Specific Square & Multiplication by Vedic Method

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Squares of number ending with 5 By specific squre & multiplication of Vedic mathematics

Here we solve specific square & multiplication by Vedic method in very easy and fast way. First we apply Vedic  sutra Number 1 “Ekadhikena purvena” to find square of number ends with 5 . Ekadhikena purvena means ‘one more than the previous’.

Read first example carefully as I explain it step by step, once you understand the steps you can also find the specific square in 3 to 4 seconds.

Ex. i) Find the square of 15-

We will find our answer in two parts

Like, Ans= LHS/RHS

Step 1: Find  L.H.S.

a. For LHS multiply the previous number by 1 more than itself

b. In 15 previous digit is 1 and one more than previous is 1+1=2

so, 1×2=2

Hence, LHS = 2

Step 2: Find R.H.S.

As our last digit is fixed i.e. 5, so our RHS will be remain same

       i.e.  52 = 25

Hence, RHS = 25

Answer   152 = LHS/RHS = 225

 

Ex. ii) Find the square of 45-

LHS= 4 x (4+1) = 4×5 = 20

RHS= 25

Answer, 452 = 2025

Similarly,

iii) 552 = 5 x (5+1)/25 = 3025

iv) 952 = 9 x (9+1)/25 = 9025

v) 1152= 11 x (11+1)/25 = 13225

vi) 3052= 30 x (30+1)/25= 93025

Multiplication of Number whose sum of last digit is 10 and the previous digits are remains same- 

Here sub sutra Antyayor Dasakepi works.  It is sub-sutra of Ekadhikena Purvena. It works for specific multiplication; whose last digit sum is 10 & previous number is same. Like 53 x57, 62 x 68, 128 x 122 etc.

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Ex. i) 52 x 58

we can see sum of last digit is 10 (2 + 8), and previous digits are also same i.e. 5

We divide our answer in two parts,

ANS = LHS/RHS

Step 1)  Find LHS

a. In (52 x 58) previous digit is 5 and one more than previous is 5+1=6

b. For LHS multiply the previous number by the number one more than itself

i.e. 5×6=30

Hence, LHS = 30

Step 2) Find  RHS

Our RHS is simply multiplication of last digit,

i.e.2×8 =16

Hence, RHS = 16

Answer, 52×58 = LHS/RHS = 3016

Ex. ii) 74 x 76

LHS= 7 x (7+1) = 56

RHS= 4×6 = 24

Answer 74×76 = 5624

Similarly,

iii) 37×33 = (3x(3+4))/(7×3) =1221

iv) 61×69 =(6x(6+1))/(1×9) =4209(Note: Here we add 0 because, the right hand should always be filled in with a two-digit number)

v) 84 x 86 = (8x(8+1))/(4×6) = 7224

The same rule is applicable if the sum of last two digits, three digits, four digits…..equal to respectively 100,1000,10,000 etc.

Note: If the sum of last digit is 100, we need to raise LHS by 0 & RHS should have 3 digit,  If the sum of last digit is 1000, we need to raise LHS by 00 & RHS should have 4 digit & so on. 

 Ex. vi) 392×308  (Here (92+08=100) & previous digit is same i.e. 3)

So, we solve like above sum

LHS =3 x (3+1) = 12

As the sum of last digits is 100 so, we raise our LHS by 0

Hence, LHS = 120

RHS = 92×08 = 736

Answer 392×308 = 120736

Ex.vii) 297 x 203       (Here (97+03=100) & Previous digit same i.e. 2)

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LHS = 2 x (2+1) = 6

Now, we raise our LHS by 0

Hence, LHS = 60

RHS = 97×03 = 291

Answer, 297×203  = 60291

Ex. viii) 4301 x 4399

(Here (01+99= 100) and previous digits are same i.e. 43)

LHS = 43 x (43+1) = 1892

Now, we raise our LHS by 0

Hence, LHS = 18920

RHS = 01×99 =99 =099 (As our RHS should have 3 digit if Addition of last

digit is 100)

Answer 4301 x 4399 = 18920,099

Ex.ix) 2989 x 2011

(Here (989+ 011= 1000) and previous digit is same i.e. 2)

LHS = 2 x 3 = 6

Now, we raise our LHS by 00

Hence, LHS = 600

RHS = 989 x 011 =10879 = (As our RHS should have 4 digit if Addition of last

digit is 1000, so we carryover 1  extra digit in the LHS )

Answer, 2989 x 2011 = 6010879

Try This:

i) 75 x 75    ii) 125 x 125    iii) 82 x 88

iv) 191 x 109  v) 35 x 35   Vi) 3993 x 3007

vii) 5389 x 5311  viii) 61 x 69