Squaring Trick -: 5
The square of , one (1^2 = 1 ), 11^2 = 121
In this blog we will discuss about the trick used in squaring of 1 , 11,111,11111 etc.

Squares of various number are -:
1^2 = 1
11^2 = 121
111^2 = 12321
1111^2 = 1234321
11111^2 = 123454321
111111^2 = 12345654321
111111111^ 2 = 12345678987654321
Procedure -: To find the square of number like 11,111,11111 e.t.c we apply following step

Step(1). First of all saw how many times
One (1) repeat in the number like
(a). 11 -:  in 11 , one repeat two times
(b). 111111-: in 111111, one repeat 6 times.

Step(2).let 'N' times one repeat in the given number i.e. the the number whose square is to be find  , write down successors from one to  'N'. like
(a). 1111^2 ?
And -: N = 4
Then  successors= 1234

Step(3). Now write predecessors from
'(N - 1)' to one . Like
Since N = 4
N - 1 -: 4 - 1 = 3
Now predecessors  from 3 to 1 -: 321

Step(4). Now write down the number which comes in step 3rd after the number comes in step 2nd , this number is the square of the given number.like

1111^2 = 1234321 will be the answer.

Example (2). -: 1111111^2 ?
Answer-: step(1). As we saw the given number , one comes seven time.
Step(2). Now , N = 7 , then successors from 7 to 1
Then successors = 1234567

Step(3).( N - 1 )= 7 - 1 = 6 .
Predecessor's from 6 to 1 = 654321.

Step(4). 1111111^2 = 1234567654321 will be the answer.

Note -: Similarly we find the square of any number in some second.

                       Squaring Trick -: 4


11^2 = 121   , 101^2 = 10201
In this blog we discuss squaring  tricks  for two types of number
(1). The number whose multiplication of  first digit and last digit  with 2 have a single digit .

Example -:  (a) take a number 13
If we multiply the first digit and last digit with 2 (1×3×2 = 6) then it is single digit .

(b). 12 , if we multiply first & last  digit with 2 (1×2×2 = 4) then we will get single digit

(2). The number whose multiplication of first digit & second digit with 2 have more than single digit.

Example -: (1). Let take a number 18 , if we multiply its first & second digit with 2 ( 1×8× = 16 ) which have more than one digit.

Let us start -: (1) if 11^2 = 121 , then
(a). 101^2 = 10201.
(b). 1001^2 = 1002001
(C). 100001^2 = 10000200001 and so on
What did we apply tricks  in these question?

Answer -:  step(1). In these if we palace zero between the digit of number 11,then these numbers like 101, 1001, 100001 create.
Step(2). If we want the value of 101^2 ?
Then place one zero (because if we palace one zero between the digit of 11 it will create)  between the digits of square of 11.

Like
Questions (1). 11^2 = 121 then
101^2 = 10201.
Similarly -
1001^2 = 1002001.
1000^2 = 100020001.

Question (2). (A).10002^2 ?
Answer -: step (a) 12^2 = 144
Step(b). 10002^2 = 100040004 will be the answer.
Similarly -
(B). 102^2 = 10404 and so on.

Question(3). (A). 10003^2 ?
Answer -: step (1). 13^2 = 169
Step(2). 10003^2 = 100060009.

(B). 103 ^ 2 = 10609 will be the answer.
Question (4). 201^2 ?
Answer -: step(1). 21^2 = 441
Step(2).201^2 =  40401 will be the answer.
Similarly -
2001^2 = 4004001
200001^2 = 40000400001 and so on.

Question (4). 31^ 2 = 961 then (30001)^2 = 900060001 and so on .
Similarly you find the value of
(1). 100002^2 ?
(2). 2000002^2 ?
(3). 4001^2?

Type (2).
Question (1) . (A). 2008^2 ?
Answer -: Step (a). Square the first digit, 2^2 = 4
Step (2). Multiply first & second digit with 2
-: {2×8}×2 = 32
Step (3). Square the last  digit ,8^2 = 64
Step(3). Now collect these numbers from step a to c & place one zero between every step to before collection i.e. 4032064.
Hence 2008^2 = 4032064 will be the right answer.
Note -: we put (n - 1) zero before collecting the numbers. Where ''n' is the number of zero in the given number.

Similarly
Question (2). 20008^2 = 400320064.

