Bulletin of the Marathwada Mathematical Society
Vol. 10, No. 2, December 2009, Pages 8–15.
MATHEMATICAL JUSTIFICATION OF SOME
NON-TRADITIONAL METHODS OF MULTIPLICATION
Y. D. Deshpande,
19 ‘Shraddha Apartment, Behind Kohinoor Arcade,
Nigdi Chowk, Nigdi, Pune-44,(M. S.), India.
Abstract
This paper deals with mathematical justification of the three non-traditional
methods of multiplication. (A) The Urdhva Tiryakbhyam method. (B) The
Ganesh method. (C) The Russian peasant multiplication method.
1
INTRODUCTION
There are various methods of multiplication that give quick answers. These
methods appear to be strange for want of mathematical reasoning behind them. In
this paper we discuss three methods of multiplication, with mathematical justification. They are the Urdhva Tiryakbhyam method, the Ganesh method and the
Russian peasant multiplication method. These methods are also illustrated with the
help of examples.
2
NOTATIONS AND PRELIMINARIES
Let N be the set of positive integers. If p ∈ N consists of (n + 1) digits,
a0 , a1 , · · · , an then p is expressed as follows:
p = 10n an + 10n−1 an−1 + · · · + 10a1 + a0
(2.1)
Conventionally, the positional weights of the digits i. e. 100 , 101 , · · · etc are dropped
and p is written as follows
p = an an−1 · · · a1 a0
(2.2)
If q ∈ N is such that q = bn bn−1 · · · b1 b0 then p × q is operated as follows
p × q = bn (an an−1 · · · a1 a0 ) + bn−1 (an an−1 · · · a1 a0 )
+ · · · + b1 (an an−1 · · · a1 a0 ) + b0 (an an−1 · · · a1 a0 )
8
(2.3)
MATHEMATICAL JUSTIFICATION OF · · · MULTIPLICATION
9
The identity (2.3) is obtained by using the distributive property in algebra
(i. e. the multiplication is distributive over addition). A careful study of equation
(2.3) clarifies that p is multiplied by every digit of q and partial products are obtained
and added together to get the final answer.
But in each partial product there are products with equal positional weights,
e.g.; (an ) × (bn−4 ); (an−1 ) × (bn−3 ); (an−2 ) × (bn−2 ); (an−3 ) × (bn−1 ) and (an−4 ) × (bn )
are the products with their weights being 102n−4 each.
In any method of multiplication , may be traditional or non-traditional, the
following operations are inevitable.
(1) Every digit of multiplicand is multiplied by every digit of multiplier.
(2) All products with equal positional weights are collected in one proper column
and added together.
(3) All digits in the multiplication are arranged as shown in equation (2.2).
The success of a method depends upon the technique that enables us to perform
the above operations easily and quickly.
3
THE URDHVA TIRYAKBHYAM METHOD
In Vedic Mathematics the Urdhva Tiryakbhyam method is useful for quick
multiplications. ‘The Urdhva Tiryak literally means ‘ vertically and cross-wise’[2].
3.1
The Modus Operandy[2]
The following simple example clarifies the procedure of the method.
Example 3.1.1
12 × 13 =?
12
× 13
1 : 3 + 2 : 6 → 156
(by starting from the left side)
a
c
b
d
ac : (ad + bc) : bd
(1) The left-most part of the answer is obtained by vertical product of left-most
digits :(1 × 1 = 1)
(2) The middle part is obtained by adding the cross-wise products:
(1 × 3) + (2 × 1) = 5
(3) The right most part is obtained by the product of the right most digits :
(2 × 3 = 6).
10
Y. D. Deshpande
Example 3.1.2
84 × 36 =?
(by starting from the right side)
84
× 36
2404
62
1
3024
(1) Right-most part is obtained by vertical product of right-most digits,4 × 6 = 24
4 is placed in right-most place, 2 below 4 as a carry
(2) The middle part is obtained by adding cross-wise products,8 × 6 + 4 × 3 = 60.
0 is placed to the left of 4, and 6 as a carry.
(3) Left-most part is obtained by vertical product,8 × 3 = 24.
This 24 is placed to the left of 0.
3.2
Mathematical Justification of Example 3.1.2
(1) The right-most part, 4 × 6 = 24 is expressed as 100 (4) × 100 (6) = 100 (24)
4 is placed in unit’s place, 2 is carried over to ten’s place.
(2) The middle part is , 8 × 6 + 4 × 3 + 2 = 62 i.e.
101 (8) × 100 (6) + 100 (4) × 101 (3) + 101 (2) = 101 (62).
Hence 2 is placed in ten’s place, and 6 is carried over to 102 0 s place.
