Alphabet number link

Here is an addition sum where the digits are replaced by alphabets-Each alphabet standing for a distinct digit.Find out the actual digits-
    P   L   A   Y
    P  L   A    Y
    P  L   A    Y
        A   L    L
--------------------------
S A  L   L   Y    

Kaprekar constant 5 digits-rejoinder

With reference to my earlier post date 10 sep 2012,I have to indicate the following details-
There is no constant of 5 digits like 74943 and 82962 as mentioned by me.
However there appear 2 series in which these 2 numbers are included.
For instance,we can start with the digits 2,3,5,8 and 9
The difference between the highest number 98532 and the smallest number 23589 formed with these digits is
of course 74943.Using these new set of digits viz 3,4,4,7 and 9 we do not get back 74943 ,in which case only we can treat 74943 as a Kaprekar constant like 6174 you get with 4 digit numbers.
In fact you get 97443-34479=62964.
You repeat now and get 96642-24669=71973
Again repeat you get 97731-13779=83952.
Repeat again,only now you get back 74943.as the same digits 2,3,5,8 and 9 are now involved.
Thus 74943 forms part of the series 74943,62964,71973,83952,74943...repeated back.
Similarly the other number 82962 forms part of the series 82962,75953,63954,61974,82962 repeated back.
You can in fact say if you like that all the 8 numbers involved are Kaprekar constants,instead of only one number like 6174 involved with 4 digits. 

Sequences-4 EMIRPS sequence

The next sequence in my list is the EMIRP sequence of numbers
If you guess correctly these are PRIME numbers which remain prime even when you read from right to left
Some examples are 13,31,17,71,37,73,79,97,101,113,149,157 etc.
The sequence has the number A 006567 in the OEIS website.

Sequences-3 Look and say sequence.

My next sequence is Look and say"" number sequence.The sequence can start with any number say 1.You look at it and say one 1. The next number is thus 11.You look at this and say two one.The next number is thus 21.You look at this and say one two one one.The next number is thus 1211.You proceed accordingly and make more numbers.The list will thus be 1,11,21,1211,111221,312211,13112221,1113213211 ..etc
This comes under the OEIS website number A 005150..

Sequences-2 RATS sequence

Let me now tell you about another sequence.
This also has a peculiar name called  RATS sequence of numbers.
I believe it is based on the procedure of  starting  with a number Reversing it,Adding it to make a Total and Sorting it.
For example if 668 is a number in the sequence,the procedure gives us 668+866=1534,which when sorted to give a number in ascending order gives us 1345,which is the next number in the sequence.Next to this comes from 1345+5431=6776,which when sorted gives us 6677.Similarly the next can be found to be 13444.
The full list of numbers in the sequence is 1,2,4,8,16,77,145,668,1345,6677,13444,55778,  133345
666677,1333444,5567777,1233345,66666677,133333444 and so on.You can see repetition of digits like 3, 6 etc
In the website OEIS referring to the online encyclopedia of integer sequences,this sequence has the number  A004000

Sequences-1 EBAN number sequence

Let me introduce some sequences
The one I first wish to introduce has a funny name.
It is called eban number sequence.
When you write down the names of the numbers in English in this sequence the alphabet E
will be found  banned
The numbers are 2,4,6,30,32,34,36,40,42,44,50,52,54,56,60,62,64,66 and so on.

First attempt to find Kaprekar constants-5 digits

This refers to my earlier blog relating to the 4 digit Kaprekar constant 6174.
To recall the details the number is arrived at by the following process.
Take any 4 digits say 8,3,2,7.
Work out the highest number and smallest number and the difference.
You get 8732-2378=6354
Repeat the procedure now with 6,3,5 and 4 and repeat again as many times as needed.
You will always end up with 6174.
I started similar procedures with 5 digits
I worked out 47 cases.
I got two results 82962 and 74943 in almost equal number of cases
I am trying to work out whether any other result is possible.I feel that no other result could come.
Anyway only a computer programme can show the answer. 

