(48351).....69702(34851).....etc etc
You can similarly have sets where numbers from the first could be triple or one-third the numbers from the second-Examples below-
50382(16794)......53082(17694)......20583(61749)
Can you find more such cases?

using a set of scales fornweighment-rejoinder-2

In case the weights are placed on only one side of the scales,the weight of the item being weighed will be the total of the weights placed on that side.If the weights are placed on both sides,the weight will be the sum of weights placed on opposite side less the sum of weights placed by the side of the item.
You will be surprised the weights required will be the numbers formed by repeatedly multiplying 1 with 3,viz 1,3,9,27,81,243,729 ......etc
Thus if the item weighs 60 gm,you will have to work out how you can get 60 by adding or subtracting those numbers.You will find 60=81-27+9-3.Thus weights 27 and 3 are placed by the side of the item weighed and 81 and 9 placed on the opposite side.
Similarly for any item weighed.
For item weighig 301 gm,you will find 301=243+81-27+3+1.So you  have to place weight 27 by the side side of the item weighed and all others on the opposite side.
Try finding what and how the weights are to be placed for items weighing 400 and 368.A tricky exercise.

Using a set of scales for weighment-rejoinder

I have not received any response to my earlier blog-The same related to the minimum number of weights required to weigh materials with weights upto 80 kg using a set of scales.
The procedure is based on the property relating to numbers formed by repeatedly multiplying 1 wth 2.
The numbers are 1,2,4,8,16,32,64,128,256   etc etc
These numbers could be added to each other without repetitions to produce any total you require and thus these will be values of the weights required for weighig any material
If you use 1,2,4,8,16 you could weigh upto 31kg
If you use 1,2,4,8,16,32 you could weigh upto 63 kg
If you use 1 2 4,8,16,32,64,you could weigh upto 127 kg and so on
For example the weighment of an item with weight  115 kg will require the weights 64,32,16,2 and 1
Similarly for 105 kg,you will require  the weights 64,32,8 and 1 and so on
Let me put a question now.Suppose you are allowed to place weights on both sides of the scales.What will be weights needed? 

Walk the talk

I got a comment recently that my numbers have not walked for some recent time past.I am now showing they could walk the talk also.This is an imaginary conversation between numbers 2 and 5.
The younger member Y(2)started the talk.
Hello can we become friends?
Elder member E(5) replied.
Dont you know .we are already friends ,whenever I regenerate myself you are close by my side like in 5*5=25 and 25*5=625 etc.Also when I regenerate myself 5 times I become yourself in disguise..Check it 5*5*5*5*5=3125...3+1+2+5=11...1+1=2
Similarly when you regenerate yourself 5 times,you become myself in disguise....Check it.2*2*2*2*2=32...3+2=5
 I will reveal to you more about our friendship in our next conversation.

Using a set of scales for weighment

Suppose you have a set of scales useful for weighing any article.Suppose also that you are allowed to place the article on one side.and that weights are allowed to be placed only on the other side.Suppose you are required to weigh articles upto  80 kg.The question is what is the minimum number of weights required to weigh any article where they can weigh from 1kg upto 80 kg.I will wait for a response before giving the answer.

Divide by 12 to get a remainder.

Take any prime number like 19.
Square it and add 14.You will get 361+14=375
Now divide by 12,you get 3 as reminder.
Even if you take any other prime number and work similarly,you will get the same remainder.
If you add 17 instead of 14,you will always get 6 as the remainder.
Even if you take any odd number not divisible by 3 instead of any prime number,you will have similar results..

,

Repeat a number

Take any three digit number like 134.
Repeat the same number by its side making a 6 digit number 134134.
Divide this now by 7,You get 19162.
Divide now by 11,You get 1742.
Divide now by 13.You will get the original 3 digit number 134.
This procedure holds good for any 3 digit number chosen.