(3) The left-most part 8×3+6 = 30 expressed as 101 (8)×101 (3)+102 (6) = 102 (30)
and hence 30 is placed in hundred’s place.
Example 3.2.1 : Evaluate 547 × 869.
Procedural Steps The following five steps are carried out.
Step 1 :
7 × 9 = 63
(3 at the unit place and carry
6 at the 10th place)
MATHEMATICAL JUSTIFICATION OF · · · MULTIPLICATION
Step 2 :
4 × 5 + 6 × 7 = 78
Step 3 :
7 × 8 + 5 × 9 + 4 × 6 = 125
Step 4 :
5 × 6 + 8 × 4 = 62
Step 5 :
5 × 8 = 40
11
(8 at the 10th place and 7 at
the 102 th place)
(5 at the 102 th place, 2 at the
103 th place and 1 at the 104 th
place)
(2 at the 103 th place and 6 at
the 104 th place)
(0 at the 104 th place and , 4 at
the 105 th place )
The final answer is obtained by adding all the digits at the corresponding places.
If one is perfect in performing addition of two or more numbers, the final answer
can be written in one line without writing the carried digits in the procedure.
3.3
Multiplication by Urdhva Tiryak A General Formula
If
p = (an an−1···a1 a0 ) and q = (bn bn−1···b1 b0 ) then
p × q = 102n (an bn ) + 102n−1 (an bn−1 + an−1 bn )
+ 102n−2 (an bn−2 + an−1 bn−1 + an−2 bn )
+ · · · + 102 (a2 b0 + a1 b1 + a0 b2 ) + 10(a1 b0 + a0 b1 ) + (a0 b0 ).
All the three, inevitable operations, discussed in section 2 are performed during the
very first attempt and that too, mentally. Thus we get one line, quick answer.
4
THE GANESH METHOD OF MULTIPLICATION
This method is almost the same as the traditional one, except that no partial
products are obtained as we do in traditional method.
4.1
Modus Operandy
For multiplication, 547 × 869 , a specially constructed square ABCD is divided
into nine small squares (9 is the product of the No of digits in both the numbers).
Each small square is divided into two triangles by its diagonal.The number 547 is
written horizontally and 869 vertically as shown in Fig. 4.1. A product of digits, one
from the multiplicand and one from the multiplier, is placed in the corresponding
small square, e.g.; the square for 5 × 6 = 30 is shown by arrows in the left part of
Fig. 4.1. As 30 is a two digit number, 3 is placed in the upper triangle and 0 in the
lower one. If the product is a one-digit number, it is placed in the lower triangle
and the upper one is filled with a zero.
12
Y. D. Deshpande
When all the small squares are filled we get trapeziums, T0 , T1 , · · · , T6 which
contain the digits with positional weights, 100 , 101 , · · · 106 respectively. Thus we
have all the digits with equal positional weights in one proper trapezium. We add
them together. The carry, if any, is added to the next, left or upper trapezium.
4.2
Mathematical Justification (See Fig. 4.1)
All the digits in each trapezium are added together as they are with equal
positional weights. The carry, if any, is added to the next, left or upper trapezium.
Thus we get the final answer. Now, lets verify if the digits are placed in proper
trapezium. 5 × 6 = 30. The place values of 5 and 6 are 102 (5) and 10(6) respectively.
Hence, the place value of 30 is 103 (30). Therefore, 0 is placed in T3 and 3 in T4 .
Similarly we can verify all the digits in the square ABCD.
5
THE RUSSIAN PEASANT MULTIPLICATION METHOD [1]
The Moscow Papyrus (1850 BC) having 25 problems and the Rhind (OrAhmes)
Payrus having 85 problems are the main sources of Egyptian Mathematics. The
Rhind Papyrus was acquired in 1858 AD and, now, placed in the British Museum.
One of the important topics discussed in the Papyri is a method of multiplication which is, now, known as The Russian Peasant Multiplication.
5.1
Modus Operandy [1]
To multiply 225 by 17, 225 is continuously halved, neglecting the remainders;
and simultaneously the number 17 is continuously doubled. The figures in front of
the odd numbers in the first column are added to get the final answer.
MATHEMATICAL JUSTIFICATION OF · · · MULTIPLICATION
13
Table 5.1
(Halving) ↓
∗225
112
56
28
14
*7
*3
*1
5.2
Remainder
(Doubling) ↓
·
1
0
0
0
0
1
1
(All numbers in the right most column
with * at their left are added).
∗17
34
68
136
272
* 544
* 1088
* 2176
3825
Binary Number system [3, 4]
In this method two number systems are used. (1) Decimal (2) Binary.
The binary number system is briefly introduced below.
Binary number system is exactly the same as the decimal one, except that the
base is 2 instead of 10.Only two digits, (0 and 1) are used in this system. The base
of the number system is indicated as subscript. For example, 110012 = 2510 .