Cyclic number-16 digits- Finding multipliers

In my blog posted on 28 feb 2012,I had mentioned about the following multipliers which make the digits at the end move to the front in the cyclic number  0588235294117647-
1,12,8,11,13,3,2,7...16,5,9,6,4,14,15,10
The second number 12 can be used in another way to find the subsequent multipliers-
The digits 1 and 2 in this number 12 have to be used as follows-
Using any two consecutive numbers in the list say m(1) and m(2),the next number can be found by working out 1*m(1)+2*m(2)
Thus taking m(1)=1 and m(2) =12 we find the next number as 1*1+2*12=25 which reduces to 8 after deducting 17 the generator of the cyclic number.
Again using m(1)=12 and m(2)=8,we find the next number as 1*12+2*8=28 which reduces to 11 after deducting 17 as before.
The next number 13 comes out in a similar way viz 1*8+2*11=30 which reduces to 13 after deducting 17
Next number 1*11+2.13=37 which reduces to 3 after deducting 17 twice
Next number is 1*13+2*3=19 which reduces to 2 as before.
All numbers are found similarly
For example using 4 and 14 the next number will be 1*4+2*14=32 which reduces to 15. 

Remainders -4 Procedure for finding the numbers to get the remainders

Regarding the problem mentioned in my blog Remainder-3 the solution is found as follows-
For remainder 12 in respect of division by 17,the numbers are 29,46,63...
The number 63 satisfies the condition-remainder 11 for division by 13.
For remainder 4 for division by 9,we add  17*13=221 continuously to 63 and test.
The numbers are 63,284,505,726,947,1168,1389,1610,1831..
The number 1831 satisfies remainder 4 for division by 9.
To consider division by 5(Remainder3),we add 17*13*9=1989 continuously to 1831.
The numbers are 1831,3820,5809,7798...
The number 7798 satisfies a remainder 3 for division by 5.
Hence 7798 is the number which satisfies all 4 conditions as required.
The method adopted can be simplified taking into account the fact below-
Instead of the remainder in each case,we consider how much the numbers are short for an even division by the dividing number.
For 13,the remainder should be 11 and for 5 the remainder should be 3-in both case the number should be 2 short for an even division by 13 and 5 and so the smallest such number is 13*5-2=63.
For 17 and 9 ,similarly the number should be short by 5(to give remainders 12 and 4) and so  the smallest such number is 17*9-5=148
This number 148 also gives remainder 3 for division by 5.
We now only have to satisfy the condition of remainder 11 for division by 13.
We hence add 17*9*5=765 continuously to 148 and test.
The numbers we get by adding 765 are 913,1678,2443,3208,3973,4738,5503,6268,7033 and finally7798 which is what we found earlier.

Remainders-3

You can have problems with 4 conditions also-
Example-Remainder 3 on division by 5
                  -do-      4      -do-          9
                  -do-      11     -do-         13
                  -do-      12      -do-        17
Try doing this as per method given by me.
Do you find anything in the problem which will enable you to simplify the method?         

Remainders-2

There has been no response to my previous blog.I desired to know the smallest number which gives remainder 2 when divided by 3,remainder 3 when divided by 5 and  remainder 5 when divided by 7.
There is a method known as chinese remainder theorem.,But I will give you another method
First take the largest divider(7) and the remainder(5).The first number which satisfies the requirement is 7+5=12.,Further numbers can be found by adding 7.They are thus 12,19,26,33,40,47,54,61,68,75,82.....etc
Check these whether you get remainder 3 when divided by 5.Only 68 will satisfy this requirement.
This number 68 surprisingly satisfies the third condition of giving reminder 2 when divided by 3.
So 68 is the required number.
Suppose I had altered condition that you must have a remainder 1 when divided by 3.You then have to add 7*5=35 to 68 repeatedly to check whether you get remainder 1 when divided by 3.The first number you get by adding 35 to 68 is 103.This satisfies the condition that you get remainder 1 when divided by 3
So 103 is the required number which satisfies all the 3 conditions.. .
To get further numbers you have to add 7*5*3=105 repeatedly to 103 .You will have the numbers 208,313,418,523 .....etc.
All these will satisfy all the 3 conditions.
To see whether you have understood the method,I will give you a more difficult problem.
You must find the smallest number which gives remainders as follows-
Remainder 5 when divided by 13
....do-       8     ...do-.             17
...do.....    11.....do-................23
You must get the answer 2150.
Further numbers obtained by adding 13*17*23=5083 repeatedly to 2150.
They are 7233,12316... etc

