User:Madhukar12134

The test for the divisibility of 11:-

To test the number to be divisible by 11 we have to add the last two numbers of the given number and we have to continue like that till it comes to two digits. If the comming last two digit number is divisible by 11 then the whole number is divisible by 11

EX:- 50688 >> 506+88=594 >> 5+94=99

as 99 is divisible by 11,50688 is divisible by 11

EX:80685 >> 806+85=891 >> 8+91=99 10879 >> 108+79=187 >> 1+87=88   869   >> 8+69=77

Multiplication by 11(i.e <99):-

To know the product of a number multiplying with 11 we have to write the number in such a way that we have to leave a gap between two numbers and add the two numbers and write the resultant number in the gap.

EX:-  11 * xy = x_y         (_=x+y) therefore the product is >> x(x+y)y

EX:- 11 * 26 = 2_6  >>   the product is 2(2+6)6 = 286

exeption: If the sum of the two numbers is greater than or equal to ten we have to write the ones digit in the gap and the tens digit we have to add to number before it.

EX: 11 * 69 =6(6+9)9 =(6+1)59 because 6+9 equal to 15, 5 is in ones place, 1 is in tens place so we have to write the 5 in the gap and add 1 to 6 therefore the product is 759.

Test for the divisibility of 99:

It is same as the divisibity of 11 if the coming the last two digits are divisible by 99 then the whole number is divisible by 99

EX: 10979892 >>109798+92=109890 >> 1098+90=1188 >>11+88=99 The last two digits are 99 therefore the number is divisible by 99.

Multiplication with 99: (i.e<99)

If we multiply a two digit number with 99 the product will be 4 digited number. In this we have to write the first digit(thousands place) of the product as the tens digit of the multiplying number then leave a gap and the third digit should be written in such a way that the sum of first and third digits is equal to 9 then again the 4th, 2nd digits in the product is empty. We have to fill this two gaps by multiplying the ones digit of the given number with 9 then we can get two digits the gaps should be replaced by the above two digits

EX: 99 * 65 1st step : 6_(9-6)_ 2nd step : 6_3_ 3rd step : 9*5= 45 (5 is ones digit of multiplying number) 4th step: product 6435

Simple Large and Medium numbers(SLM numbers):

According to the numbers are divided into 3 types they are small, large, medium numbers Small numbers: If a number is divided by 3 then the remainder is 2 the the given number is called small number.

EX:2,5,8,11,1097,1196,1394 etc....

Medium numbers: If the number is exactly divided by 3 then the give number is called medium numbers. EX: 3,6,9,12,15,18,12342, 178239,2312322 etc......

Large numbers: When the number is divided by 3 remainder is 1 the the  number is called large numers. EX:- 1,4,7,10,9610,2312323,etc...... points: 1. Sum of two small numbers is a large numbers EX:- 2+5=7 (7 is a large number)

2.The sum of two large numbers is a small number. EX: 4+13=17 (17 is a small number)

3. The sum of two medium numbers is a medium number. EX:- 6+9=15 (15 is a medium number)

4. The sum sum of small and large number is medium number. EX:- 2+4=6 (6 is a medium number)

5.The sum of medium and small number is small number. EX:6+2=8(8 is a small number)

6.The sum of medium and large number is large number. EX:3+4=7 (7 is large number)

Importance of zero and in divisibility of 11:-

The number zero has no value as actully but when it is placed in the right side of the given number the value atomatically increases. It has great role in the 11 it is possible if in the case of even digited numbers (greater than 4) The number should be same it means in a four digited number there should be two 2 digited numbers. and in six digit 2 three digit numbers etc...... In all these cases the zero is placed in such a way that it divides the number exactly two parts.

Let us consider first with four digit numbers let the number be 7373 Generally the number is not divisibile by 11 but if we place a zero between 73 and 73 the number will be divisible by 11. that is 73073 is divisible by 11.

Let us consider 6 digit number that is 108108 This number is divisible by 11 but if we place a zero between 108 and 108 the number is not divsible by 11. that is 1080108 is not divisible by 11.

Lets take a number divisible by 11 if we placed a zero between the last two numbers then we have to exchange the two numbers in order to make it divisible by 11 EX: 2486 is divisible by 11. 24086 is not divisble by 11. exchanging the last two digits we get 24068 is divisible by 11 EX: 2475 is divisible b y 11, by placing zero we have to change it to 24057 EX:198297,1982079 both are divisible by 11

The linking of small large and medium numbers with 99:

99 number table contains minimum 2 to 3 digits there will be one small and one large and one medum number EX: 198 >> 1 is large 9 is medium and 8 is small if we take 4 digit numbers there will be one small and one large or 2 medium numbers EX: 4356 >> 43 is large and 56 is small if a number contains 6 digits it is devided into 3 parts exactly. then the three parts are