Question(3)200008^2 = 40003200064. And so on.
Question (4) -: 40008^2 = 1600640064

Question (5). 600008^2 = 360009600064.

Note -: If we practice these two tricks carefully then we solve these types of questions within a minute.

                                 SQUARING TRICK -: 3rd

Note -: when we want to square any number then we also apply another method which is as follows :
 Take an example -: 116^2 ?

Step (1).  Multiply unit digit of the given number with 2 . (6×2 = 12)

Step (2). Now multiply the given number with the number which is two times of unit digit. (116×12)

Step (3). Find the square of unit digit of the given number. (6^2 = 36 = 6×6)

Step (4). Now subtract the square of unit digit from the number which comes in step second . {( 116×12 )- 36} = 1356.

Step (5). Square the number which is  multiple of 10 comes before the given number . if 116 is given then the multiple of 10 before it is 110. (110^2 = 12100)

Step (6). Now add the numbers which we find in step '4' and step '5' , which is the answer of square of the given number .
(1356+12100 = 13456).

So , 116^2 = 13456 will be the answer..

Question(1) -: 118^2 ?

Answer -:
step(a) . unit digit is 8 in this question so ,multiply  it with two  -: 8 × 2  = 16

Step(b) . multiply the double (the answer which comes in step a) of unit digit with the
 given number -: 118 × 16= 1888

Step(c) . square the unit digit of the given number  which is 8 -: 8^2 = 64

Step(d) . now subtract the answer of step c (64) from step b (1888) -: 1888 - 64 = 1824 .

Step(e) . square the number which is multiple of 10 which comes before the given  number. 110 is the number which is multiple of 10 and come before 118 -: 110^2 = 12100.

Step( f). Now add the resultant numbers of step 'd' (1824) and 'e'(12100) -:
1824 + 12100 = 13924 .

Question -: 239 ^ 2 ?

Answer -: step (a) -: 9×2 = 18.
Step (b). 239 ×18 = 4302
Step (c). 9^2 = 81
Step (d). 4302 - 81 = 4221.
Step(e .  230^2 = 52900
Step ( f) . 52900 + 4221 = 57121
Answer willbe 57121.

Question -: 523×523(523^2) ?
Answer -: step(a). 3×2 = 6
Step(b). 523×6 = 3138
Step(c). 3×3 (3^2) = 9
Step(d). 3138 - 9 = 3129
Step(e). 520^2 = 270400
Step(f). 270400 + 3129 = 273529
Hence ans will be 273529.

Question -: 2782 ^ 2 ?
Answer -: step(a).2×2 = 4 (here unit digit number is 2 which is multiply with 2)
Step(b). 2782×4 = 11128
Step(c). 2^2 = 4
Step(d). 11128 - 4 = 11124
Step(e).2780^2 = 7728400
Step(f). 7728400 + 11124 = 7739524
Answer will be 7739524.

                                       SQUARING TRICK

Note -: As we see below , the symbol  " ^ " comes every where .what is the use of this symbol ?
Answer -: The use of this symbol (" ^ " ) is to raise the power of any number like
26^2 = 26×26 ( 26 raise to the power 2)
48^2 =  48×48
579^2 = 579×579
6978^ 2 = 6978 × 6978
6978^3 = 6978×6978× 6978 ( 6978 raise to the power 3)
6978^5 = 6978×6978×6978×6978×6978 (6978 raise to the power 5)

Type 1st -:

(1) Find the answer of  -: 36^2

Procedure  -: step (1) . square the one's digit of the given number  .
6^2 = 36 . (36^2 =     6)

Step(2). Multiply one's and tenth digit of the given number to each other and then multiply it with 2.
(3×6)×2 = 36 .
Now add carry over 3 from 36 to 36
= 36 + 3 = 39.    (36^2 =      96)

Step (3) . square the tenth digit of the given number .
3^2 = 9.
Now add 3 (carry over from 39) to 9
= 9 + 3 = 12 .   (36^2 = 1296)
Answer -: will be 1296 .

Note -: By applying this method we can solve the square of any number without written work within seconds

Question (2) -: Find the result of 582^2.

