Vedic Math
MUTIPLICATION
Using “Vertical and Crosswise”
Multiplication of large numbers becomes easy when we use this sutra. We will
explain it with an example.
Step 1
Step 2
Step 3
Step 1: multiply the two digits on the right side
Step 2: cross-multiply digits of both columns, and add together
Step 3: multiply the two digits of the left side
Example 1: 21 X 32
Step 1:
2 1
Step 2:
X 3 2
Step 3:
6 7 2
Step 4:
1 x 2 = 2; write 2 for the right column
2 x 2= 4
3 x 1 = 3; add 3 + 4 = 7, for the middle column
2 x 3 = 6; write 6 for the left column
Example 2: 72 X 56
Step 1:
7 2
Step 2:
X 5 6
Step 3:
4 05312
Step 4:
2 x 6 = 12 write 2 and carry over 1
5 x 2 = 10
7 x 6 = 42 add both products = 52 + 1 = 53, carry 5
7 x 5 = 35 plus 5 carry over = 40, for the left column
Exercise 6: Solve the following using the sutra “vertical and crosswise”
1) 24 x 56
2) 43 x 29
3) 76 x 24
4) 13 x 43
5) 23 x 75
6) 62 x 81
7) 19 x 25
8) 15 x 79
9) 32 x 65
10) 71 x 29
Vedic Math
Three Digit Multiplication
Vertical and crosswise sutra can be extended to any number of digits. Let’s first look
at 3 digit multiplication.
Step 1
Step 2
Step 3
Step 4
Step 5
Example 3: 123 X 645
1 2 3
Step 1: 3 x 5 = 15; write 5, carry 1
X 6 4 5
Step 2: 2x5=10 + 3x4=12; 10 + 12 = 22+1= 23, write 3 carry 2
719332315
Step 3: 1x5=5 + 6x3=18 + 2x4=8; 5+18+8 = 31+2=33, write 3 carry 3
Step 4: 4x1=4 + 6x2=12; 4 + 12 = 16 + 3 = 19, write 9 carry 1
Step 5: 6x1=6 + 1 = 7
Example 4: 214 X 23
2 1 4
Step 1: 4 x 3 = 1 2
X 0 2 3
Step 2: (3x1) + (4x2) = 3 + 8 = 11 + 1 = 12
4 9 12 12
Step 3: (3x2) + (4x0) + (1x2) = 8 + 1 = 9
Step 4: (2x2) + (1x0) = 4
Step 5: 2 x 0 = 0
Exercise 7: Solve the following using the sutra “vertical and crosswise”
1) 123 x 215
2) 236 x 213
3) 721 x 72
4) 281 x 19
5) 982 x 124
6) 724 x 362
7) 289 x 727
8) 473 x 67
9) 926 x 94
10) 836 x 492
Vedic Math
Four Digit Multiplication
Vertical and crosswise sutra can be extended to any number of digits. Let’s now look
at 4 digit multiplication. This process can be extended to any number of digits
Step 1
Step 2
6
Step 7
Step 3
Step 4
Step 5
Exercise 8:
1) 1023 x 2132
2) 3125 x 126
3) 5216 x 29
4) 27 x 3271
5) 521 x 6215
6) 5612 x 2135
7) 752 x 1920
8) 9210 x 28
9) 617 x 5611
10) 6403 x 45
Step
Answers
Exercise 6:
1) 1344
2) 1247
3) 1824
6) 5022
7) 475 8) 1185
4) 559 5) 1725
9) 2080
10) 2059
Exercise 7:
1) 26445
2) 50268
3) 51912
4) 5339
5) 121768
6) 262088
7) 210103
8) 31691
9) 87044
10) 411312
2) 393750
3) 151264
4) 88317
5) 3238015
8) 257880
9) 3461987
10) 288135
Exercise 8:
1) 2181036
6) 11981620 7) 1443840
Vedic Math
More Multiplication
Multiplication using compliments
The “all from 9 and last from 10” sutra allows us to multiply large numbers that are
close to a base easily. This also applies when numbers are near different bases.
Example 5: 88 X 98
88 - 12
98 - 2
86 / 24
Example 6: 9 x 8
9 - 1
8 - 2
7 / 2
Example 7: 97 x 98
97 - 3
98 - 2
95 / 06
Both numbers are close to the base 100. We write the
numbers under each other with compliments on right.
the – sign is used because the numbers are below the
base. The right side is obtained by multiplying the
compliments. The left side by cross-subtraction (88-2 =
98 – 12 = 86)
We use the base of 10.
