Cyclic number-16 digits-More information

In an earlier blog I had shown a cyclic number of 16 digits reproduced below-
0588235294117647
I had shown that multiplication of this number successively by 12,8,11,13,3,2,7,16,5,9.6,4,14,15 and 10  makes the last digit at right move to the left side successively.
Let me now tell you how this sequence of multipliers could be found.
The cyclic number is actually found by working out the reciprocal of 17,which could be treated as a generator of the same.The first number in the sequence viz 12 is the most important multiplier.Every subsequent multiplier is found by repeatedly multiplying the previous one by 12.From every result of multiplication you have to however deduct 17 as many times as necessary.
Thus the second multiplier is 12*12-17*8=                                         144-136=8
Third multiplier is                  8*12-17*5=                                            96-85= 11
Fourth                                11*12-17*7=                                          132-119=13
Fifth                                   13*12-17*9=                                          156-153=3 etc etc
You can go upto the multiplier 7 like this .The remaining multipliers are those found by deducting successively those found earlier from 17...Viz17-1=16,   17-12=5,    17-8=9     17-11=6 etc etc
I will inform you in another blog how the first number 12 can be used to find the subsequent multipliers in another way

More surprises-magic squares-3/3 cells

I had shown some data relating to 3/3 cell magic squares in my first blog.
I am now revealing more information
The digits in the rows and columns were as follows-
4,9,2
3,5,7
8,1,6
The sum of the squares of the numbers formed by the digits in each row and each column expressed in two different ways were shown by me to be equal..
The same position is revealed even when the digits in the diagonal and non-diagonal cells are taken into account as shown below-(S means the square of the number)
Relating to diagonal 4,5,6-,...S(456)+S(978)+S(231)=S(654)+S(879)+S(132)=1217781.
Similarly relating to diagonal 2,5,8-     S(258)+S(693)+S(714)=S(852)+S(396)+S(417)=1056609 

Number 1089-Rejoinder

I had mentioned in an earlier blog dealing with number 1089 that we should choose the 3 digit numbers such that in the first step of subtraction we do not get a 2 digit number.In fact you will find that the only 2 digit number which occurs is always 99.This should be treated as the 3 digit number 099 and so on reversal of digits you get 990.The next step then gives you 990+099=1089 as in all other cases.There is thus no exception.

Number 6174

Let me show how the number 6174 occurs and what is strange about it.
The number 6174 is somewhat similar to 1089 discussed in an earlier blog.
Take any 4 distinct digits like say 2,8,3,5
Form the highest number 8532 and the smallest number 2358
Deduct the smaller one from higher one...8532-2358=6174
The process needs several repetitions  many times.However always you will end up with 6174- like a  black hole
Second example-1,8,3,5.......8531-1358=7173......Repeat...7731-1377=6354....
Repeat  6543=3456=3087....Repeat    8730-0378=8352...Repeat....8532-2358=6174
  

Number 1089

Let me show how the number 1089 gets formed by operating on a 3 digit number
Choose any 3 digit number say  731.Reverse the digits and subtract the new number from the one chosen earlier.We get 731-137=594
Now reverse the digits again and add...We will find the result as 1089.
594+495=1089.
You will get this number 1089 in all cases whatever the number chosen earlier-
Exceptions-For subtraction,the smaller number should be subtracted from the larger number.
The numbers should be so chosen ,so that the subtraction produces a 3 digit number.
Readers could try and see how the system works with 4 digit numbers. 

Case of a triangle sides of length 13,14 and 15

Let me bring to the notice of readers a property found by me.
Consider a triangle formed with sides of length 13,14 and 15 units.
Suppose you draw an altitude line on the side having length 14 units from the opposite vertex. Will you be able to visualise what its length will be.You will be surprised to know that it will be of length 12 units.This is perhaps the only such case involving lengths of 4 consecutive units in the form of 3 sides and one altitude.

