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MENTAL vs. Vedic Mathematics. (2) Examples
 MENTAL vs.
VEDIC MATHEMATICS
(2) EXAMPLES


Notation used

We use the symbol "|" to separate the component numbers of a power-of-10 representation. There may be n symbols in a row, which can also be represented by "|n". The symbol "|1" is equivalent to "|". In general, Examples: Negative numbers can also be used. For example,
Subtract

Sutra 2 applies: All of 9 and the last of 10.

If we want to subtract two numbers a and b (a > b), the complementary of b, C(b), is calculated first and added to a: a + C(b). The result is subtracted to the nearest power of 10 greater than b.

The complement of a number n, C(n), is the 9's complement of each of its digits plus one unit, or 10 raised to the number of digits of n minus n. For example, Formula: a − b = a + C(b) − 10n (n is the length of b)

For example:
Product of numbers close to the same power of 10

Sutra 2 applies: All of 9 and the last of 10.

Formula: a1a2 = a1 + e2 |n e1e2 ≡ a2 + e1 |n e1e2

(e1 and e2 are the differences from the nearest base)

Justification: Schematic:

a1e1 = a1 − 10n
a2e2 = a2 − 10n
a1 + e2 = a 2 + e1e1e2
a1 + e2 |n e1e2

Examples:
  1. 96×93 = 8928

    9696 − 100 = −4
    9393 − 100 = −7
    93 − 4 = 96 − 7(−4)×(−7)
    8928
    89 |2 28 = 8928

  2. 1002×997 = 1002 − 3 |3 2×(−3) = 999 |3 −6 = 999 |3 −6 = 998 |3 994 = 998994

  3. 99976×9999998 = 99976 − 2 |5 −24×&−2 = 99974 | 00048 = 9997400048

  4. 998×1025 = 998 + 25 |3 −2×25 = 1023 |3 −50 = 1023050 = 1022950

Product of numbers close to a working base

A multiple or submultiple is taken as the working base (b).

Formula: a1a2 = f(a1 + a2) |n e1e2 siendo b = f×10n,

Justification: Schematic:

a1e1
a2e2
a1 + e2e1e2
a1 + e 2 |2 e1e2

For example, in 61×53 we take as working basis 50 = 100/2.

6111 (61 − 50)
533 (53 − 50)
61 + 3 = 53 + 1111×3
6433
64×50 = 64×100/2 = 3233
32 |2 33 = 3233


Square of a number close to a working base

Formula: a2 = (a + b)(a − b) + b2 for any b.

Justification: Examples:
  1. 972 = (97 + 3)(97 − 3) + 32 = 100×94 + 9 = 9409 (we have chosen b = 3)

  2. 572 = (57 + 7)(57 − 7) + 72 = 64×50 + 49 = 3249 (we chose b = 7)

Using the mean value to multiply

When two numbers are multiplied, the mean value of both can sometimes be used to calculate their product.

Formula: (a − e) (a + e) = a2 − e2

Examples:
  1. 57×63.
    The average value is 60, 602 = 3600, 57×63 = 602 − 32 = 3591

  2. x + y = 10
    xy = 24

    The mean value of x e y is 5, xy = (5 + e)(5 − e ) = 25 − e2 = 24, e2 = 1, e = 1 and −1. Therefore, x = 5 + 1 = 6 and y = 5 − 1 = 4.

Multiplication by 9...9

9...9 = 10n − 1

Formula: a(10n − 1) = a − 1 |n 10n − a

Justification: Examples:
  1. 7×9 = 7 − 1 | 10 − 7 = 6 | 3 = 63

  2. 18×99 = 17 |2 100 − 18 = 17 |2 82 = 1782

  3. 79×9999 = 79 − 1 |4 10000 − 79 = 78 |4 9921 = 789921

Square of a number close to a power of 10

Formula: a2 = 10n + 2e |n e2 (where e = a − 10n)

Justification: Examples:
  1. 142 = 10 + 2×4 | 42 = 18 | 16 = 18×10 + 16 = 196

  2. 1082 = 100 + 2×8 |2 82 = 116 |2 64 = 11664

  3. 9882 = 1000 − 12×2 |3 122 = 976 |3 144 = 976144

  4. 10082 = 1000 + 2×8 |3 82 = 1016 |3 064 = 1016064

Square of a number

The so-called "Dvanda Yoga" (duplex method) is applied. The duplex function (D) of a number is illustrated with an example:

a = a1a2a 3a4 = 1234, a2 = 1552756 The result is:
Square of an expression

The duplex method is also applied. Examples:
  1. (x + 5)2
    D(x) = x2
    D(x + 5) = 2×x×5 = 10x
    D(5) = 52
    Result: x2 + 10x + 5

