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.