More information-Tricky addition sums

There has been no response from readers-I am hence explaining the trick
In each case the digits in the first row are the same as the digits in the first column.Similarly second row/second column,third row/third column,fourth row/fourth column.etc.Except for this there are no other repetitions and thus only the 10 digits from 0 to 9 are in use in every case.Because of this restriction a computer program is needed to get the required values. 

Generator 271-Cyclic number 00369

In an earlier blog under my account Nadamadum numbers,I had dealt with generator 41 and the corresponding cyclic number of 5 digits viz 02439.
I am now presenting details about another 5 digit cyclic number 00369 from 271 operating as the generator..
(You will observe that 9 times 41 is 369 and 9 times 271 is 2439)
Cyclic number 02439 (5digits) had with it 7 other subsidiary cyclic numbers of 5 digits .All these  8 cyclic numbers  made it possible to have cyclic formations of 02439 due to multiplication of the same by all the 40 numbers from 1 to 40.
In the case of 271,there are 270 numbers from 1 to 270  for multiplication.So you need 54 cyclic numbers each of 5 digits and hence there are 53 additional (subsidiary)  cyclic numbers all multiples of the main one viz 00369. Showing all of them is not going to be easy.So I will show only a few examples-with the suffix M relating to the number by which 00369 is multiplied.
M-1 is 00369.This has cyclic formations of 03690,36900,69003 and 90036 involving M-10,M-100,M-187 and M-244.
This has relationship with cyclic formations 99630,96309,63099,30996 and 09963 involving M-270,M-261,M-171,M-84 and M-27.By relationship I mean the cyclic numbers add up to 99999 and the multiplying numbers add up to 271..-For example 69003+30996=99999 and 187+84=271etc
Another pair of cyclic numbers and cyclic formations----08856(M-24)...88560(M-240),85608(M-232),56088(M-152) 60885(M-165) and...91143(M-247),11439(M-31),14391(M-39),43911(M-119),39114(M-106)...60885+39114=99999 and 165+106=271 etc.
We will thus have 27 pairs having such relationships.
Another pair relates to M values 17,170,74,198,83 and 254,101,197,73 ,188.-..M-17 is 06273 and M-254 is 93726.
Readers may if they like find the other 24 pairs.

Magic squares-3/3 cells-another method for formation

Let me show another method for formation of a 3/3 cells magic square.I would have liked to show the procedure with a diagram where one can easily understand but it is difficult for me to show diagrams now. The method is being shown specifically for the reason that the same can be adopted for forming a 5/5 or 7/7 magic squares
You first write down the digits 1,2, 3  in the first row followed with digits 4 ,5,6 in the second row and digits 7,8,9 in the third row..
Next you draw parallel lines in a diamond like formation,where you will get a square with 3/3 cells with the digit 2 in the top single cell, and the digit 8 in the bottom single cell.Just below the top single cell there will appear   two empty cells and just above the bottom single cell there will similarly appear 2 empty cells.In between these two you will have 3 cells with the digits 4,5 and 6.The digits 1,3,9 and 7 will appear outside the grid of 3/3 cells.It will then be necessary to transfer these 4 digits into the 4 empty cells inside.The transfers have to be made in the diametrically opposite direction.After this the digits 9 and 7 will come in the two cells below the top cell containing digit 2 and the digits 3and 1 will come in the 2 cells just above the bottom cell containing the digit 8.
With this you will get the final 3/3 magic square,though in a diamond like formation..
I have shown a diagram here- to make it easy for readers to understand the steps...

Divisibility test for 19

In one of my earlier blogs I had shown the procedure for testing numbers for divisibility by 17.This was based on the fact that 17 is a factor of 1003.On a similar basis you can formulate a procedure for testing numbers for divisibility by 19 based on the fact that 19 is a factor of 1007.
You have to form groups of 3 digits each like A,B,C,D etc commencing from right as before.The only difference is that you have to work out one seventh of the number in Group A and deduct it from the number in Group B and follow the same procedure further on.As before you can add a multiple of 19 like 19,38,57,76,95 or 114 in case the number to be divided  does not permit even division.
You can try testing the following numbers
28415015603
238815028
415920368715796
As  53 is also a factor of 1007 you can follow similar procedure for testing divisibility by 53.Try testing the number 7629781579 for divisibility by 53.,. 