  2. (2x + 3y + 4)2
    D(2x) = (2x)2 = 4x2
    D(2x + 3y) = 2×2x×3y = 12xy
    D(2x + 3y + 4) = 2×2x×4 + (3y)2 = 16x + 9y2
    D(3y + 4) = 2×3y×4 = 24y
    D(4) = 42 = 16
    Result: 4x2 + 12xy + 16x + 9y2 + 24y + 16

Square of a number ending in 5

Sutra 1 applies: For one more than the previous one.

Formula: (a | 5)2 = a(a + 1) |2 25

Justification: Schematic:

a5
aa + 125
a(a + 1)25
a(a + 1) |2 25

Examples:
  1. 752 = 7×(7 + 1) |2 25 = 7×8 |2 25 = 56 |2 25 = 5625

  2. 1352 = 13×(13 + 1) |2 25 = 13×14 |2 25 = 182 |2 25 = 18225

  3. 36152 = 361×(361 + 1) |2 25 = 130682 |2 25 = 13068225

Square root

The procedure is the inverse of the square, the duplex method being used in reverse. For example, √(38837824).

It is noted that the number consists of 4 pairs of digits, so the result must have 4 digits: a1a2a3a4.
Product of two numbers whose sum of the last digits is the base (10) and the previous parts are equal

Sutra 1 also applies: For one more than the previous one.

Formula: a | d1 × a | d2 = a(a + 1) | d1d2

Justification: Examples:
Cube of a number close to a power of 10

Formula: a3 = a+2e |n 3e2 |n e2

Justification: Examples:
Multiplication of 3 numbers close to a power of 10

Formula: abc = 10n + a' + b' + c' |n a'b' + a'c' + b'c' |n a'b'c'

Justification: Outline:

a a'
b b'
c c'
a − b' − c' = b − a' − c' = c − a' − b' − b' a'b' + a'c' + b'c' a'b'c'
a − b' − c' |n a'b ' + a'c' + b'c' |n a'b'c'

Examples:

988 −12 (988 − 1000)
996 −4 (996 − 1000)
995 −5 (995 − 1000)
988 − 4 − 5 = 996 − 12 − 5 = 995 − 12 − 4 = 979 (−12×−4) + (−4×−5) + (−5×−12) = 128 −240 (−12×−4×−5)
979128240 = 979127760


General multiplication procedure

The product of two numbers, a and b, is a number formed by sums of combinations of products. The order number i is the sum of the products of the digits of both numbers such that the sum of the exponents of the powers of 10 add up to i. We call P(i) the powers of order i.

Example: In multiplication, sutra 3 applies: vertically and crosswise. Examples:
  1. 12×13

    1 2
    1 3
    1×1 1×3 + 2×1 2×3
    1 5 6
    156

  2. 34×27

    3 4
    2 7
    3×2 3×7 + 4×2 4×7
    6 29 28
    6 31 8
    9 1 8
    918

  3. 123×456

    1 2 3
    4 5 6
    1×4 1×5 + 2×4 1×6 + 3×4 + 2×5 2×6 + 3×5 3×6
    4 13 28 27 18
    4 13 28 28 8
    4 13 30 8 8
    4 16 0 8 8
    5 6 0 8 8
    560088

Single digit division

Illustrated with an example: 671/4. The steps are as follows:
  1. 6/4 = 1 (remainder 2)
    Add 1 to the quotient of the result.
    Put the remainder (2) in front of the second digit (7) of the dividend.

  2. 27/4 = 6 (remainder 3)
    6 is added to the quotient of the result.
    Put the remainder (3) in front of the third digit (1) of the dividend.

  3. 31/4 = 7 (remainder 3)
    Add 7 to the quotient of the result.
    The result is: 167 (remainder 3).
    If we wanted more decimals, we would add zeros to the dividend and continue operating in the same way.

Scheme:

4671
6/4 = 1 (remainder 2)
1
127/4 = 6 (remainder 3)
1631
1631/4 = 7 (remainder 3)
1673 (resto)
1673 (remainder)

General scheme:

d a1 a2
a1/d = c1 (resto r1) r1a1/d = c2 (resto r2)
c1 c2
c1 | c2


Division 1/19
  1. First method. Multiplication is used. It proceeds from right to left.