Arithmetical progressions-2

In my previous blog,I had introduced the subject.
I referred to a progression with the first term 'a' =32 and common difference 'd'=5.
Here is how one can find subsequent terms-
For the second term you have to add the value 'd=5' once to 'a=32' giving you 32+5=37.
For the third term you have to add twice the same value giving you 32+10 =42.
For the fourth term you have to add thrice giving you 32+15=47 and so on.
Hence for the 16th term you have to add the value 15 times to give you 32+15*5=32+75=107
For the 45th term you have to similarly add the value 44 times to give you 32+44*5=32+220=252.
Now for finding the sums-
If the total no of terms is odd ,there will always be a middle term ,double the value of which will equal the total  of the value of the first term and the value of the last term.The total of the value of the second term and the penultimate term will be the same amount..This process continues throughout.
When there are 45 terms in all ,the middle term will be the 23 rd term viz 32+22*5=32+110=142.Double this value is 284.
The total value of the first  and 45th terms is  (32+252)=284..Similarly the total of the values of  the second term (37) and 44th term (247) is (37+247)=284..Similarly you get the same result considering the third term(42) and 43rd term (242) viz (42+242)=284 and so on..
The total of all the 45 terms will hence be 45 times the value of the middle term viz 45*142=6380.
You can check the correctness of this method by first  considering 3 terms then 5 terms,then 7 terms and so on
When the total no of terms is even there will be two middle terms and hence you have to take the mean value of those two terms.
When there are 16 terms,the 8th term(67) and the 9th term(72) will be the middle terms and hence you have to take their mean and multiply by 16,to get the total of all the 16 terms.You will have 16*(67+72)/2=8*139=1112 .

Arithmetical Progressions

I wish to familiarise the topic Arithmetical Progressions
Any sequence of numbers starting with a particular number and progressively increasing with equal additions can be stated to be forming such progressions.
The start number can be called the first term-generally denoted by 'a'
The additions in equal values can be called the 'common difference-generally denoted by 'd'
'a' can be anything like 5,    100      ,3/7    ,4.05,    6^2   etc
'd' can also be similarly anything-can also have negative values.
A few examples shown below-
100,103,106,109,112......
(4.5)   .(5.2)    (5.9)    (6.6)....
7/45......9/45    11/45...13/45....15/45    17/45
1000,992,984,976,968...  etc
Given the first term  like 32 and common difference like 5,we have methods to find out the 16th or 45th etc terms.Also the sums of the various terms upto the 16th term or 45th term.
I will explain the methods in my next blog.

Remainders

I am posing a question for you to solve.
I want you to find the smallest number which gives a remainder 2 when divided by 3,a remainder 3 when divided by 5 and a remainder 5 when divided by 7., 
How can you find more such numbers?

Clock face-Times..Rejoinder.

No one has yet given me a reply.The solution is based on the following facts-
After 12 noon,the hands first merge a little after 01-05
Next a little after 02-10...Next a little after 03-15 etcetc.
In fact,they will be together again several times and for the 11th time they will be at 12 midnight.
Thus in 12 hours,11 intervals occur
Each interval will be {1 +(1/11)}hours or 1 hour and {5+(5/11)} minutes.
The times of merger are thus {5+(5/11)} minutes after 1 PM...{10+(10/11)}minutes after 2PM,{16+(4/11}minutes after 3PM..,{21+(9/11)}minutes after 4PM  etc etc .

Forming fractions

Suppose we have 2 fractions like 2/7 and 4/9.
We may like to find more fractions with values between these two.There is a method for this.In general if a/b and c/d are 2 fractions,then the fraction (a+c)/(b+d) will have values between those two.You can thus find as many fractions as you want between 2/7 and 4/9.
I have shown some of them below.
Let A=2/7 and B=4/9.The first fraction in between C=(2+4)/(7+9)=6/16=3/8
Next D between 2/7 and 3/8=5/15=1/3.Next E between 3/8 and 4/9=7/17and so on.The following is a more detailed list-
2/7,5/17,3/10,4/13,1/3,5/14,4/11,7/19,3/8,5/13,2/5,9/22,7/17,18/43,11/26,3/7,4/9.You can expand the list to whatever length you wish.