Step (1). Square the ones digit of the given number  = 2^2  = 4 = 2×2
Step (2) . multiply ones and tenth digit of the given number and then multiply it by 2
= (8×2)×2 = 32   { 582^2 =         24}

Step (3) .  multiply ones and hundredth digit of the given number to each other, and then multiply it with 2 ,  and then add square of tenth digit to it   -
= { (5×2)×2 + ( 8^2) } = { 20 + 64 } = 84 .
Now add carry over 3 from 32 to 84  -
84 + 3 = 87 .    { 582^2 =      724}

Step (4) . Multiply tenth and hundredth digit of number to each other and then multiply it with 2 -
(5×8)×2 = 80 .
Now add carry over 8 from 87 to 42
= 80 + 8 = 88 {582^2 =   8724 }

Step ( 5) . square the value of hundredth of the given number
 5^2 = 25
Now add  8 from 88 to 25 .
 25+8 = 33 .
The answer become -: 338724.
.
Question (3) -:Find the value of  4248^2.

Answer -:
step (1) -: 8^2 = 64         {4248^2 =       4}

Step (2) -: (4×8)×2 = 64
Add 6 ( carryover ) from 64 to 64= 64+6 = 70
{4248^2 =      04}

Step(3) -: {(2×8)×2 }+ (4^2) = 48
Add carryover 7 to 48 = 48+7= 55.
 {4248^2 =    504}

Step(4) -: {(4×8)×2} +{ (2×4)×2} = 80
Add 5 (carryover from 55) to 80 = 5+80 = 85
{4248^2 =   5504}

Step (5) -:{ (4×4)×2}+(2^2) = 36
Add 8 (carryover from 85) to 36 = 36+8 = 44
{ 4248^2 =   45504}

Step(6) -: (4×2)×2 = 16
Add 4 (carryover from 44) to 16 = 16+4 = 20
{ 4248^2 =    045504}

Step(7) -: 4^2 = 16
Add 2 (carryover from 20) to 16 = 16+2 = 18
{4248^2 = 18045504}
So the answer of 4248^2 = 18045504.

                      MULTIPLY TRICk
Note-; when we multiply two number , Remembered that every number multiply to each other at ones .
 Note -: this is also vedic method .

Type (1) -:

(a) . Find the answer -:   26×12

Answer-:
step 1. Multiply ones digit of these numbers , mean to say -
 6×2 = 12.    { 26×12=       2}

Step 2 . multiply ones digit of 26 to tens digit of 12 and tens digit of 26 to ones digit of 12 and add the resultants -
(6×1)+(2×2) = 10
 Now add  1(carry over from 12) to 10
1+10=11 .  {26×12=        12}

Step 3. Multiply tenth digit of these number
              2×1= 2
 Add 1 ( carry over from 11) to this result
 2+1 = 3 .  { 26×12 = 312}

Answer -: will be 312.

(b) . find the answer -: 79×13
Answer -:
Step 1. 9×3 = 27   {79×12 =     7}
Step 2. (9×1) + (7×3) = 30
Now , add 2 (carry cover from 27) to 30
30+2= 32 .   { 79×13=    27}
Step 3 . 7× 1 = 7
Now add 3 (carry over from 32) to 7
7+3=10 .  { 1027}

Answer -: will be 1027.

Type (2) -: find the result -: 378×27 .

Answer -: step 1 . 8×7 = 56.   { 378 ×27 =     6}

Step 2 . (7×7) + ( 8×2) = 65
Now ,add 5 ( carry over from 56) to 65
65+5 = 70 .   { 378 × 27 =            06}

Step 3. (7×3) + (2×7) =  35
now, add 7 to 35
35 + 7 = 42 .    { 378×27 =      206}

Step 4 . 2×3 = 6
now add 4 to 6
 4+6 = 10 .  { 378 × 27 = 10206}

Answer-: will be 10206

Type (3) -: Find the result of 376×482

Answer -: step (1) . 6×2 = 12 .  (376×482 =    2)

                  Step(2). (6×8) + (7×2) = 62
Now add 1(carryover from 12) to 62
= 62 + 1 = 63 .   (376×482 =          32)

                  Step (3). (3×2) + (6×4) +(7×8) = 86
Now add 6 (carry over from 62) to 86
      = 6 + 86 = 92. (376×482 =        232)

               Step(4). (3×8) + (7×4) = 52
Now add 9 (carry over ) to 52
    = 52 + 9 = 61 . (376×482 =       1232)

         Step(5). 3×4 = 12 .
Now add 6 (carry over from 61) to 12
         = 12+6 = 18 .  (376×482 = 181232 )

Answer -: 376 × 482 = 181232.

