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 .