### Quick Divisibility tricks

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.

### Vedic Math - Division of large numbers (Part-II)

In previous article, we learn the procedure of doing 'division of large numbers' with divisor's first digit as '1'. In this article, we learn the procedure with divisor's first digit has all numbers except 1.

In this article too, we first go through with 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

### Vedic Math - Division of large numbers (Part-I)

In this article, we shall learn the procedure to divide the large numbers. To make it better understandable, lets go through with the division of polynomials. It will help us to understand the procedure better.

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
- (6x- 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 )

The second step is to rearrange the expression of the dividend, leaving space after last two expressions from right. The last two expressions are for the remainder. (Here question arises, how do we know from where we have to split. The answer is : one less than the length of the divisor expression (i.e. number of coefficients + constant). In above expression, length of divisor expression is 3, therefore we shall split after two expressions from right (3 - 1 = 2).

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, 6x2=12, 6x9=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, 25x2=50, 25x9=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.

IMPORTANT: In this article, we consider only those divisors whose first digit is unity i.e. '1'. The divisors whose first digit is not '1', will be taken in the next article.

### Vedic Math - Divide bigger numbers by more than two digit number near base

In the last article, we learn to 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

### Vedic Math - Dividing Three and Four Digit Numbers by any Two Digit Numbers Near Base

In last article, we discussed about division of numbers by any one digit number (near base). In this article, we shall discuss about division of three and four digit numbers by any two digit numbers (near base).

If we are dividing by 2 digit number than we need to have 2 places in the right column.

Dividing three digit numbers by any two digit numbers

To divide abc by mn, we apply the same rule as before:
1) Put a, b and c in the first row (as shown below)
2) List p x a in the second column of the second row. (p = base- mn)
a | b c
| p x a
----------------
a |(p x a) + bc

Quotient: a, Remainder: (p x a) + bc

Example:
To divide 102 by 75, the nearest base is 100, so p = 100-75 = 25

102 divided by 75
1 | 02
| 25
--------------
1 | 27
Quotient= 1, Remainder= 27

234 divided by 73
p = 100 - 73 = 27
2 | 34
| 54
-------------
2 | 88    (Remainder(88) is greater than 73, Add 1 to quotient i.e. 2+1=3 and subtract divisor(73)                        from remainder i.e.(88-73=15))
3 | 15
New Quotient = 3, New remainder = 15

Dividing four digit numbers by any two digit numbers

abcd divided by mn

This gets a little more complicated and one must be very careful with the places in the columns.
p = base - mn = 100 - mn

### Vedic Math - Dividing numbers by any one digit numbers (DIVISION BY 8 ETC.)

In the last article, we discussed division by 9. And in this article, we discuss divisors below base value 10, 100 etc. In these, we do the pairing from right.

Easy way to divide by 9 can be extended for 8, 7 etc.

ab divided by m
The rule we use is:
1) Put a and b in the first row
2) Write a as quotient in third row
3) List p x a (a the quotient) in the second column of the second row. (p= 10 - m)
a | b
| p x a
----------
a | b +(p x a)
Quotient: a, remainder: b + (p x a)
Clearly this is easiest for 9, p= 10 - 9= 1

8) 3 | 1               7) 2 | 0                6) 1 | 1
| 6                       | 6                       | 4
---------              ----------              ----------
3 | 7                    2 | 6                    1 | 5

In the above examples with divisor 8, 7, and 6, we get quotient 3 reminder 7 , quotient 2 reminder 6 and quotient 1 reminder 5 respectively.
We have to multiply the quotient by 2 in the case of 8, by 3 in the case of 7, by 4 in the case of 6 and so on. What this means, that we have to multiply the quotient-digit by the Divisor's complement from 10.

For example, Divide 31 by 8
Step 1: We bring the first digit 3 into the answer as quotient.
Step 2: Then instead of adding this to the 1 as we do when dividing by 9, we add double of 3
(i.e. 3 * (10-8)) to the 1 and get 7 for the remainder.
We double the 3 because 8 is 2 below 10.