If p2 ∈ N consists of n digits, a1 , a2 , · · · , an then p2 is written as follows
p2 = [an an−1 · · · a2 a1 ]2
(5.1)
p2 can be expressed in its expanded form as follows.
¤
£
p2 = 2n−1 an + 2n−2 an−1 + · · · + 2a2 + a1 10
(5.2)
a1 and an are the least significant digit (LSD) and the most significant digit (MSD)
respectively.
In a sense, the sum of the place values of all bits in p2 is its decimal equivalent.
In binary number system only two digits 0 and 1 are used. Place value of
0 is always 0 and place value of 1 is equal to its corresponding positional weight.
Therefore, the decimal equivalent of a binary number is the sum of the positional
weights of the bit 1 in the number, e.g.;
32, 16, 8, 4, 2, 1 Positionalweights
(1 0 0 0 1 0)2 = 1 × 32 + 0 × 16 + 0 × 8 + 0 × 4 + 1 × 2 + 0 = 3410
(5.3)
14
5.3
Y. D. Deshpande
Division Remainder Technique (DRT) [3]
2
Table 5.3 (A)
Remainder
39
19
1
9
1
4
1
2
0
1
0
0
1
Table 5.3 (B)
Remainder
39
1
19
1
9
1
4
0
2
0
1
1
392 = 1001112
To find binary equivalent of a decimal number, division remainder technique (DRT)
is most commonly used. Suppose we want the binary representation of 39. For this
purpose 3910 is continuously divided by 2 till the last quotient is 0. The remainders
are recorded and collected in reverse order. (See Table 5.3 (A))
5.4
Modified Division Remainder Technique (M-DRT) (See Table
5.3(B)).
Without loss of generality we make a small change in DRT, for the convenience of
the present discussion.We shift the remainders in Table 5.3 (A) one step upward.
We get each remainder just in front of each corresponding quotient as shown in
Table 5.3 (B). Hence the modified division remainder technique (M-DRT) reads as
follows.
The given decimal number is continuously halved till the last quotient is 1
instead of 0. The bits 0 and 1 are placed in front of the even and odd quotients
respectively. Thus we get the binary equivalent of 39.
5.5
Mathematical Justification
Table 5.5
Col.1
1) 225
2) 112
3)
56
4)
28
5)
14
6)
7
7)
3
8)
1
Col.2
* 17
34
68
136
272
* 544
* 1088
* 2176
Col.3
17 = 17 × 20
34 = 17 × 21
68 = 17 × 22
136 = 17 × 23
272 = 17 × 24
544 = 17 × 25
1088 = 17 × 26
2176 = 17 × 27
In this table, col. 3 gives the meaning of the figures continuously doubled in column
2. In col. 1, 225 is continuously halved and we get 8 numbers in this column.
MATHEMATICAL JUSTIFICATION OF · · · MULTIPLICATION
15
Now, it is clear that the binary equivalent of 225 is an 8-bit number. With 8 bits
256(28 ) different patterns of 0s and 1s are possible.In column 2, 17 is continuously
doubled and 8 figures are obtained. (See Col.3 Table 5.5)
The sum of all these figures is the product of 17 and 255(28 − 1). 255 is the
greatest 8-bit number containing all 1 bits and no zero bit.
The given number 225 < 255. Therefore, the 8-bit pattern, representing 225
must contain some zero bits. In Col. 2 the products with zero bits are to be deleted
and those with 1 bits are to be added to get the product of 17 and 225. (as per
equation 2.3).
As per M-DRT, we have 22510 = 111000012 . Hence we have;
1710 × 22510 = 1710 × 111000012
= 1710 (27 + 26 + 25 + 20 ) → as per equation (2.3)
= 17 × 27 + 17 × 26 + 17 × 25 + 17 × 20 .
These four products i. e. ; 2176, 1088, 544, 17 are already available in steps 8, 7, 6 and
1 respectively.Therefore, to multiply 225 by 17 , the figures infront of odd numbers
are added and the final answer is obtained.
References
[1] Mohan Apte, Brief History of Number Theory, Kaprekar Felicitation
Number — 62,Dec 2004.
[2] Jagadguru Swami Sri Bharati Krsna Tirthaji Maharaja, Sankaracarya
of Goverdhan Matha, Puri, Vedic Mathematics, Motilal Banarasidas
Publishers,Pvt. Ltd.,Delhi,1992.
[3] Pradip K. Sinha & Priti Sinha, Computer Fundamental, 3 rd Edn,
3 rd Chapter,2005.
[4] V.Rajaraman, Fundamentals of Computer, IV th Edn, Chapter 6, Binary
Arithmetic,New York, USA, April 2000.