    Sutra 1 is used: by one more than the previous one. The predecessor of 19 is 1. By one more than the predecessor = 1 + 1 = 2 (multiply by 2).

    We take the first digit, which is the dividend (1) and multiply it successively by 2 as follows:

      1. First digit of the result.
      1×2 = 2. The result is put in front of the 1: 21.
      2×2 = 4. The result is put in front of the previous one: 421.
      4×2 = 8. Result: 8421.
      8×2 = 16. The 6 is placed, with 1 carry: 168421.

    Analogously, we obtain:

      1368421
      7368421
      147368421
      947368421
      18947368421
      178947368421
      1578947368421
      11578947368421
      31578947368421
      631578947368421
      12631578947368421
      52631578947368421

    After this number, the decimals are repeated.

    The final result is 1/19 = 0.052631578947368421

    The period consists of 18 digits, with the second 9 being the complementary of the first 9. The denominator minus the denominator (19 − 1 = 18) indicates the length of the period.

    Note that only multiplication (×2) and addition (+1) have been used in the division.

  2. Second method. It is the dual method of the previous one. Division by 2 is used instead of multiplication. The period is calculated from left to right.

    The first digit of the dividend (1) is divided by 2, which gives quotient 0 and remainder 1. The most significant digit is 0, with carry 1: .10
    Dividing 10 by 2 = 5 (there is no remainder): .05

      0.1 = 0.10
      10/2 = 5 (remainder 0) 0.05
      5/2 = 2 (remainder 1) 0.0512
      12/2 = 6 (remainder 0) 0.00526
      6/2 = 3 (remainder 0) 0.005263
      3/2 = 1 (remainder 1) 0.00526311
      11/2 = 5 (remainder 1) 0.005263115
      15/2 = 7 (remainder 1) 0.0052631517
      17/2 = 8 (remainder 1) 0.0052631578

    The second cycle is the complements to 9. This second method is preferable because the most significant figures are obtained first and also because only one operation is used: division by 2.

Division by 9
  1. 2 digits (a1a2/9).

    Formula: a1a2/9 = a1 (cociente), a1+a2 (resto)

    Justification: a1a2 = 10a1 + a2 = (9 + 1)a 1 + a2 = 9a1 + (a1 + a2)

    Schematic:

    a1a2
    a1 (cociente)a1 + a2 (resto)

  2. 3 digits (a1a2a3/9).

    Formula: a1a2a3/9 = 11a1 + a2 (quotient), a1 + a2 + a3 (remainder)

    Justification: a1a2a3 = 100a1 + 10a2 + a3

    100a1/9 = 11a1 (resto a2), 10a2/9 = a2 (remainder a2)

    a1a2a3/9 = 11a 1 + a2 = 10a1 + a1 + a2 = a1 | a1 + a2 (remainder a1 + a2 + a3)

    Outline:

    a1 a2 a3
    a1 a1 + a2 a1 + a2 + a3
    a1 | a1 + a2 (quotient) a1 + a2 + a3 (remainder)

    Example: 743/9 = 11 (quotient) and 4 (remainder)

    743
    77 + 47 + 4 + 3 (remainder)
    7 | 11 = 7×10 + 1 = 8114 (remainder)

  3. 4 digits (a1a2a3a4/9).

    Schematic:

    a1a2a3a4
    a1a1 + a2a1 + a2 + a3a1 + a2 + a3 + a4
    a1 | a1 + a2 | a1 + a2 + a3 (quotient) a1 + a2 + a3 + a4 (remainder)

    Example: 8745/9 = 971 (quotient) and 6 (remainder)

    8745
    88 + 78 + 7 + 48 + 7 + 4 + 5
    8151924 = 9×2 + 6
    815216
    81716
    9716
    9 | 7 | 1 = 971 (quotient)6 (remainder)
In general, in division by 9, the first digit of the quotient is the first digit of the dividend. From this first digit, the digits of the dividend (minus the last digit) are added successively to obtain the remainder of the digits of the quotient. The remainder is the sum of the digits of the dividend.

Other examples:
  1. 212/9 = 2 | 2 + 1 = 2 | 3 = 23, remainder = 2 + 1 + 2 = 5

  2. 1204/9 = 1 | 1 + 2 | 1 + 2 + 0 = 1 | 3 | 3 = 133, remainder = 1 + 2 + 0 + 4 + 7 = 7

  3. 132101/9 = 1 | 1 + 3 | 1 + 3 + 2 | 1 + 3 + 2 + 2 + 1 | 1 + 3 + 2 + 1 + 0 = 1 | 4 | 6 | 7 | 7 = 14677, remainder = 1 + 3 + 2 + 1 + 0 + 1 + 1 = 8

Division by 99

It is similar to division by 9, but grouping the digits 2 by 2, from the right.