Envelope trick

I wish to present a trick which you can play with your friends/relatives.
It is better to have at least 4 persons before you for this as otherwise the trick is likely to be discovered,though not always.You will write a number say 43 on a piece of paper without showing it to anybody,place it in an envelope and close it shut.
You will then present a sheet of paper containing 16 numbers in 4 rows/4 columns.You will give one person a coloured pencil say green asking him to score out one row and one column..Another person with a red pencil to score out another row/column.A third person with blue pencil to score out another row/column.The last 4th person to score out the last row/column with say an yellow pencil..
The green colours meet at a particular number in the grid.Similarly the other colours will meet each at a different number in the grid.You will then ask the persons to add up all the 4 numbers where each set of coloured pencils meet .They will be surprised to find that the total is 43,which you have written and placed in the closed envelope.
The 16 numbers which you have written in the sheet of paper presented for scoring out will be as follows-
14,16,11,10
 9, 11, 6,  5
 8, 10, 5,  4
17,19,14,13.
How do you find the trick?
I have given below two more grids below for numbers 51 and  103-
Number 51-Numbers given row-wise--(8,11,17,13)..(12,15,11,17)...(14,17,13,19)..(10,13,9,15)
Number 103--(26,24,35,32)....(24,22,33,30)...(19,17,28,25)...(21,19,30,27)   

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. 

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

More cases-Digits being the same

This has reference to my earlier blog title-Digits being the same.
I am giving below another group of 5 digits from among the digits from 0 to 9-
8,6,5,4,1-
You can frame 4 different numbers from these 5 digits,which on doubling will give you 4 numbers formed out of the digits 9,7,3,2,0.
One example is 14865 which on doubling will give you 29730.
Can you find the remaining 3 numbers?

Question N-1(Black hole number 123)

Let me pose a question reg black hole number 123.,referred to in my earlier blog.
Suppose you take a number containing  all odd digits or all even digits..
Do you  think the situation will hold good?
Supposing also there are one or more zeros.How should you count these zeros viz whether as even or odd?
Test a few cases and find the answers.

Two sets of 6 digit numbers

Let me present two sets of 6 digit numbers-with 3 numbers in each set-which show a surprising feature.
The numbers are
123789,561945,642864...........242868,323787,761943.The totals of the numbers in each set is the same.
Now remove the first digit in each no in each set.The totals of the 5digit numbers in each set will be the same.
Continue the process of removing the first digits till you reach single digit numbers.The total of the numbers in each set will continue to be the same.
This is not the only surprising feature.
You can remove the last digit in each number(instead of the first digit).The totals of the new 5digit,4digit...single digit numbers will continue to be the same.
You want another surprise! Instead of adding the numbers,you can add the squares of the numbers The totals will continue to be the same.I will only show for example the position when the numbers are brought down to 2 digit numbers and single digit numbers.
89+45+64=198=68+87+43.....9+5+4=18=8+7+3.
12+56+64=132=24+32+76....2+6+4=12=4+2+6.
For squares the totals are  122 and 56  considering only single digit cases.

Joining digits and making numbers

Let me present two groups of 3 digits each which can be used for making numbers having a surprising property.
First group the digits are 4,5 and 6.They make a total of 15 and their squares make up a total of 77.
Second group -the digits  are  8,3 and 2 giving the total of 13 ,but giving the same total 77 relating to the squares.
You can join one digit from the first group (say 4) with a digit from the second group (say 8) making the numbers 48 or 84.You can similarly join other digits one from each group.- making 6 more 2 digit numbers.The surprising fact is that you will have again two groups of 3 numbers each,each number consisting of 2 digits and the squares of the numbers each group making the same total.-shown below for ready perusal-
The notation S refers to the squares of the numbers involved-
S48+S53+S62=S84+S35+S26=8957
S48+S52+S63=S84+S25+S36=8977
S43+S58+S62=S34+S85+S26=9057
S43+S52+S68=S34+S25+S86=9177
S42+S58+S63=S24+S85+S36=9097
S42+S53+S68=S24+S35+S86=9197