Lets extend it for three digits Dividend

abc divide by m, p=10-m
a | b          | c
| p x a    | p x ((p x a) + b)
--------------------------------------
Quotient: a / (p x a) + b, Remainder: (p x ((p x a) + b)) + c

Similarly, for 311 divided by 8

8) 3  | 1 |  1
| 6
--------------
3  | 7 | 14
--------------
37   |  15
---------------
38   |   7
• We bring down the first digit i.e. 3,
• Add 2 x 3 (i.e. p x a) to 1 (i.e. b) in the next column and put down 7 (i.e.(p x a) + b),
• Then add 2 x 7 (i.e. p x ((p x a) + b)) to the 1 in the last column and put down 15 as the remainder.
• Since this remainder is bigger than 8, hence we compute answer further.  As 15 > 8, so add 1 to quotient and subtract 8 from remainder. Final answer is quotient = 38 and reminder = 7

### Vedic Math - Divsion by 9

Let us learn, how to do 'Division' operation using 'Vedic Math'. Conventionally, we do it like following:

Divisor ) Dividend ( Quotient
---------
---------
_________
Remainder

However, in the Vedic process, the format is
Divisor ) Dividend
--------
__________________
Quotient | Remainder

Let us first start with one of the special case of division i.e. Division By 9, a very interesting and simple technique.

When dividing by 9, the remainder is always the digit sum of the original number.

For 2-digit number divided by 9
To divide ab by 9 : Rewrite ab as a | b . The quotient is a, and the remainder is simply a + b.

a  |  b
|  a
---------
a  | a + b

Examples:

12 divided by 9
Here quotient = 1 and remainder = 1+2 = 3

23 divided by 9
Here quotient = 2 and remainder = 2+3 = 5

70 divided by 9
Here quotient = 7 and remainder = 7+0 = 7

Now, let us discuss the cases when remainder is greater than 9 :-

### Vedic Math - Cube Roots of more than 6-Digit Number - Part III

In this article, we are going to learn an interesting Mathematical technique to find, if the given number is a perfect cube or not. It is very important step while computing cube roots. Infact, before applying any method to find the cube root, we have to check whether it is perfect cube or not and then accordingly we choose the technique. For example, following scenario tells us the importance of finding perfect cube step while computing the cube root.

Example : 1728 has cube root 12 since two groups are 1 and 728. From 728, we derive last digit as                 2 from 1 (first group), we derive first digit as 1.
So, cube root of 1728 is 12.
But now, if number is 1278, which again has two groups: 1 and 278. It can derive the same last digit as 2 and first digit as 1 , which implies that cube root of 1278 is 12, which is not true because technique stands true for perfect cube root only.

There is a simple technique to check whether the number is perfect cube or not. For this, we add the digits of the number. See the below chart in which we add the digits of cubes from 1 to 10.

Above example shows that sum of digits of a perfect cube is either 1, 8 or 9. However, it is not true that all numbers which sum to 1,8 or 9, will be perfect cube.

For example,
Sum of digits of 1728 and 1278 are same i.e.(1+7+2+8) = (18) = 9 . But 1278 is not a perfect cube.

Hence if sum of digits of a number is not 1,8 or 9, we are very sure that the number is not a perfect cube. However, a number may not be perfect cube root even if sum of digits is 1,8 or 9. To scrutinize that, we need to apply factorisation. If number is small like 1278, factorisation is good method. See below:

For bigger numbers, factorisation could be time consuming technique. Hence, for large numbers, we shall apply general method of finding the cube of root.

Case 2 : Cube root for all the cubes, whether perfect cubes or not.    (Case 1 discussed in last two articles)
From last two articles, we conclude about the sequence of digits (a+b+c)³ as:
(1) The first place by a³
(2) The second place by 3a2b
(3) The third place by 3ab2+3a2c
(4) The fourth place by 6abc+b³
(5) The fifth place by 3ac2+3b2c
(6) The sixth place by 3bc2
(7) The seventh place by c³ ; and so on.

In 'General Technique', we find Dividends(D), Quotients(Q), and Remainders(R). Steps involved as:
(1) First determine D, Q and R
(2) From the second dividend, no deduction is to be made.
(3) From the third, subtract 3ab2
(4) From the fourth, deduct 6 abc+b³
(5) from the fifth, subtract 3ac2+3b2c
(6) from the sixth, deduct 3bc2
(7) from the seventh, subtract c³. ; and so on.

(a) Quotient(Q) is closest minimum exact cube to the first cube i.e. 'F' term used in last two articles.
(b) And, Reminder(R) is the difference between the first group and closest minimum exact cube.
(c) Dividend(D) is found by multiplying the 'Square of Quotient(Q)' by 3 (Q2*3)

Lets take an example to make it more clear.
Example 1 : 248858189

(1) First arrange the numbers in groups i.e. 248,858,189
Here, N = 3
First group(248) has closet minimum exact cube (216) which is 6³. So, First Quotient(Q) = 6
First Reminder(R) = 248-216 = 32
First Dividend(D) = 62*3 = 108

(2) The second Gross Dividend is 328. We don't subtract anything at that point. We only divide it by 108 and write down 2 and 112 as Q and R.
[Important Note : Here we are taking quotient to be 2 instead of 3; because if we take it as 3, the reminder comes out 4 (328-108*3) and third dividend turns out 45 which is absurd and will not be dividable by 108]

(3) The third Gross Dividend is 1125. Subtract 3ab2 (here, a=6, b=2, first two quotients) i.e. 3*6*22 = 72 from 1125 (i.e. 1125-72=1053) Therefore, Third Actual Dividend is 1053; divide 1053 by 108 gives 9 and 81 as Q and R.