Examples:
  1. 121314/99 = 12 |2 12 + 13 = 12 |2 25 = 1225, remainder = 12 + 13 + 14 = 39

  2. 1213141/99 = 1 |2 1 + 21 |2 1 + 21 + 13 = 1 |2 22 |2 53 = 112253, remainder = 1 + 21 + 13 +41 = 76

Division by 999

It is similar to the two previous divisions, but grouping the digits 3 by 3, from the right.

Examples:
  1. 123456/999 = 123, remainder = 123 + 456 = 579

  2. 1234567/999 = 1 + 1234 = 1235, remainder = 1 + 234 + 567 = 802

  3. 8999/999 = 8, remainder = 8 + 999 = 1007 > 999, new remainder = 1007 − 999 = 8, new quotient = 8 + 1 = 9.

Division of a number by a digit d > 0
  1. 2 digits (a1a2/d).

    Formula: a1a2/d = a1, remainder = d'a1 + a2, where d' = 10 − d

    Justification: a1a2 = 10a1 + a2, 10 = d + d', 10/d = 1 (remainder d'), 10a1/d = a1 (remainder = d'), a1a2/d = a1 (remainder d'a1 + a2)

    Outline:

    a1 a2
    d'a1
    a1 (cociente) a2 + d'a1 (resto)

    Example: 85/6 = 14 (remainder 1), d' = 4

    85
    84×8 = 32
    832 + 5 = 37
    837 = 6×6 + 1
    8+6 1
    14 (quotient) 1 (remainder)

  2. 3 digits (a1a2a3/d).

    Schematic:

    a1 a2 a3
    c1 = a1 d'c1d'c2
    c1 c2 = a2 + d'c 1a3 + d'c2
    c1 | c2 (quotient) a3 + d'c2 (resto)

    Example: 385/4 = 96 (remainder 1), d' = 6

    3 8 5
    3 6×3 = 18 2×6 = 12
    3 8 + 18 = 26 5 + 12 = 17
    3 26 = 4×6 + 2 17 = 4×4 + 1
    3 + 6 = 9 2 + 4 1
    961
    9 | 6 = 96 (quotient) 1 (remainder)

  3. 4 digits (a1a2a3a4/d).

    a1 a2 a3 a4
    c1=a1 d'×c1 d'×c2 d'×c3
    c1 c2 = d'×c1 + a2 c3 = d'×c2 + a3 d'×c3 + a4
    c1c2c3/ (cociente) d'×c3 + a4 (resto)

    Example: 1342/8 = 167 (quotient), 6 (remainder). d' = 10 − 8 = 2.

    1 3 4 2
    1 2×1 = 2 2×5 = 10 6×2 = 12
    1 3+2 = 5 4+10 = 14 2+ 12 = 14
    1 5 14 = 8+6 14
    1 5+1 = 6 6 14 = 8 + 6
    1 6 6+1 = 7 6
    167 (quotient) 6 (remainder)

General division procedure

The procedure of division between two numbers is a straightforward procedure. Tirthaji called it the "crown jewel" of VM. It allows you to divide numbers of any size on one line. We illustrate it with an example: 40342/173 = 233 (remainder 33).

340342000
17664 16413← restos sucesivos
2331907← cocientes sucesivos

We calculate 40/17 = 2 (remainder 6).

The diagonal formed by the next digit of the divisor (3), the initial remainder (6) and the initial quotient (2) is initially created: 3 − 6 − 2. With each diagonal a cycle of 3 steps is performed:
  1. The last calculated quotient (2) is multiplied by the unit of the divisor (3): 2×3 = 6.

  2. Subtract this number (6) from the first two numbers of the diagonal (63): 63 − 6 = 57.

  3. The result (57) is divided by the first part of the divisor (17) and the quotient (3) and the remainder (6) are written down. The new diagonal is 4 − 6 − 3. And the same procedure is repeated.
Steps:
  1. Diagonal 3 - 6 - 2.
    1. 2×3 = 6.
    2. 63 − 6 = 57.
    3. 57/17 = 3 (remainder 6).