1st: three parts are small numbers 2nd: three parts are large numbers. 3rd: one small, one large, one medium 4th: 3 medium numbers

EX:383526 	>>   38,35,26 are small numbers EX:465894    >>    46,58,94 are large numbers EX:367191    >>    36-medium,71-small,91-large numbers EX:275418    >>    27,54,18 are large numbers

Double number present in the 99 theory: We have to write it like this only because if we add the last two numbers of the given number we can get 99 EX:99*1=99, 99*2=198 We have to write it like this only because if we add the last two numbers of the given number we can get 88 EX:187=1+87=88 similarly 286=2+86=88 and so on

table of 3= 3,102 ,201 ,300 ,399 ,498, 597 ,696 ,795 ,894 ,993 etc............................ We can write it up to 99 because if we take 100 that is 100=1+00=1 there fore  we can write the table of 100 as a table of 1 In this way we can find the simplest numbers i.e simplest number of 1241is 53 therefore instead of writing the table of 1241 we can write the table of 53 and 53 is called the simplest value of 1241 simplest value of 697=103 Like this we can write the table for every number. Now we are coming to find the double numbers  in then given table Ex:50886-table with 88 >>>>,   52866,     53856,     54846,     55836 ,    56826,   57816 ,   58806 ,   59796

1.   50886, 2. 52866, 3. 55836, 4. 58806.

In the first number 50886 >> 88 is repeated similarly in second number 66, in third number 55, and in fourth 88 id repeated so in the total series that is 50886 to 59796 there are four repeated numbers. Hence this series is called quarterkar series in this quarter defines the number of doubled digits and kar inclueds my name. That is madhukar. Similarly we can have when pentakar, triakar, diakar, hexakar and decakar series. Pentakar series:- 60786, 61776, 62766, 63756, 64746, 65736,66726, 67716, 68706, 69696 1st 61776, 2nd 62766, 3rd 66726, 4th 67716, 5th 69696 It contain 5 double numbers therefore these series is pentakar series. Triakar:- 20988, 21978, 22968, 23958, 24948, 25938, 26928, 27918, 28908, 29898 1st 20988, 2nd 22968, 3rd 29898. It contains three doubled numbers. Therefore it is triakar series.

Diakar: 6039, 6138, 6237, 6336, 6435, 6534, 6633, 6732, 6831, 6930 1st 6336, 2nd 6633 It contain two doubled numbers therefore it is diakar series.

Hexakar series: 520344, 521334, 522324,523314,524304, 525294, 526284, 527274, 528264, 529254 1st 520344, 2nd 521334, 3rd522324 , 4th 523314 , 5th 525294 , 6th 527274. It contains six doubled numbers. Hence it is hexakar series. (Note: i think this is the only number having six doubled numbers when i found this number i was really shocked.)

Decakar series: 11088, 11187, 11286, 11385, 11484, 11583, 11682, 11781, 11880, 11979. 1st 11088, 2nd 11187, 3rd 11286, 4th 11385, 5th 11484, 6th 11583, 7th 11682, 8th 11781, 9th 11880, 10th 11979. It contains ten doubled numbers. Hence it is decakar series.

Monakar series: 36, 135, 234, 333, 432, 531, 630, 729, 828. 927. It contains only 1 doubled number that is 333 Hence it is monokar series.

Fun with 11: For some numbers if we add the last two numbers with the remaining digits, we can get the remaining digits or the simplest numbers of the remaining numbers. (Note: In these the last number should be 9)

series: 1.109 >>> 1+09=10 2.219 >>> 2+19=21 3.329 >>> 3+29=32 4.439 >>> 4+39=43 5.549 >>> 5+49=54 6.659 >>> 6+59=65 7.769 >>> 7+69=76 8.879 >>> 8+79=87 9.989 >>> 9+89=98

In this we can observe that by adding the last two numbers with the first number we can get the first two numbers. If we observe clearly we can notice that differance between two adjacent numbers is 110. If we continue by addind 110 to 989 we can get a new series. That is 1099, 1209, 1319, 1429, 1539, 1649, 1759, 1869, 1979 EX: 1759 >>> 17+59= 76 (76 is the simplest number of 175)

If we take 6,4 digited numbers and divided them into 2 or 3 parts and if we reverse that number we can get another  number  and that particular number is also divisibile by 11 (note: the taken number should be divisible by 11 ) ex:191763 If we divide it in to 3 parts we get 19, 17, 63 this number 191763 is divisibile by 11 .If we reverse the 3 parts the getting number is also divisibile  by 11.That is by reversing 19,17,63 we get 91, 71, 36. If we unite them we get a number i.e 917136 is also divisibile by 11 EX: 2486 by dividing we get  2 parts i.e 24,86. by reversing it we can get 4268 is also divisibile  by 11.