Note -: we see that every number from 376 and 482 multiply to each other at ones time .

          Perfect square of numbers up to 100

Note -: if we learn perfect square of numbers up to 100 then it is easy for us in various competitive exams .

1×1 = 1
2×2 = 4
3×3 = 9
4×4 = 16
5×5 = 25
6×6 = 36
7×7 = 49
8×8=64
9×9 = 81
10×10=100
11×11=121
12×12 = 144
13×13 = 169
14×14=196
15×15 = 225
16×16 = 256
17×17 = 289
18×18 = 324
19×19 =361
20×20=400
21×21=441
22×22=484
23×23=529
24×24=576
25×25=625
26×26=676
27×27=729
28×28 = 784
29×29=841
30×30=900
31×31=961
32×32=1024
33×33=1089
34×34=1156
35×35=1225
36×36=1296
37×37=1369
38×38=1444
39×39=1521
40×40=1600
41×41=1681
42×42=1764
43×43=1849
44×44=1936
45×45=2025
46×46=2116
47×47=2209
48×48=2304
49×49=2401
50×50=2500
51×51=2601
52×52=2704
53×53=2809
54×54=2916
56×56=3136
57×57=3249
58×58=3364
59×59=3481
60×60=3600
61×61=3721
62×62=3844
63×63=3969
64×64=4096
65×65=4225
66×66=4356
67×67=4489
68×68=4624
69×69=4761
70×70=4900
71×71=5041
72×72=5184
73×73=5329
74×74=5476
75×75=5625
76×76=5776
77×77=5929
78×78=6084
79×79=6241
80×80=6400
81×81=6561
82×82=6724
83×83=6889
84×84=7056
85×85=7225
86×86=7396
87×87=7569
88×88=7744
89×89=7921
90×90=8100
91×91=8281
92×92=8464
93×93=8649
94×94=8836
95×95=9025
96×96=9216
97×97=9409
98×98=9604
99×99=9801
100×100=10000

                                   SQUARING
Note-:
            We must consider the importance of squaring because many problems we solve easily if we learn squaring . some basic methods of squaring is given below

 (A) . Square of the number whose last digit is 1 or ending with 1 -:

To get the square of any such number , we write the square of previous number and then add the previous number and the number whose square being asked .

Let an example -: The square of 61

61^2 (61×61) -: 60^2 + 1(60+61) = 3600 + 121 = 3721
 Similarly, 91^2 = (90^2) +1(90+91) = 8100 + 181 = 8281
and  41^2(41×41) = (40^2) +1(40+41) = 1600 + 81 = 1681

And   161^2 = (160^2) + (160+161) = 25600+ 321 = 25921

And.    1111^2 = (1110^2) + 1(1110 + 1111) = 1232100+2221 = 1234321.

NOTE-:  by using sign "^" we write power of any number like 61^3 means cube of 61 or (61×61×61) ,118^5 means 118 to the power 5  or (118×118×118×118×118).

NOTE -: We see above that 1 is multiply by addition  {1(1110+1111)} here 1 is the difference between the number whose square is required and the previous number . {1111 -1110=1}

(B). Square of the numbers whose last digit is 2 or 8 -:
Let an example
(a) 18^2  = (20^2) - 2(18+20) = 324

 (b) 112^ 2 = (110^ 2) + 2(110+112)  = 12100 + 444 = 12544

 (C) 38^ 2 = (40^2) - 2(38+40)  = 1600 - 156 = 1444 .

 Note -: here we see that 2 is multiply which is difference difference between the number whose square is required and the previous or next number .like
112 - 110 = 2
40 - 2 = 38

(C) . square of the number whose unit digit is 3 or 7 -:

Let an example -:
(a). 63^2 = (60^2) + 3(63+60) = 3600 + 369 = 3969 .

(b). 137^ 2 = (140^2) - 3(137+140) = 19600 - 831 = 18769

(C). 47^2 = (50^2) - 3(50+47) = 2500 - 291 = 2209

(d) . 2347^ 2 = (2350^2) - 3(2347+2350) = 5522500 - 4691 = 5517809 .