  2. The diagonal 4 - 6 - 3 is created.
    1. 3×3 = 9.
    2. 64 − 9 = 55.
    3. 55/17 = 3 (remainder 4).

  3. The diagonal 2 - 4 - 3 is created.
    1. 3×3 = 9.
    2. 42 − 9 = 33.
    3. 33/7 = 1 (remainder 16).

  4. The diagonal 0 - 16 - 1 is created.
    1. 1×3 = 3.
    2. 160 − 3 = 157.
    3. 157/17 = 9 (remainder 4).

  5. The diagonal 0 - 4 - 9 is created.
    1. 9×3 = 27.
    2. 40 − 27 = 13.
    3. 13/17 = 0 (remainder 13).

  6. The diagonal 0 - 13 - 0 is created.
    1. 0×3 = 0.
    2. 130 − 0 = 130.
    3. 130/17 = 7 (remainder 11).
The result is 233.1907....


Other Examples of Application of Sutras

Transposing and applying (sutra 4)

This sutra refers to the property of mathematical operations to be reversed or reflected. For example, −1 on one side of the equation becomes +1 on the other side. Examples:
  1. (3x2 + 10x + 12)/(x + 2) = 3x + 4 (remainder 4)

    x + 23x2 +10x+12
    −2−6−8
    3x10 − 6 = 412 − 8
    3x + 4 (quotient)4 (remainder)

    • We transpose the +2 of the divisor as −2 in the dividend.

    • We divide the first term of the dividend by the first term of the divisor: 3x2/x = 3x. And we write it down in the result. The coefficient of the term is c = 3.

    • We multiply the coefficient c (3) by the transposed number (−2): 3×(−2) = −6. This result is put under the second term (+10x). Add 10 and −6 = 4.

    • Multiply 4 by the transposed number (−2) = 4×(−2) = −8. Add 12 and −8 = 4.

    • The result is 3x + 4 (remainder 4).

  2. 1364/112 = 12 (remainder 2)

    1121364
    −121−1−2
    13 − 1−2−4
    1224 − 4
    12 (quotient)2 (remainder)0

    • We transpose the +12 of the divisor as −12 in the dividend.

    • We divide the first term of the dividend (1) by the first term of the divisor (1): 1/1 = 1. And we write it in the result. The coefficient of the term is c = 1.

    • We multiply the coefficient c (1) by the two components of the transposed number (−1 and −2): 1×(−1) = −1 and 1×(−2) = −2. These results are put under the second and third terms (3 and 6). Add 3 and −1 = 2 and put it under the second term.

    • Multiply 2 by the transposed number (−2) = −4. Add 4 and −4 = 0.

    • The result is 12 (remainder 2).

When the sum is the same, the sum is equal to zero (subtra 5)

This sutra admits many interpretations, being applicable in different contexts. Examples:
If one is proportional, the other is zero (sutra 6)

This sutra is special. It is only applicable to certain types of equations. Example: We see that there is proportionality between the coefficients of x and the independent terms: 9/3 = 18/6 = 3. Therefore, y = 0.


By addition and subtraction (subtraction 7)

In many mathematical problems and processes, we use addition and subtraction, simultaneously or sequentially. Examples:
  1. x + y = 47
    x − y = 19

    By addition, 2x = 66, x = 33.
    By subtraction, 2y = 28, y = 14.

  2. (a2b3 × a4b5)/a 2b = a2+4−2b3+5−1 = a4b7

By mere observation (sub-sutra 12)

Sometimes, just observing the data of the problem is enough to solve it immediately. Examples:
  1. Find the equation of a line passing through the points (10, 5) and (18, 9).
    We observe that 5×2 = 10 and 9×2 = 18. Therefore, y = 2x.

  2. Solve the equation x + 1/3 = 10/3.
    We observe that 10/3 = 3 + 1/3. Therefore, x = 3.

Proportionally (sub-subtra 1)

Proportion is fundamental in mathematics. Calculations are simplified when proportionality can be applied. Examples:
  1. 435/5 = 435×2 / 5×2 = 870/10 = 87

  2. 63/25 = 63×4 / 25×4 = 252/100 = 2.52

  3. 715/125 = 715×8 / 125×8 = 5720/1000 = 5.72

Only the last terms (sub-subtra 9)

This sub-sub-subtra is useful for solving simple equations of the following type: when the numerator and denominator on the left-hand side of the equation, excluding the independent terms, are in the same proportion to each other as the numerator and denominator on the right-hand side. For example, the equation We see that the first two terms of the numerator on the left side have the same proportion as those on the right side: Therefore, And only the last terms count: The solution is x = −1.