Clock face-times when the hour hand and minute hand merge

This is a problem for you.
Both the hands on a clock face merge at 12noon or midnight.
After 12noon,they again merge a little later than  01-05 and again a little later than 02-10.
Can you find the actual times in each case?
Also subsequent stages.

Number 153

I am now introducing a new black hole number 153.
Take any number divisible by 3 like 423.
Find the cubes of digits in 423 and add.You get 99.
Repeat the same process with 99.You get 1458.
Repeat the processes again.
1458 gives you 702...702 gives you 351...351 gives you 153..and finally 153 gives you again 153.
The same result will be arrived at with any number divisible by 3
Second example   63.The repeat processes give you 243,99 and again as in previous case 153.
Third example 168.The results are 729,378,882,1032,36 and again 153 like for 63
You can consider even larger numbers divisible by 3 also.More repetitions may be required.

Recurring decimals

You must have seen decimals with some digits occurring in them like those shown below-
0.23232323.....
0.236236236236....
0.238383838...
0.2538383838...
The first two examples show all digits recurring while the next 2 show only some of latter digits recurring.
I now wish to show how you can convert these into ordinary fractions.
The first will be 23/99.The second will be 236/999
The third will be 2/10+38/990 which will get reduced to 118/495
The fourth will be 25/100+38/9900 which will get reduced to 2513/9900.
Try understanding the process.
Try working out 0.437878787878...etc

Walk the talk-2

Here is an imaginary conversation between me(M) and a number (N)....N starts the conversation talking about itself-
N       -Hello    Let me describe myself    I want you to find me out using the data proposed to be given
M-      Proceed
N-      I am a 10 digit number,in which each digit starting from the first tells its story
First digit-          This shows the number of'zeros' I have in me
Second digit-     The number of 'ones' I have
Third digit-        The number of 'twos I have
Fourth digit-       The number of 'threes' I have
Fifth digit-          The number of 'fours' I have
Sixth digit-         The number of 'fives' I have
Seventh digit-     The number of 'sixes'I have
Eighth digit-        The number of 'sevens' I have
Ninth digit-         The number of 'eights' I have
Tenth digit           The number of 'nines' I have
Can you find me out?
M-Give me some time for finding out
Can you believe me? I found the number finally..It is 6210001000

Right and left-rejoinder

I mentioned about the number 21978 in a previous blog, which when multiplied by 4 becomes 87912,with the digits reversed.

Another similar number is10989 which when   multiplied by 9 becomes  98901  in a similar fashion.

Right and left

The number 21978 is formed out of the 5 digits 2,1,9,7 and 8
If the digits are taken from right to left ,the number formed will be 87912
You will find that 87912 is four times 21978.
Such numbers are not definitely very easy to find. 

Reverse a magic square-Corrigendum

There was a small mistake in my previous blog-The actual position is as follows-
There are 3 magic squares,formed with the numbers 11,16,18,19,61,66,68,69,81,86,88,89
91,96,98,99
First is the normal one,Second one formed by reversing or inverting the first,the third is formed by rotating the first by 90 degrees.All three give the same total 264 all around.They are shown below-
96,11,89,68
88,69,91,16
61,86,18,99
19,98,66,81
Second one-
16,68,99,81
91,89,18,66
88,96,61,19
69,11,86,98
Third one-
18,99,86,61
66,81,98,19
91,16,69,88
89,68,11,96

Reverse a magic square

I  show below a magic square with totals 264 all around-
96,11,89,68
88,69,91,16
61,86,18,99
19,98,66,81
Now reverse the grid.,and make a small change,you will get a new magic square having the same totals all around-
18,99,86,61
66,81,98,19
91,16,69,88
89,68,11,96

Inflation number 72

This is an issue involving inflation.
Suppose you find that the cost of a product or service uniformly goes up every year by 10%.
Will you be able to find out when the cost or service becomes double?
The answer is 7.2 years,the period 7.2 being one tenth of 72.
If the cost goes up every year by 5%,the period will be 14.4 years,the same being one-fifth of 72
Using the number 72,you can find the period for any percentage of increase.
Will you be able to work out how this number 72 is involved?.