(D) . square of the numbers whose unit digit is 4 or 6 -:

(a) 46^2 = (50^2) - 4(50+46) = 2500 - 384 = 2116 .

(b) . 46^2 = (40^2) + 6(40+46) = 1600 + 516 = 2116.

(C). 54^2 = (50^2) + 4(50+54) = 2500 + 416 = 2916 .

(d) . 124^2 = (120^2) + 4(120+124) = 14400+ 976  =15376.

NOTE -: Generalized formula for this method is given below
Let ," N"is the number whose square is to be calculated and "B" is the base and "d" is the difference between "N"and "B" then
 (1) . N^2 = (B^2) + d(B + N)        , when B<N .
(2) . N^2 = (B^2) - d(B + N)          , when B>N .

(E) Square of the number whose unit digit is 5 -:

Note -: any number whose unit digit is 5 . we write the 25 in rightmost in place of 5 , then we multiply the rest number with its successive number ( i.e one more than the previous one ). Look at some examples

Example -: square of 45
Solution -: 45^2 =  step 1. 25
                                 Step 2. 4(4+1) = 20
                                 Step 3. 2025
So , (45^2) = 2025 Answer.
 (2). 125^2 = step 1. 25
                       Step 2. 12(12+1) = 156
                       Step 3. 15625
So , (125^2) = 15625 answer.
(3). 345^2 = step1. 25
                      Step2. 34(34+1) = 1190
                       Step 3. 119025
So, (345^2) = 119025 .

Note -:
(a) . the number whose unit digit is 0,1,5 and 6 always give the same unit digits respectively ,on squaring.
(b). 2,3,7and 8 never appear as unit digit in the square of a number.
(C). PERFECT SQUARE -: The square of any natural number is known as perfect square e.g.,4,9,16,25,81,121,625,3600,2401,256,196,    324,225,289,361,400,441, etc.

                      Method of Multiplication (B)

In our previous blog we known about vedic method of multiplication  . In this blog we will study about vedic method of multiplication I another way .

NOTE -: when will We see the examples below we see that each number multiply with the digits of another number at one time .
 NOTE -: if we acquire this method we will solve multiplication of any two numbers in few seconds .

Let an example -:  solve 79 × 13 .

Solution -:   79 × 13  = 1027
                   
              7×1 = 7  | {(7×3) + (9×1)} | 9× 3=27
                             | = 30                    |
Now we find the required number from these three numbers {7,30,27}  , let see below
         (a) 7 + 3 = 10 | (b) 30+ 2 = 2.| (c)  7

    In (a) -:
 { 3 from tenth digit of 32 }

In (b) -:
 { we add 2 from tenth digit of 27 }

In (c) -:
{unit digit of 27 }

Now the required number is -: 1027 Answer .

Example (2) -:  solve 79 × 326 .

Solution -:  79 × 326 = 25754
7 × 3 | {(7 × 2) + (9 × 3)| {(7 × 6) + (9 × 2)|9×6
= 21 |           = 41.          |      = 60         |= 54

Now we get the required number from these four numbers { 27,41,72,54}  , let see below
   
(A)  21 + 4| (B) 41 + 6| (c) 60 +5 |(d) 4

   = 25        | = 7.           | = 5.         | 4

So Answer will be -: 25754

Example (3) -: Evaluate 325768 × 289 .

Solution -:  325768 × 289 = 94146952.

(A) 9×8 = 72
(B) (8×8)+(9×6) = 118
(C) (2×8)+(8×6)+(9×7) = 127
(D) (2×6)+(8×7)+(9×5) = 113
(E) (2×7)+(8×5)+(2×9) = 72
(F)  (2×5)+(8×2)+(9×3)= 53
(G) (2×2)+(8×3) = 28
(H) 2×3 = 6

OR

2×3| {(8×3)+(2×2)}|{(9×3)+(8×2)+(2×5)}|
(H).         (G).                       (F)
{(9×2)+(8×5)+(2×7)}|{(9×5)+(8×7)+(2×6)}
               (E).                          (D)
{(9×7)+(8×6)+(2×8)}|{(9×6)+(8×8)}|9×8|
          (C).                            (B).              (A)

Now we find the required number from these number (72,118,127,113,72,53,28,6) -

From :
(A)    (2)    unit digit from 72
(B) 118+7 = 125  (5) ,  unit digit from 125
 (C) 127+12 = 139 (9) , unit digit from 139
(D) 113+13 = 126 (6) , unit digit from 126
(E) 72+12 = 84 (4) , unit digit from 84
(F) 53+8 = 61 (1) , unit digit from 61
(G)28 + 6 = 34 (4) , unit digit from 34
(H) 6+3= 9

So the required answer will be 94146952

Note -: we write the digit from bottom to top , like from H to A in above question.