(4) The fourth Gross Dividend is 818. Subtract 6abc + b³ (here, a=6, b=2, c=9) i.e. 6*6*2*9+2³ = 656 from 818 (i.e 818-656=162) So, Fourth Actual Dividend is 162. Divide this again by 108 and write down 0 and 162 as Q and R.

(5)  The fifth Gross Dividend is 1621. Subtract 3ac2+3b2c (here, a=6, b=2, c=9) i.e. 1458+108 = 1506 from 1621 (i.e 1621-1506=55) So, Fifth Actual Dividend is 55. Divide it by 108 and write down 0 and 55 as Q and R.

(6) The sixth Gross Dividend is 558. Subtract 3bc2 (here, b=2, c=9) i.e. 486 from 558 (i.e 558-486=72) So, Sixth Actual Dividend is 72. Divide this by 108 and write down 0 and 72 as Q and R.

(7) The last / seventh Gross Dividend is 729. Subtract c³ (here, c=9) i.e. 729 from 729 (i.e 729-729=0) So, Seventh Actual Dividend is 0 and write down 0 and 0 as Q and R.

Put decimal after 3 digits (N=3). After decimal there are all zeros. This means that the given number is a perfect cube and the cube root is 629.
(After putting decimal, if there are still numbers except 0's than the number is not perfect cube)

Steps involved in finding dividend, quotient and reminder for (a+b+c+d)³ are:
(1) First determine D, Q and R
(2) From the second dividend, no deduction is to be made.
(3) From the third, subtract 3ab2
(4) From the fourth, deduct 6abc+b³
(5) from the fifth, subtract 6abd+3ac2+3b2c
(6) from the sixth, deduct 6acd+3bc2+3b2d
(7) from the seventh, subtract 6bcd+3ad2+c2
(8) From the eighth, subtract 3bd2+3c2d
(9) From the ninth, subtract 3cd2
(10) From the tenth, subtract d³; and so on.

OR We can convert (a+b+c+d) into (a+b+c). By considering first two groups into one group.
For example: 12278428443 can be written as
12, 278, 428, 443 (a+b+c+d)
12278, 428, 443   (a+b+c)
But, in second, we shall get the first quotient bigger.

Lets take one more example of an imperfect cube (not a perfect cube).
Find cube root of 417 upto 3-decimals place.

So, Cube root of 417 is 7.471

Hope these methods will help you all in computing cube root of the number.

In the upcoming articles, we will discuss about various techniques of division using Vedic Math.

### Vedic Math - Cube Roots of more than 6-Digit Number - Part II

In last article, we have discussed the method to find cube root of more than 6-digit numbers; especially the odd numbers. Today we shall discuss the procedure for even numbers. In this procedure, only two extra steps are added, one in the beginning and other at the end. Rest all is same.

Procedure: As first added step, we keep on dividing the number by 8 till we get an odd cube. Following it, same method of successive elimination of the digits will apply. At the end, multiply the cube root by 8 to obtain the cube root of the original number.

Example :  2840362499528
First, we continue without using those two additional steps, which will help you to understand the problems arises while dealing with even cubes.

The cube root of the cube 2,840,362,499,528      (say, F + J + H + M + L )
Here,  N=5       (means that cube root will be of 5 digits number)
L=2                  (i.e. 2³=8, matching with the last digit of the last group '528')
and F=1            (i.e, 1³=1, nearest cube of first group '1')

Step1 : L=2 & L³=8. Subtracting this,
Step2 : 3L2M=12M (substituting L = 2)
Hence, 12M = Number ending with 2
Here M is either '1' or '6'      (ambiguous values)
Lets take 6  (pure gamble)

Now, Deducting 3L2M = 12M = 72
Step3 : 3LM2+3L2H = 12H + 216 (substituting L = 2, M = 6)
12H + 216 = Number ending with 8
12H = Number ending with 2
Here H is either '1' or '6'
Lets take 1 (again gamble)

Now, Subtract 3LM2+3L2H = 12H + 216
= 228
Step4 : 3L2J+6LMH+M³ = 12J+12+216
= 12J+228
12J+228 = Number ending with 6
12J = Number ending with 8
Here J is either '4' or '9'
Lets take J = 4