adding zeros on the right side
For a base of 100 there must be 2 digits on the right side
Example 8: 88 x 89 extra numbers on the right side
88 - 12
Base of 100
89 - 11
132 is a 3-digit number, so 1 is carried to the left side
77 / 132
78 / 32
Example 9: 67859 x 99998 multiplying very large numbers easily
67859 – 32141 Here the base is 100,000
99998 – 2
The compliment is easy to find using “all from 9 last from 10
67857 / 64282 = 6,785,764,282
Example 10: 1123 x 1003 numbers above the base
1123 + 123
Above the base we cross-add instead of subtracting
1003 +
3
1126 / 369 = 1,126,369
Example 11:
10004
103
10304
10004 x 103 multiplying numbers near different bases
+ 0004
Here the smaller number and its complement as far to
+ 03
the left as possible. The right side is obtained as usual,
/ 12
but since the smaller number only has 2 digits on the
right so does the answer. On the left side the answer is
Obtained by adding to the far left side.
Example 12:
99998
96
95992
99998 x 96 different bases with numbers below the base
- 00002
- 04
/ 08
Example 13: 1004 x 98 different bases, above and below
To share the same base 98 becomes 980, so it is in
The position shown.
1004 + 004
Here the cross-wise operation is 98(0) + 4 = 984, or
98 - 02
1004
984 / -08
-02
983 / 92
984
Example 14: 9111 x 9900
9111 – 0889
9900 – 0100
9011 / 88900
9019 / 8900
When the numbers being multiplied are both below the
base there will be as many digits on the right side as on
the left. We add zeros to the deficiency to maintain the
symmetry.
Exercise 9 Try to find the answer mentally
1) 87 x 97
2) 78 x 98
3) 69 x 97
4) 98 x 86
5) 92 x 93
6) 73 x 98
7) 98 x 99
8) 88 x 96
9) 12 x 13
10) 13 x 11
11) 14 x 13
12) 112 x 112
Exercise 10
1) 103 x 92
2) 107 x 91
3) 1003 x 998
6) 103 x 104 7) 107 x 106 8) 1012 x 1011
11) 997 x 96 12) 9996 x 96
13) 977 x 97
4) 101 x 99
5) 14 x 9
9) 14 x 12
10) 1122 x 1007
14) 97 x 9
15) 9876 x 96
Answers
Exercise 9
1. 8439
2. 7644
3. 6693
4. 8428
5. 8556
7. 9702
8. 8448
9. 156 10. 143
1. 9476
2. 9737
3. 1000994
7. 11342
8. 1023132
9. 168
13. 94769
14. 873
15. 948096
1. 572
2) 319
3) 3454
4) 4686
5) 57343
6) 80282
7) 23606
8)651387
9) 2390916
10) 3934579
1. 672
2) 504
3) 936
4) 1488
5) 7668
6) 25608
7) 61944
8) 854556
9) 350016
10) 32522580
1) 4554
2) 7326
3) 6435
4) 693306
5) 373626
6) 617382
7) 68413158 8) 9824501754
10) 92907
11) 749924
11. 182
6. 7154
12. 12544
Exercise 10
4. 9999
5. 126
10. 1129854
11. 95712
6. 10712
12. 959616
Exercise 11
Exercise 12
Exercise 13
15) 256397436
9) 654971345028
12) 5429457 13) 32399676
16) 36539346 17) 288999711
19) 432
20) 603
21) 1098
24) 5112
25) 12517447
18) 2684499973155
22) 35541
26) 59671269
14) 87199128
23) 455301
27) 25973001
28) 2561031
Exercise 14
1) 624
2) 2352
3) 3024
4) 2484
5) 158004
6) 479526
7) 280896
8) 252384
9) 109722
10) 613800
Vedic Math
Multiplication by Specific Numbers
Multiplication by 11 using the sutra “Antyayoreva” – “Only the last two digits”
By using this sutra we do addition instead of multiplication, adding only the last two
digits.
Example 15: 4523 x 11
0 4523 0
Step 1: We place a zero at both ends of the number
0 4523 0
Step 2: Just continue to add the last 2 digits
3
045230
Step 3: Just continue to add the last 2 digits
5 3
045230
Step 4: Just continue to add the last 2 digits
7 5 3
045230
Continuing to add the last 2 digits we get the answer: 49753
497 5 3
Example 16: 35612 x 11
0356120
3911732
(6+5=11, so write 1, carry 1)
Exercise 11: Multiply the following numbers by 11
1) 52 x 11
2) 29 x 11
3) 314 x 11
4) 426 x 11
6) 7462 x 11 7) 2146 x 11 8) 59217x11 9) 217356x11
5) 213 x 11
10) 357689 x 11
Vedic Math
Multiplication by Specific Numbers
Multiplication by 12 using the sutra: Sopantyadvayamantyam” – “The ultimate and
twice the penultimate”. This sutra also uses addition instead of multiplication. As in
the previous example we make a zero sandwich. Then we add the last digit and
twice the penultimate (second last) digit.
