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# India.

Squaring around 75.

Let's do 81^2 = ?

1) Find the deficiency/excess from 75. ( 81 - 75 = 6 )
2) add ( 3 / 2 ) times 1) to 56. ( 56 + ( 3 / 2 )* 6 = 65 )
3) square the answer from 1 ) and add it to 25 . ( 25 + 6^2 = 61 )

so the answer is 81^2 = 6561.

Let's do it again for 71^2 = ?

71 - 75 = - 4
( 3 / 2 ) * ( - 4 ) + 56 = 50
( - 4 )^2 + 25 = 41

so 71^2 = 5041

Vedic Math - Cube Roots

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) = 36*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 , ) 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.

Flag Method ( Division)

Direct Flag Method is a General Method of Vedic Mathematics is used to carry division of ANY types of numbers.
Single Digit Flag :
#1. 1234/12

Dividend = 1234 and Divisor = 12.
Split divisor (12) in 2 parts (1 and 2) where division will be carried using ONLY 1(new divisor) and 2 is called as flag.
As flag is single digit, Split dividend in 2 parts such that 2nd part will have same number as that of flag i.e. 1 digit.

Process (see the example for each step):
Division of 1 by 1 (Q=1 and R= 0). Write Q=1 and carry forward the R=0(written in white under and between 1&2).
Multiply the new Q(1) with the flag(2) and subtract this product from 02 = 0 and divide this subtraction by 1. It gives Q= 0 and R= 0(Carry forward R=0). (i.e. Multiply, Subtract, Divide)
Follow same above process, So multiply new Q(0) with flag(2) and subtract this product from 03 = 3 and divide this subtraction by 1. It gives Q= 3 and R= 0(Carry forward R=0).
For remainder we carry same process EXCEPT we don’t divide. (i.e. Multiply, Subtract) So multiply new Q(3) with flag(2) and subtract this product from 04 = -2. As we get negative subtraction, we reduce the quotient by 1 and increase the remainder by the 1st multiplier of new multiplier (1X1 =1). So new Q = 2 and new R =1. We carry this method till we don’t have negative subtraction.
Now Multiply the new Q(2) with the flag(2) and subtract this product from 14 = 10(positive) and put it down as it is.
So final answer: Quotient = 102 and Remainder = 10 (Remainder should always < Divisor