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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 **VedicΒ sutra Number 1 βEkadhikena purvenaβ** to find square of number ends with 5 . **Ekadhikena purvena means βone more than the previousβ.**

**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: FindΒ 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, 1Γ2=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**

**Β Β Β Β i.e.Β 5 ^{2} = 25**

Hence, RHS = 25

**Answer**Β Β **15 ^{2} = LHS/RHS = 225**

**Ex. ii) Find the square of 45-**

**LHS**= 4 x (4+1) = 4Γ5 = 20

**RHS**= 25

**Answer,** **45 ^{2} = 2025**

**Similarly,**

**iii) 55 ^{2}** = 5 x (5+1)/25 = 3025

**iv) 95**^{2 }= 9 x (9+1)/25 = 9025

**v) 115 ^{2}**= 11 x (11+1)/25 = 13225

**vi) 305 ^{2}**= 30 x (30+1)/25= 93025

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

Here sub sutra Antyayor Dasakepi works.Β 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.

**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)**Β **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. 5Γ6=30

Hence, LHS = 30

**Step 2) FindΒ RHS**

Our RHS is simply multiplication of last digit,

i.e.2Γ8 =16

Hence, RHS = 16

**Answer, 52Γ58 = LHS/RHS = 3016**

**Ex. ii) 74 x 76**

**LHS=** 7 x (7+1) = 56

**RHS=** 4Γ6 = 24

**Answer** **74Γ76 = 5624**

**Similarly,**

**iii) 37Γ33** = (3x(3+4))/(7Γ3) =1221

**iv) 61Γ69** =(6x(6+1))/(1Γ9) =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))/(4Γ6) = 7224

**The same rule is applicable if the sum of last two digits, three digits, four digitsβ¦..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,Β If the sum of last digit is 1000, we need to raise LHS by 00 & RHS should have 4 digit & so on.Β **

**Β ****Ex. vi) 392Γ308 **Β (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** = 92Γ08 = 736

**Answer** **392Γ308 = 120736**

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

** LHS** = 2 x (2+1) = 6

Now, we raise our LHS by 0

Hence, LHS = 60

**RHS** = 97Γ03 = 291

**Answer**, **297Γ203Β = 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** = 01Γ99 =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 1Β extra digit in the LHS )

**Answer, 2989 x 2011 = 6010879**

**Try This:**

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

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

vii) 5389 x 5311Β viii) 61 x 69