Repeating sequence

I am giving below a few initial terms of a sequence-
0,0,0,1,0,2,1,0,3,2,1,4,0,5,3,,2,6,1,7,4,0,8,5.........
The surprising feature of this sequence is that if  you delete the first occurrence of each digit,the original sequence will show up again.
I am showing below the position after deleting the first occurrence of each digit  ,the  deleted digits shown in brackets for ready check-
(0),0,0,(1),0,(2),1,0,(3),2,1,(4),0,(5),3,2,(6),1,(7),4,0,(8),5.....
Can you find out the further digits/numbers in the original sequence so that this property remains in force?

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   . 

More cases-Magic Squares.

This has reference to my earlier blog title-Magic Squares.
I had given therein an example of a magic square formed out of the 9 digits from 1 to 9 giving a total of 15 in each row,column and both diagonals.
I am now asking you to frame Magic Squares with the following properties.-
1)Magic square with zero at the central cell
2) Magic square with  6 at the place similar to that in a clock
3)Magic square with  6 at the place similar to that in a clock and with the totals 42 in each row,column and both diagonals..

Question F-1

A reader has suggested that I could put some questions for getting replies -so that  readers can participate more in connection with the blogs.
I am now therefore putting my fist question-
In my earlier blog I had shown the procedure for testing  numbers for divisibility by 17.I had mentioned that the procedure was worked out by me as per the fact that the number 1003 was divisible by 17.
In fact the number 1003 is the product of 17 and 59.
So the same procedure could with any alteration necessary could be applied to test the divisibility of any number by 59.
I am now giving the numbers    50364080202  and 5036402781
I suggest readers may test whether these numbers are divisible by 59.

Divisibility test for 17

Under my blog account Nadamadum numbers I had given details about a divisibility test for 7,11 and13.This was based on the fact that the product of 7,11 and 13 is 1001.As the product of 17 and 59 is 1003,I have discovered a test for divisibility for 17 on a similar basis.Details are shown below-
Three examples were taken for the test-numbers 159327944,..65482725...45327029331.
I will give full step by step details for the first example 159327944.
Step 1-The number is divided into groups of 3 digits starting from right.They are A-944,B-327,C-159
Step2-You have to divide the number in group A by 3 and subtract the result from group B.If the number in group A is not evenly divisible by 3,you can add a small multiple of 17 to it before division.Thus we add 34 to 944 in group A,making it 978 and the result of division by 3 is thus 326.Subtracting from 327 in group B,we have the result 001.
Step3-We have now to divide the new number in group B by 3 and similarly subtract the result from
Group C..We add 17 to 001 making it 018 and the result of division by 3 being 006,we subtract 006 from 159 in group C making in 153.
Step4-We now test the final number in group C for divisibility by 17.If it is so divisible the original start number would be divisible by 17.In fact 153 is divisible by 17 and hence the test number 159327944 is divisible by 17.The actual quotient is 9372232.
Example 2-Number 65482725.
In this case the number in group B(the penultimate group) will come as 229.If we add 17 to it for making it divisible by 3,the number to be subtracted fromC(the final group) comes as 82 but the number in C is less viz 65.We therefore have to change the procedure and instead of subtracting one-third of B from C and testing the same,we subtract 3 times the number in C from B and test the same.Thus we subtract 3*65=195 from 229 giving us the result 34.This result is divisible by 17 and hence the start number 65482725 will be divisible by 17.The quotient is actually  3851925.
Example 3-Test number 45327029331,-Here we will have a fourth group D=45.
The changed procedure shown in Example 2 is adoptable only when the number to be subtracted from the next group is the penultimate group.Otherwise like in a normal subtraction sum we can borrow 1 from the next group before subtraction..Thus for subtracting (331+17)/3=116 from 029 in group B we borrow 1 from group C and subtract.The new number in group B will be 1029 and the subtraction will give us the result 913 in group B while the number in group C will now be 326 after the borrowal of 1 from 327.The rest of the procedure is the same.The final figure in group D for testing comes as 45-11=34 and as this is divisible by 17 the test/start number viz 45327029331 would be divisible by 17.In fact the actual quotient is 2666295843.
You have to read the steps carefully and make the calculations.Read the steps a couple of times if neceesary..