Since we already know 'F' , so no need to know the expansion of (F+J+H+M+L)³
Therefore, cube root is 14162        (F=1, J=4, H=1, M=6, L=2)

In the above example, we see that there are so many ambiguous values like in step 2, 3, 4. To solve this problem, we divide the number by 8.
So,
After first division by 8, if we get the last digit as odd number; we shall consider this number. If last digit is still even, we shall again divide by 8 till the last digit comes out to be an odd number. Remember one more thing, we will multiply the final output by 8 only once and not the number of times we divide it by the number (like, after dividing 3 times '792 994 249 216' by 8, we get '1 548 816 893', an odd number. Then we calculate the cube root of this odd number. Finally we shall multiply the cube root with 8 only once,
output*8 = Final Cube Root)

For, 355,045,312,441
N=4,
L=1     (1³=1)
F=7     (7³=343 < 355)

Step1 : L=1 & L³=1. Subtracting this,
Step2 : 3L2M=3M (substituting L = 1)
3M = Number ending with 4
So M should be 8

Now, Deducting 3L2M = 3M
= 24
Step3 : 3LM2+3L2H = 3H + 192 (substituting L = 1, M = 8)
3H + 192 = Number ending with 2
3H = Number ending with 0
Hence H = 0

Therefore, Cube Root is 7081    (F=7, H=0, M=8, L=1)

The Cube Root of the original number = 7081*8
= 14162

In next article, we shall learn to find the cube root for all cube numbers (whether perfect or not). Thanks for visiting the blog. Please keep sharing the knowledge by posting in 'comments' section.

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### Vedic Math - Cube Roots of more than 6-Digit Number - Part I

Back after a long break. In previous article, we learn how to find the cube root of 4 or 5 or 6 digits perfect cubes. Let us continue it further and discuss how to find the cube root of perfect or imperfect cubes.

In this article, we shall learn to find the cube roots for:
1. Cube root of perfect cubes, for any number of digits.
2. Cube root for all the cubes, whether perfect cubes or not.

To summarize what we have learned till now for cube root:

Arrange the given number in three-digit groups, starting from right to left. A single digit, if any left over at the left hand side, is counted as a simple group itself. The number of  digits in the cube root will be the same as the number of digit-groups  in the given  number itself.

• 169 will count as 1 group
• 1 258 will count as 2 groups
• 43 781 will count as 2 groups
• 2 154 890 will count as 3 groups

If the given number has 'n' digits, its cube root will be having n/3 or (n+1)/3 digits. Also remember few other points from previous article:

The Cubes  of the first nine natural numbers
1³ = 1        2³ = 8        3³ =27        4³ = 64        5³ = 125        6³ = 216        7³ = 343        8³ = 512         9³ = 729        10³ = 1000
From it, we understand that

• 1,4,5,6,9,0 numbers repeat themselves in the ending of their cubes
• 2,3,7 and 8 have their complements from 10, at the end of their cube

Let us start with actual technique now. Any number can be written in an algebraic expression. For example, if arithmetical number is 'dcba', it can be written in algebraic form as:
Algebraic Expression is: a + 10b + l00c + 1000d.

Now if we need to find the cube of a number 'cba', algebraically we can expand it like  (a+10b+102c)3. Let us expand it:
(a+10b+102c) 3 =  a3 + 10 (3a2b) + 102 (3ab2+3a2c) + 103 (b3+6abc) + 104 (3ac2+3b2c) + 105 (3bc2) + 106 (c3)

Now removing the powers of ten and putting the result in algebraic form, it tells us the formation of cube as:
(1) The units' place is determined by a³.
(2) The tens' place is contributed by 3 a2b
(3) The hundreds' place is contributed to by 3ab2 + 3a2c
(4) The thousands' place is formed by b³ + 6abc
(5) The ten thousands' place is given by 3ac2 + 3b2c
(6) The hundred thousands' (lakhs') place is constituted of 3bc2 ; and
(7) The millions' place is formed by c³.

The number of zeroes in the various coefficients of the expanded Algebraic Expression are the basis of the formula / analysis.

Case1 :  Cube root of perfect cubes for any number of digits

Suppose we have a cube number n of any number of digits. To find its cube root, find following:
- The number of groups (N) in cube (as we discussed above to make the sets of 3 digits)
- First digit of cube root denoted as 'F' (Nearest cube root of first group from left)
- Last digit of cube root denoted as 'L' (Cube root of last group from left)
- Middle numbers of cube root(i.e. 'M' or 'H' or 'J'....), we shall find using the procedure.

