VEDIC MATHEMATICS

 Home Tables Squaring Finding Squares Basic Operations Gallery Credits AboutUs

# FINDING SQUARES,

cube roots and navashesh.

This is the most important area in any examination or competition.

Students do everything  but when they encounter a square in the last step , they leave it.

To counter this demon Vedic Maths gives us Efficient technique .

For finding square of a number near 100

Say you want to find the square of a number near 100 or multiples of 100.

98²= 98 – 2 / 02²= 96 / 04 = 9604 [100 – 98 = 2]
97²= 97 – 3 / 03²= 94 / 09 = 9409[100 – 97 = 3]

92²= 92 – 8 /8²  = 84 / 64 = 84 64[100 – 92 = 8]

89²= 89 – 11 / 11²=78 /121 = 7921 [100 – 89 = 11]

103²= 103 + 3 /03² = 106 / 09  = 10609 [103 -100 =3]

107²=107 + 7 / 7²=114 /49 =11449 [107 – 100 =7 ]

 Cube roots Finding cube roots require some background.   Background             Last digit           1³                         1                1 2³                         8                8 3³                        27              7 4³                        64              4 5³                        125           5 6³                        216            6 7³                        343            3 8³                        512            2 9³                        729            9   From the above example we can make out that the last digit of 2³ is 8  ,  3³ is 7 and the last digit of 7³  is 3  and  8³ is 2. All other digits repeat themselves. Procedure of finding a cube :-   Start from the right and put a comma when three digits are over.             EXAMPLES:-     9,261 1,728 32,768 175,616                After putting the comma .see the last digit of the number ; compare that with the table provided earlier .You get the last digit.   Now see the first group of numbers and ascertain cube of which number is less than the group. That number is your first Digit.   You have thus found the first digit and the last digit .   Let us take an example:- 9,261 2     1   Steps:-   Counting from the last , we put  a comma after 9.   By looking at the last digit , we ascertain that the last digit of the cube root will be 1.   Now  we  see 9 and ascertain that 2³ = 8 is less than 9 and 3³= 27 is more than 9.   Our first digit thus comes to 2,and the answer is 21.    For example :- 32,768 3      2   By looking at the last digit,we find that the last digit of the cube root is equal to 2.   By seeing 32 we put 3, as our first digit as 3³ = 27 which is less than 32 and 4³ = 64 is more than 27 .   Our answer is 32.

Result verification Technique

There are various methods of verifying the results obtained after completion of a mathematical operation. Vedic mathematics offers a very simple method of result verification, which can be used in day to day activity with great ease.

Navashesh means ‘nine and its reminder’

What does this mean?

Finding Navashesh

Any number can be converted into a single digit (Navashesh) after adding the digits. Let me explain this with the help of a few examples.

1. Find the single-digit equivalent for 32.

Just add 3 and 2 You will get 5.

The single –digit equivalent (Navashesh) is 5.

2. Find the single-digit equivalent for 342.

In this case you will be required to add them all

3+4+2=9

The single-digit equivalent (Navashesh) is 9.

3. Find the single-digit equivalent for 372

In this case you will be required to add them all 3+7+2=12=1+2=3

The single-digit equivalent (Navashesh) is 3.

4. Find the single-digit equivalent for 2367.

In this case you will be required to add them all 2+3+6+7=18=1+8=9

The single-digit equivalent (Navashesh) is 9.

5. Find the single-digit equivalent for -732.

In this case you will be required to add them all –(7+3+2)=-12=-(1+2)=-3=(-3+9)=6

In this case of a negative number, add9 to it. The single-digit equivalent (Navashesh) is 6.

After going through these examples, you can find out the single-digit equivalent Navashesh of any large number.

The Navashesh methodology gives you a quick way of verifying the results. How?

In addition, subtraction and multiplication you can use this to verify the results.

This formula simply states that the Navashesh remains unchanged. In other words, Navashesh of the digits before operation and after operation will remain unchanged. Let me explain this with the help of a few examples: (I will use NV to denote Navashesh of the number).

Example-6

67+34=101

 Navashesh of LHS (Problem) Navashesh of RHS (Answer) NV (67) =4, NV (34) =7 NV (101) =2 NV (67) + (NV (34) = NV (4+7)= NV (11) =2

Example-7

3673+2341=6014

 Navashesh of LHS (Problem) Navashesh of RHS (Answer) NV (3673) =1, NV (2341) =1 NV (6014) =2 NV (3673) + (NV (2341) = NV (1+1)= NV (2) =2

Example-8

3251+6242+845=10338

 Navashesh of LHS (Problem) Navashesh of RHS (Answer) NV (3251) =2, NV (6242) =5, NV (845) =8 NV (10338) =6 NV (3251) + (NV (6242) + NV (845) = NV (2+5+8)=NV (15)=6

Example-9

854+564+3254+12+6524=10938

 Navashesh of LHS (Problem) Navashesh of RHS (Answer) NV (854) =8, NV (564) =6, NV (3254)=5, NV (12) =3, NV (6254) =8 NV (10938) =3 NV (854) + (NV (564)+ NV (3254) =+ NV (12) + NV (6254) NV (8+6+5+3+8)= NV (30) =3

Navashesh and Subtraction

In all the above examples, we have seen that the addition operation does not change the Navashesh. The Navashesh of the LHS, ie problem side is equal to the Navashesh of the RHS, ie, answer side. Navashesh remains unchanged.

Let us see what happens in subtraction.

Example-10

88-31=57

 Navashesh of LHS (Problem) Navashesh of RHS (Answer) NV (88) =7, NV (31) =4 NV (57) =3 NV (88) -(NV (31) = NV (7-4)= NV (3) =3

Example-11

5283-2312=2971

 Navashesh of LHS (Problem) Navashesh of RHS (Answer) NV (5283) =9, NV (2312) =8 NV (2971) =1 NV (5283- (NV (2312) = NV (9-8)= NV (1) =1