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.
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)
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