Example 18: 4523 x 12
0 4523 0
Step 1: We place a zero at both ends of the number
0 4523 0
Step 2: add the last digit plus twice the second to last
6
045230
Step 3: add the last digit plus twice the second to last
7 6
045230
1
Continue to add the last digit plus twice the second to last
2 7 6
045230
Continuing in the same way we get the answer: 54276
5 14 12 7 6
Example 19: 713624 x 12 = 8563488
Example 20: 3176214 x 12 = 38114568
07136240
031762140
8156131488
38112114568
Exercise 12: multiply the following numbers by 12
1) 56 x 12
6) 2134 x 12
2) 42 x 12
3) 78 x 12
7) 5162 x 12
4) 124 x 12
8) 71213 x 12
5) 639 x 12
9) 21968 x 12
Vedic Math
Multiplication by Specific Numbers
Multiplication by 9 using the sutras
“Ekanyunena Purvena” – “By one less than the one before”
“Nikhilam Navatascaramam Dasatah” – “All from 9 and last from 10”
First type: when all the digits of the multiplier and multiplicand are the same and all
the digits of the multiplier are 9’s.
Example 20:
342 x 999 = 341658
Step 1: We divide our answer in 2 parts, the left side (LS) and the right side (RS).
LS:
341 /
Step 2: We write 1 less than the multiplicand: 342-1=341
RS:
341 / 658
Step 3: We write the complement of the multiplicand using “all
from 9 and last from 10”. Compliment of 342=658
OR, subtract each digit of the LS of the answer from 9:
(9-3=6), (9-4=5), (9-1=8) = 658
Example 21: 5642 x 9999 = 5641 / 4358 = 54614358
Example 22: 71496 x 99999 = 71495 / 28504 = 7149528504
Second Type: the digits in the multiplier are more than the multiplicand.
This is as easy as the previous example. We simply have to add zeros before the
multiplicand and make the number of digits equal to the digits of the multiplier.
Example 23: 23 x 999 = 22977
Step 1:
Since the multiplicand has 2 digits and the multiplier 3 digits we make
023 x 999
the multiplicand 023 so that they both have the same number of digits.
22 /
Step 2: LS: we write one less than the multiplicand
22 / 977
Step 3: RS is the complement of 023 = 977
Example 24: 548 x 99999 = 00548 x 99999 00547 / 99452 = 54799452
Example 25: 7825 x 9999999=0007825 x 99999 7824/9992175 = 78249992175
Third Type: Here the number of digits in the multiplier are LESS than the digits in
the multiplicand, and the multiplier is all 9’s.
Example 26: 436 x 99 = 43164
Step 1: We divide our multiplicand into 2 parts so that the RS has the same number
of digits as the number of 9’s in the multiplier.
There are two 9’s in the multiplier so there are two digits on RS
4 : 36
separated by a colon.
4: 36
- 5
LS: we add 1 to the left part; 4 + 1 = 5, and subtract it from the
RS: 436 – 5
4 : 31
431 / 64
RS: write the complement of the right part next to the left part.
(complement of 36 = 64).
Example 27: 20462 x 999 = 20441538
20 : 462
LS: add 1 to the left part and subtract it
20 : 462
- 21
20441 /
20441 / 538 RS: is the complement of 462
Exercise 13: Solve the following. (See if you can find the answer mentally):
1) 46 x 99
2) 74 x 99
7) 6842x99
8) 98246x99999
12) 543 x 9999
3) 65 x 99
9) 654972 x 999999 10) 93 x 999 11) 76 x 9999
13) 324 x 99999
16) 3654 x 99999 17) 289 x 999999
20) 67 x 9
21) 122 x 9
25) 1253 x 999
4) 694 x 999 5) 374 x 99 6) 618 x 99
14) 872 x 99999
18) 26845 x 99999999
22) 359 x 99
26) 59731 x 999
15) 2564 x 99999
19) 48 x 9
23) 4599 x 99
24) 568 x 9
27) 25999 x 999
28) 25869 x 99
Vedic Math
Working Base Multiplication
We have demonstrated multiplication of numbers close to any one base, but when
both numbers are not close to a convenient power of 10 we use what we call a
“working base” which is a multiple or sub-multiple of 10, 100, 1000, etc. We then do
our multiplication as we have already learned taking the working base into account.
When using working bases we apply the sutra: “Anurupyena” – “Proportionality”
For example, is we want to multiply 46 x 53, both numbers are far from 10 or 100,
but close to 50, so we use 50 as our working base. This can be done in two ways:
10 x 5 = 50; here 10 is the theoretical base, and 50 is the working base, or
100/2 = 50; here 100 is the theoretical base, and 50 is the working base
The theoretical base is important when we have to decide the number of digits on
the right side of the answer. With a theoretical base of 10 there is only one digit on
the RS; with a theoretical base of 100 there will be two digits on the RS.
