Vedic Maths
FUN WITH MATHEMATICS
squaring numbers
Vedic Mathematics- Squaring a Number
Yavadunam
– Whatever the deficiency subtract that deficit and write alongside the square of
– Squaring numbers close to a base
Ekadhikena Purvena
– Squaring numbers ending in 5
DwandaYoga
– Duplex Combination Process
VEDIC MATHEMATICS - SUB-SUTRA
3. Yavadunam Tavadunikrtya Varganca Yojayet
– Whatever the deficiency subtract that deficit and write alongside the square of
EXAMPLE 1.
Square of 8
10 – 8 = 2, square of 2 is 4
8 – 2 = 6
Thus, Square of 8 = 64
EXAMPLE 2.
Square of 6
10 – 6 = 4, square of 4 is 16
6 – 4 = 2
Thus, Square of 6 = 2(1)6=36
RULE:
To find squares of numbers close to base 10,
we subtract the number from base 10 and take a square of the result.
Then we subtract the result from the number and cross the results.
VEDIC MATHEMATICS - SUB-SUTRA
3. Yavadunam Tavadunikrtya Varganca Yojayet
– Squaring numbers close to a base
Examples: 96²
= 92/16 { LHS 96 is 4 less than 100. so reduce it by 4 (96 – 4 = 92)
lessen 96 still further by same number
{ RHS square the deficiency (4 x 4 = 16)
103²
= 106/09 { LHS 100 + 3 increase by 3 / / 103 + 03 = 106
{ RHS square the increment 03 x 03 = 09)
1011²
= 1022/121 { LHS 1000+11=1022 / RHS square the increment (11 x 11 = 121)
VEDIC MATHEMATICS - SUB-SUTRA
3. Yavadunam Tavadunikrtya Varganca Yojayet
– Squaring numbers close to a base
Examples:
991²
= 982/081 { LHS 991-9/ RHS 009 x 009 }
93²
= 86/49 { LHS 93-7/ RHS 07 x 07 }
88²
= 76/144 { LHS 88-12/ RHS 12 x 12 }
= 77/44
VEDIC MATHEMATICS - SUB-SUTRA
Ekadhikena Purvena
– Squaring numbers ending in 5
Examples:
35²
= 3 x (3+1) / 5 x 5
= 3x4 /25
= 1225
85²
= 8 x (8+1) / 5 x 5
= 8x9 / 25
= 7225
850²
= 8 x (8+1) / 50 x 50
= 8x9 / 2500
= 722500
VEDIC MATHEMATICS - SUB-SUTRA
Dvanda Yoga
– Duplex Combination Process
Duplex is denoted by D.
D(3) = 3² = 9
D(43) = 2x4x3 = 24
D(567) = 2x5x7 + 62 = 70 + 36 = 106
D(3456) = 2x3x6 + 2x4x5 = 36 + 40 = 76
D(34567) = 2x3x7 + 2x4x6 + 52 = 42 + 48 + 25 = 115
It goes on….
SQUARE OF 1221 = 1490841
L R & similar to urdhva tiryagbhyam sutra
Calculate
D(1st digit) -----(1) = 1 x 1 = 1
D( 1st 2 digits)----(12) = 2 x 1 x 2 = 4
D(1st 3 digits)----(122) = 2 x 1 x 2 + 2 x 2 = 4+4=8
D ( All 4 digits)---(1221) = 2 x 1 x 1+2 x 2 x 2=2 + 8=10
D(Last 3 digits)---(221) = 2 x 1 x 2 + 2 x 2 = 4+4=8
D(last 2 digits)---(21) = 2 x 1 x 2 = 4
D(last digit) -----(1) = 1 x 1 =1
Add the carry forward
Answer is 1490841
Vedic Mathematics
Squaring a 2 digit Number beginning with 1
Example 1:
17²
= 17 x 17
= 1x1 / 2x 7 / 7 x 7
= 1 / 14 / 49
= 1 / 14 +4 / 9
= 1+1 / 8 / 9
= 289
18²
= 18 x 18
= 1x1 / 2x 8 / 8 x 8
= 1 / 16 / 64
= 1 / 16 +6 / 4
= 1+2 / 22 / 4
= 324
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