     

                              Method of multiplication(A)

VEDIC METHODS OF MULTIPLICATION :

EXAMPLE :
(1) . solve  23×27.
Solution : step (a) -: 7×3 = 21
                   Step(b)-: 7×2+2×3 = 14 + 6 = 20
                    Step(c)-: 2×2 = 4
                     Step(d)-: 1{unit digit from 21}
                     Step(e)-: 20+2 = 22 { here addition of 2 from tenth digit of 21 }
                      Step(f)-: 4+2=6 {here addition of 2 from tenth digit of 22}
And will be    -:          
 23
×27
---------
621.    
---------

(2) . solve 3456× 89 .
Solution : 3456
                   × 89
                 ---------------------
                                      4.        {9×6 = 54}
                ---------------------
3456
   ×89
--------------
            84.       {9×5+(8×6)} + 5 = 9
--------------
3456
 × 89
---------------
          584.      { 9×4 +(8×5)} +9 = 85
----------------

3456
 × 89
----------------
          7584.          {9×3+(8×4)} + 8 = 67
----------------

3456
 ×89
------------------
307584.             { 8×3} + 6 = 30
------------------
 Answer is -: 307584

Note -:  By using this method we will get answer of any two numbers.

          

                       TESTS OF DIVISIBILITY

(J). Divisibility by 6 :

A number is divisible by 6 only when it is divisible by 2 and 3 both. So first of all we see that the number is even or not then we check for the divisibility by 3 .

Example ; (1) check whether 216 is divisible by 6 or not .
Solution : step (a). First of all we check the given number is divisible by 2 or not
Since unit digit is divisible by 2 then the given number is divisible by 2 .

Step (2). We see that the given number is divisible by 2 so now we check whether the given number is divisible by 3 or not .
Sum of digits of the given number is (6+1+2) 9 , which is divisible by 3 so the given number is divisible by 3 also .

Step(3). Since the given number is divisible by 2 and 3 both so the given number is divisible by 6 also .

(K). Divisibility by 12 :

 A number is divisible by 12 only when it is divisible by 4 and 3 both at the same time . so first of all check the divisibility by 4 then by 4 .

NOTE : Thus we can conclude that any number which is divisible by a composite number( The natural numbers which are not prime , are called composite numbers) ,as mentioned above , must be divisible by all its factors whose L.C.M is the given number divisior . 

                     TESTS OF DIVISIBILITY

(f).  Divisibility By 10 :

A number is divisible by 10 only when its unit digit is 0(zero).
Example : (1) 68720 is divisible by 10 because its unit digit is 0 .

(2) 29850 is divisible by 10 because its unit digit is 0.

(3) 687405 is not divisible by 10 ,since its unit digit is 5 not 0 .

(G).  Divisibility by 5 :

A number is divisible by 5 only when its unit digit is 0 or 5 .

Example ; (1) 6785 is divisible by 5 because its unit digit is 5 .
(2) 96750 is divisible by 5 because its unit digit is 0 .

(H). Divisibility by 7,11,13 :
A number can be divisible by 7,11 or 13  if and only if the difference of the number formed by the last three digits and the number formed by the rest digits is divisible by 7,11 or 13 respectively .

Example ; check whether 139132 is divisible by 7 or not .

Solution; step (H.1)
 number formed by the last three digits is 132 .
Step (H.2) : number formed by rest of digits is 139
Step (H.3) ; we take difference between these two numbers
139 - 132 = 7
Since the difference is divisible by 7 . Hence the given number is also divisible by 7.

(2) check whether 12478375 is divisible by 13 or not.

Solution;  step1 . 12478 - 375 = 12103
                  Step 2. 12 - 103 = - 91
Since the difference is divisible by 13 .Hence the given number is divisible by 13 .

(3) check whether 29435417 is divisible by 11or not.