Example 28: 52 x 54 = 2808
Step 1: We will use 10 as the theoretical base and 10 x 5 = 50 as the working base.
10 x 5 = 50
Step 2: we place the complements of the working base next
54
+4
to them. 54 is more than base 50 and a + indicates this
52
+2
52 is greater than base 50, indicated by +2
56 / 8
Step 3: multiply the RS and add crosswise to find the LS
Step 4: Now we proportionately multiply the LS answer to get
The theoretical base from the working base
56 x 5 = 280
280 LS and 8 RS = 2808
Example 29: 523 X 488 = 255224
Step 1: we will use 100 as the theoretical base and 500 as the working base
100 x 5 = 500
Step 2: find the complements
523
+ 23
Step 3: multiply the complements to get (MINUS) -276
488
- 12
Step 4: cross-subtract
511 / -276
Step 5: multiply the LS x 5 to return to theoretical base
Working Base Multiplication, continued
511 x 5 = 2555 – 2 = 2553 now carry MINUS 2 to the LS
Since we have -76 on the RS. To make it positive we
2553 – 1 / 24
take the complement of 76 = 24. and subtract 1 extra
= 255224
on the LS for coming out of complements.
Example 30: 62 x 68 = 4216 By two different working bases
10 x 6 = 60
62 + 2
68 + 8
70 / 16
421 / 6
We take 10 as the theoretical base, and 60 as the
working base. Next we take the complements from the
working base and get the answer in 2 parts.
For the RS we multiply, and since the theoretical base is
10 we need one digit, so write 6 and carry 1.
The LS is obtained by cross-addition, and multiply the result by
6: 70 x 6 = 420 plus 1 carry = 421. Answer: 4216
Next we will use the working base of 70. 10 x 7 = 70
62 - 8
68 - 2
60 / 16
The complements are now negative. We get our answer in 2
parts, as normal.
Now multiply the LS x 7: 60 x 7 = 420 + 1 carry = 421
Answer: 4216
Example 31: 523 x 488 = 255224
We will take 100 as the theoretical base and 500 as the working base 100 x 5=500
523 + 23
488 - 12
511 / -276
2552 / 24
Since 100 is the theoretical base we need 2 digits on the RS.
Therefore we write 76 and carry over MINUS 2 to the LS.
First we multiply 511 x 5=2555, now -2 = 2553.
On the RS we have a negative number so we must use its
complement = 24, and subtract 1 extra for coming out of
complements. Answer: 255224
Exercise 14
1) 24 x 26
6) 687 x 698
2) 49 x 48
3) 56 x 54
4) 54 x 46
5) 399 x 396
7) 532 x 528 8) 478 x 528 9) 478 x 399
10)775 x 792
Vedic Math
The Vinculum
Calculations can be considerably simplified by the use of the “vinculum”. This is the
name of a minus when it is put on top of a number. For example, -1 written in its
vinculum form is 1, also described as “bar 1”. We can use the vinculum to make big
numbers small. For example, 29 can be written as 31, meaning 30 -1. If we use 31
instead of 29 we avoid having to deal with 9.
The vinculum is an ingenious method that has many advantages, among which are:
1. it gives flexibility – we use the vinculum when we want to
2. larger numbers – 6, 7, 8, 9 can be avoided
3. numbers can cancel each other out or made to cancel
4. 0 and 1 occur twice as frequently as they otherwise would
We will find the sutra: “all from 9 and last from 10” useful in putting a number into
vinculum from, or taking it out of vinculum from.
Example 32: find the vinculum of 29878
The vinculum is found by increasing the first number to 3 and applying the sutra to
the remaining numbers:
The vinculum of 29878 = 30122
Example 33: use the vinculum to eliminate the big numbers in 387179
For this number we split it into two parts: 387 / 179 and apply the sutra to each part:
387179 = 413221
Further examples of putting a number in vinculum form are:
293 = 313, 83 = 123, 800 = 1200, 185 = 225 or 215, 909 = 1 1 1 1
817 = 8 2 3 or 1 1 8 3 or 1 2 2 3
Example 34: taking a number out of vinculum form
42132 = 37868
the 4 is decreased to 3 and the sutra applied to the rest.
Example 35: further examples of removal of the vinculum:
4 1 1 3 2 2 1 = 3892781, 703 = 697,
7 2 0 3 = 6797,
27 = -13
Simple applications of the Vinculum
Example 36:
29 x 3 we can avoid working with a large number by using the
vinculum, writing the vinculum from: 31 x 3 = 3 x 93 = 87
Example 37: 261 ÷ 9
9 ) 2 6 11
3 1 = 29