Vedic Math Divisibility Rules

Vedic Math Divisibility Rules

Divisibility by 2: If last digit of the number is even i.e. 0,2,4,6,8
Ex : 876999578 is divisible by 2.

Divisibility by 3: If the sum of digits is divisible by 3.
Ex: 327 is divisible by 3, since sum of its digits = (3+2+7) = 12 , which is divisible by 3.

Divisibility by 4: If the last two digits of the number is divisible by 4
Ex: 2648 is divisible by 4, since the number formed by the last two digits is 48 which is divisible by 4.

Divisibility by 5: If the last digit of the number ends with 0 or 5.
Ex: 20870 ends in a 0, so it is divisible by 5.

Divisibility by 6: If the number is divisible by both 2 & 3.
Ex: 558 is divisible by 6, because it is divisible by 2(number is even) as well as 3 (5+5+8=18,which is divisible by 3).

Divisibility by 7: Multiply the last digit by 5 and add the product to the remaining truncated number. Continue doing these steps until you reach a 2 digit number.
If the result is divisible by 7, we say that the original dividend is divisible by 7.
Example: 33803–> 3380+(3*5)=3380+15=3395 –>339+(5*5)=339+25 = 364 –> 36+(4*5)=36+20=56 (since this number is divisible by 7, you can say 3185 is also divisible by 7)
* There are many more approaches to check the divisibility by 7.

Divisibility by 8: If the last three digits of the number are divisible by 8.
Ex: 3652736 is divisible by 8 because last three digits (736) is divisible by 8.
Note: Rule of divisibility by 2 & 4 on the last three digit number will not be applicable here. You have to check the divisibility manually.
Ex: 516 is divisible by 2 & 4 but not by 8.

Divisibility by 9: If the sum of the digits is divisible by 9.
Ex: 672381 is divisible by 9, since sum of digits = (6+7+2+3+8+1) = 27 is divisible by 9.

Divisibility by 10: If the digit at units place is 0 it is divisible by 10.
Ex: 697420, 243540 is divisible by 10.

Divisibility by 11: If the difference of ‘sum of its digits at odd places’ and ‘sum of its digits at even places’ is either 0 or a number divisible by 11.
Ex: 4832718 is divisible by 11, since:
(Sum of digits at odd places) and (sum of digits at even places)
= (8+7+3+4)-(1+2+8) = 11

Divisibility by 12: A number is divisible by 12 if it is divisible by both 4 and 3.
Ex: 34632
(i) The number formed by last two digits is 32, which is divisible by 4
(ii) Sum of digits = (3+4+6+2) = 18, which is divisible by 3.

Divisibility by 13: Multiply the last digit by 4 and add the product to the remaining truncated number. Continue doing these steps until you reach a 2 digit number.
If the result is divisible by 13, we say that the original dividend is divisible by 13.
Example: 3185–> 318+(5*4)=318+20=338 –>33+(8*4)=33+32 = 65 (since this number is divisible by 13, you can say 3185 is also divisible by 13)

Divisibility by 14: If a number is divisible by both 2 & 7.

Divisibility by 15: If a number is divisible by both 3 & 5.

TIP: If a number is divisible by two different prime numbers, then it is divisible by the products of those two numbers.
Ex: 30 is divisible by both 3 and 5, it is also divisible by 15.

Division of polynomials.

-4x³-7x²+9x-12 divided by 2x-4
_________________
2x-4 | -4x³ – 7x² + 9x – 12 | -2x² – 15/2x – 21/2
-(-4x³ + 8x²)
————————
-15x² + 9x
-(-15x² + 30x)
————————
– 21x – 12
-(- 21x + 42)
———————
-54
Quotient = -2x² – 15/2x – 21/2 , Remainder = -54

See the above example in another form by making the first coefficient of the divisor as ‘1’. Like in following example, we divide the divisor by 2 and later divide the quotient also by 2 :

2x-4  |  -4x³ – 7x² + 9x    – 12
x-2
+2             – 8    – 30     – 42
—————————–
-4   – 15   – 21    – 54

Now, divide this by 2 (2 is the first coefficient of the divisor).
Also note that we don’t divide the remainder by 2, it will remain constant.