Tricky addition sums

I am giving below  examples of 2 types of addition sums-Are  you able to find anything strange?You write the numbers one below the other and then you will find what is strange.
First type-
1529+5837+2340=9706.
1259+2430+5387=9076
1529+5746+2408=9683
2318+3790+1956=8064
3217+2945+1406=7568
Second Type-
5481+4692=8970+1203
1405+4893=0972+5326
1560+5849=6437+0972
1634+6785=3829+4590
2680+6795=8934+0541
These cases were found by me with the use of computer programmes.I have with me data for 80 cases of the first type and 240 cases of the second type.

Magic Squares-3/3 cells-Hints

I had asked readers in one of my earlier blogs to frame magic squares in compliance with certain conditions.
I am giving below some hints-
It is merely necessary to replace the digits with new numbers/digits to comply with the conditions as indicated below-(The first row shows the digits 1 to 9 while the subsequent rows show the replacement values.The last column shows the totals in the case of each square formed with the replacement values)
1         2         3         4         5         6        7         8         9         15
4         5         6         7         8         9       10        11       12       24
0         2         4         6         8        10      12        14       16       24
-4      -1         2         5         8        11      14        17       20       24
20      17       14       11        8         5        2         -1       -4        24
6        6.5       7        7.5       8        8.5      9         9.5      10       24
8        9         10       11       12       13      14        15       16        36
8        7          6         5         4         3        2          1         0         12
8        6          4         2         0        -2      -4         -6       -8         0
8       10        12       14       16       18      20        22       24        48
The first 5 of the replacement values indicate that the central cell will have the number 8 giving the totals 24 in all those cases.The subsequent replacement values indicate that the digit 1 should be replaced by  8 leading to different totals based on the number/digit replacing 5.
You can similarly form different magic squares adopting some other number/digit instead of 8 adopted  in the above examples.

Divisibility test for 59

I had asked readers to attempt a  test for divisibility by 59 of 2 numbers based  on the procedure shown by me for testing divisibility by 17 in an earlier blog.The same procedure is applicable for both because the product of 17 and 59 is 1003 and the procedure was based on this fact.
I had suggested that a minor alteration is needed.
This alteration is to add a multiple of 59( instead of a multiple of 17) for evenly dividing by 3.I am giving below the steps for one of the numbers viz 50364080202 indicated by me as this involves a new fact not specifically explained by me.
Number under test-50364080202
Step 1--Groups  A-202...B-080...C-364...D-50
Step 2--A multiple of 59 has to be added to 202 as this is not divisible by 3.We add 59 altering 202 as 261.Division by 3 gives us 87.
Step 3-As 87 or 087 is more than 080 in Group B we borrow 1 from Group C .Thus 080 becomes 1080 and subtraction of 087 will give us 993.The number in Group C is now 363 as 1 has been borrowed from 364.
Step 4-Division of 993 by 3 gives us 331 and subtraction of 331 from 363 will give us 032
Step 5-As 032 is not evenly divisible by 3 we add 118( a multiple of 59) to it making it 150 .Division by 3 will give us 050.
Step 6-This is now  to be deducted from the number in Group D and the balance tested for divisibility by 59.But there is no balance as the subtraction of 050 from the number 050 in Group D gives us zero.I had not specifically indicated earlier what should be done in such cases.I have to state that in such cases the final figure should be treated as evenly divisible by the divisor viz 59.-as zero times 59 is zero.
The final conclusion is that the test number viz 50364080202 is divisible by 59.
The actual quotient is 853628478
Readers may now test the second number mentioned by me viz 5036402781   .