Following are the steps for the procedure:
(i) From the units' place of given number, subtract the L³ (i.e. a³, refer to algebraic expression above); and that eliminates the last digit of the number.
(ii) From the ten's place, we subtract 3L2M (i.e. 3a2b) and thus eliminate the second last digit (penultimate digit).
(iii) From the hundreds' place, we subtract 3LM2 + 3L2F (i.e. 3ab2 + 3a2c) and hence eliminate the pre-penultimate digit.
and so on

Let us understand it with an example.

Example 1 :  83453453

First arrange the numbers in groups i.e. 83,453,453
Here the number of 3 digit groups i.e. N = 3 (that also means that cube root will be of 3 digits number)
Last digit of the number is 3. The last digit of cube root of the number ending with 3 should be 7 (i.e.7³ = 343) So L = 7
The first group is 83 , the closest minimum exact cube to 83 is 64 which is nothing but 4³, therefore, F = 4

Hence N = 3, L = 7, and F = 4

And let us proceed to find the middle number:

Step1 : L=7 so L³=343
Now subtract '343' from the last digit

This eliminates the last digit of number, and give way to calculation of middle numbers

Step2 :   3L2M
= 147M        (substituting L = 7)
Now the last digit of 147M is '1' (which is the last digit of the number '8345311' after subtracting in first step).
Hence,  147M = Number ending with 1
So M should be 3 ( because 147*3 can give a number ending with 1)

Deducting 3L2M = 147M = 147*3 = 441

This eliminates the second last digit of number, and give way to calculation of further numbers.

Step3 :  3LM+ 3L2F = 147F + 189  =  ending with 7 (last digit of '834487' )
147F  =  Number ending with 8   (i.e.834487-189)
Hence, F = 4     (147*4 can give a number ending with 8)

Hence, Cube Root = 437

Note:- The last step can be skipped safely, as we were already aware about the the first digit of cube root i.e. F.

Example 2: 13824
Arrange the number in groups i.e. 13, 824
Here N=2 (which means that cube root will be of 2 digits number)
L = 4 (i.e. 4³ = 64) and
F = 2 (exact cube near to 13 is 8 i.e.2³)
Therefore,  cube root = 24

Example 3 : 76928302277
Arrange the number in groups i.e. 76, 928, 302, 277  (say, F + H + M + L )
Here N = 4 (which means that cube root will be of 4 digits number, FHML)
L = 3 (i.e. 3³ = 27, matching with the last digit of the last group '277')
and F = 4 (i.e, 4³ = 64, nearest cube of first group '76')

Step1 : L=3 & L³=27. Subtracting this,
Step2 : 3L2M=27M (substituting L = 3)
Now the last digit of 27M is '5' (which is the last digit of the number '7692830225' after subtracting in first step).
Hence, 27M = Number ending with 5
So M should be 5 ( because 27*5 can give a number ending with 5)

Now, Deducting 3L2M = 27M
= 135         (27*5)
Step3 : 3LM2+3L2H = 27H + 225 (substituting L = 3, M = 5)
27H+225 = Number ending with 9
27H = Number ending with 4   (i.e.769283009-225)
Hence,  H = 2   (27*2 can give a number ending with 4)
So,  3LM2+3L2H = 27H+225 = 27*2+225 = 279
Therefore, cube root is 4253    (F=4,  H=2,  M=5,  L=3)

Note: According to the procedure, we can determine the first digit by the same method of successive elimination of the digits. For this, we have to expand (F + H + M + L)³. And, on expanding, we find that in the 4th step i.e. thousand's place number, 'F' appears to be '3L2F + 6LMH + M³'. We need to deduct this value to eliminate the fourth digit. So,

3L2F + 6LMH + M³ = 27F+180+125
= 27F+305
27F + 305 = Number ending with 3
27F = Number ending with 7
Hence,   F = 4
Therefore,  Cube Root = 4253

[For reference, (103d+102c+10b+a) 3 = a3 + 10 (3a2b) + 102 (3ab2+3a2c) + 103 (b3+6abc+3a2d) + 104 (3ac2+3b2c+6abd) + 105 (3bc2+3b2d+6acd) + 106 (c3+3ad2+6bcd) + 107 (3bd2+3c2d) + 108 (3cd2) + 109 (d3)]

Note:- Again, in case of perfect cubes, the last step can be skipped safely, as we are already aware about the the first digit of cube root i.e. F.

Important point to notice is that above method can be applied for odd cubes only. If the cube is even, then we would found ambiguous values at each step; like we may get two possibilities of M, H,.... We shall discuss the procedure to find cube root of such even cubes in next article. Till then, practice above procedure and keep posting your queries and share any findings.

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