-2   – 15/2  – 21/2  – 54
Quotient = -2x² – 15/2x – 21/2 , Remainder = -54

One more illustration:
First, make the first coefficient of divisor as ‘1’

2x² -3x +1   |  2x⁵ -9x⁴ +5x³ +16x²    -16x +36
x²-3/2x+1/2
+3/2 -1/2            3    -1
-9    +3
-15/2     +5/2
69/4 -23/4
———————————————
2    -6     -5      23/2   15/4 +121/4
Divide by 2 )  1    -3    -5/2   23/4     15/4 +121/4

Quotient = x³-3x²-5/2x+23/4 , Remainder = 15/4x +121/4

Remember that Remainder is constant in every case

Lets take an example, 12x²-8x-32 by x-2
____________
x-2 | 12x² – 8x – 32  | 12x + 16
-(12x² – 24x)
—————–
16x – 32
-(16x – 32)
—————–
0
Here, Quotient = 12x + 16 , Remainder = 0

Take one more example, 6x⁴+13x³+39x²+37x+45 by x²-2x-9
_______________________
x²-2x-9 | 6x⁴+ 13x³+ 39x²+ 37x+ 45  |   6x² + 25x + 143
– (6x⁴ – 12x³– 54x²)
————————–
25x³+ 93x²+ 37x
-(25x³– 50x²– 225x)
————————-
143x²+ 262x+ 45
-(143x²– 286x -1287)
—————————
548x+1332
Here, Quotient = 6x² + 25x + 143 , Remainder = 548x + 1332

Now, see the above example in different form, which will actually be our procedure for division of large numbers

x²-2x-9 | 6x⁴+ 13x³+ 39x²  | +37x + 45 |
2  9           12      54
50       225
286   1287
—————————————-
6     25     143    |  548   1332
Quotient = 6x² + 25x + 143 , Remainder = 548x + 1332

Following are the steps that we done above:

Step 1: Divisor ( x²-2x-9 )

The first step is to write the divisor. In second line, write all the coefficients except the first, changing all of those to the opposite sign of coefficients. (Here, opposite signs of ‘-2’ and ‘-9’ are ‘2’ and ‘9’ respectively and first coefficient of x² i.e. ‘1’ is not being considered )

Step 2: Dividend ( 6x⁴+13x^3+39x²    +37x + 45 )

Step 3: Procedure

• Multiply the first coefficient of the dividend (i.e. ‘6’ in above example) by each digit of the rewritten divisor (i.e. 2 & 9) on the left (we get, 6×2=12, 6×9=54). Write each digit one after the other leaving the first place.
• Now see across the row, add the digits(i.e. 13+12= 25). Now repeat the same as done in Step3(i), multiply ’25’ by rewritten divisor (i.e. 2 & 9) we get, 25×2=50, 25×9=225. Write each digit one after the other leaving the first two places.
• Repeating the same as done in last step, now add (39+54+50=143). Now multiply 143 by (2 & 9) and we get (286 & 1287).
• Now add remaining columns which is the remainder, add (37+225+286=548, 45+1287=1332) and Quotient is (6 25 143)expression.

We have to repeat the process up to last digit of the dividend.

divide three- and four-digit numbers with two-digit numbers near base. In this article, we shall continue to discuss it further to divide bigger numbers by more than two digit number near base.

Procedure is the same as mentioned in the previous article. Please check that for more details about formula.

In below examples, we have deliberately taken ‘divisior’ of big digits to understand the concept.

12345 divided by 8897
Here p = 10000-8897 = 1103
1 | 2345
| 1103        (p x a)
1 | 3448
Quotient = 1 and Remainder = 3448

51235 divided by 7999
Here p = 10000-7999 = 2001
5 | 1235
|10005        (p x a)
5 |11240  (Here, remainder is greater than the divisor. Add 1 to quotient i.e. 5+1=6 and
subtract divisor from remainder i.e.(11240-7999=3241))
6 | 3241
Quotient = 6 and Remainder = 3241

12345 divided by 882
Here p = 1000-882 = 112, and a = 1
12 | 345
1 | 12      (p x a)
| 336    (p x (b+e))
13 | 801
Quotient = 13, Remainder = 801

2002002 divided by 89998
Here p = 1000000-89998 = 10002
20 | 02002
2 | 0004     (p x a)
| 20004   (p x (b+e))
22 | 22046
Quotient = 22 and Remainder = 22046

11001 divided by 88
Here p = 100-88 = 12
110 | 01
12 |         (p x a = 12 x 1 = 12)
2 | 4      (p x (b+e) = 12 x (1+1) = 24)
| 48    (p x (c+f+g) = 12 x (0+2+2) = 48)
124 | 89    (Remainder greater than divisor)
125 | 1
Quotient = 125 and Remainder = 1

11111111 divided by 99979
Here p = 1000000-99979 = 00021
111 | 11111
00 | 021
0 | 0021
| 00021
111 | 13442
Quotient = 111, Remainder = 13442