Solution ; step(1) 29435 - 417 = 29018
                   Step(2) 29 - 018 = 11
Since the difference is divisible by 11 so the number is also divisible by 11 .

(I). Divisibility by 11 :
A number is divisible by 11 if the difference between the sum of the digit at odd places and sum of digits at even places is equal to zero or a number divisible by 11.

Example ; check 29435417 is divisible by 11 or not .
Solution; step(1). Sum  of its digits at odd places is
7+4+3+9 = 23
                  Step(2). Sum of its digits at even places is
1+5+4+2 = 12
                 Step (3). Difference between these two numbers is
23 - 12 = 11
Since difference is divisible by 11 .Hence the given number is also divisible by 11 .

(2). 57945822
(Sum of digit at odd places - sum of digits at even places)
(2+8+4+7) - ( 2+5+9+5) = 21 - 21 = 0
Since the difference is zero so the number is divisible by 11 .

                          TESTS OF DIVISIBILITY
(a) Divisibility By 2 :
A number is divisible by 2 if its unit digit is divisible by 2 .
Example : (1). 7684
Solution:
The unit is 4 , which is divisible by 2 (4÷2=2)  so 7684 is divisible by 2.
(2). 2785
Solution : The unit digit is 5 , which is not divisible by 2 . so 2785 is never divisible by 2.

(b). Divisibility by 3 :
A number is divisible by 3 only when the sum of its digit is divisible by 3.
Example : (1). 695421
Solution ; in the number 695421 , the sum of the digit is 27 , which is divisible by 3.
So the number is divisible by 3 .
(2). 948655
Solution : in the number 948655 , the sum of the digits is 37 , which is not divisible by 3 .
So the number 948655 is not divisible by 3 .

(c). Divisibility by 9 :
A number is divisible by 9 only when the sum of its digits is divisible by 9 .
Example : (1). 246591
In the number 246591 the sum of its digits
= 27 , which is divisible by 9.
So the number is divisible by 9 .

(d) . Divisibility by 4 : A number is divisible by  4 if the number  formed by its last two digits is divisible by 4 .
Example : (1) 6879376
Solution : since 76 {the last two digit} is divisible by 4 . so the number 6879376 is divisible by 4 .

(e). Divisibility by 8 : A number is divisible by 8 if the number formed by hundred's ten's and unit's digit of the given number is divisible by 8  .
Example : (1) 16789352 , the number formed by last three digits , namely 352 is divisible by 8. So 16789352 is divisible by 8 .

Numbers:in hindu - Arabic system , we have ten digits , namely 0,1,2,3,4,5,6,7,8,9
Called zero,one,two,three,four,five,six,seven,eight,and nine respectively.
Numeral: A number is denoted by a group of digit , called numeral .
For denoting a numeral ,we use the place - value chart .
2. Face value : The face value of a digit in a numeral is its own value , at whatever place it may be . example
In the numeral 58932
The face value of  9 is 9
The face value of 2 is 2.

3. Types of Numbers :
a. Natural Numbers
b.Whole Numbers
c.Integers

4. Even And Odd Numbers :

5; prime Numbers : A counting number is called a prime number if it has exactly two factors namely itself and 1 that is prime number are divided by 1 and itself ,  not any other number.
All prime number less than 100 are:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.

Test For a Number to be primes :
Let p be a given number  and let n be the smallest counting number such that (n×n) greater than equal to p {(n×n)>=p} .
Now test whether p is divisible by any of the prime number less than or equal to n.
Example : (1) test which of the following are prime numbers?
(A) 61
Solution : Step (A.1)
choose a number whose square is greater than the given number (like in this question 61)
The closest number n= 8
Step(A.2):  now square of 8 is
n×n = 8×8= 64
which is greater than 61

Step (A.3) : prime number less than 8 is 2,3,5,7
Step (A.4) : now 61 is divided by  these numbers  we see that none of them divides 61
Hence 61 is a prime number

(B) : 173
n = 14
14×14 = 196 which is greater than 173
Prime numbers less than 14 is 2,3,5,7,11,13
Clearly , non of them divides 173
Hence it is a prime number

(C) : 321
n = 18
18×18 = 324, which is greater than 321
Prime numbers less than 18 are 2,3,5,7,11,13,17,
We see that 3 divides 321 hence 321 is not a prime number .
Other questions are
(D): 437, 137 , 991 etc .