## Monday, 21 May 2012

### Solve equations easily and fast

**The vedic maths sutra to be used: "Adjust After Transposing (Paravartya Yojayet)"**

When students are introduced to the concept of solving equations, the first method taught for solving basic equations is as follows:

Let us say we want to solve the equation 3x - 9 = 0

3x - 9 = 0

3x = 9 (transposing 9 to the other side of equation)

x = 9 / 3 (adjusting 3 with 9 on the right hand side of the equation)

x = 6 (Solution of equation)

This method is little lengthy and can not be used quickly for solving complex equations. Vedic mathematics has a sutra for solving equations and is based on the same concept of transposing and then applying. (Paravartya = Transpose; Yojayet = Adjust)

So let us understand this vedic math trick to solve equations fast and easily.

There is a precondition for using this formula though, and it is that there has to be a linear relationship with ratio of coefficient of one of the unknown in the equation with the constant term. For all the linear equations satisfying this condition, can be solved by simply putting the other unknown (other than the one with ratio of coefficients in proportion of the constant) as zero.

Let us just analyse our normal method of solving equations and see how it actually gives us an answer.

Let us say, we want to solve an equation

ax + b = px + q

ax - px = q - b (transposing both the sides)

x (a - p) = q - b

x = (q - b) / (a - p) (adjusted both the sides)

Thus, every equation that we want to solve using this lengthy method, can be solved easily just by using the final solution that x = (q - b) / (a - p)

*Example 1:*

5x + 8 = 3x + 16

Without getting into any lengthy calculation, let us solve this equation by using the vedic math trick, the answer would be:

x = (16 - 8) / (5 - 3)

x = 4

*Example 2:*

4x - 6 = 3x + 16

x = (16 + 6) / (4 - 3)

x = 22

Isn't it simple solving equations using this method instead of getting into length calculations.

Now let us consider solving tricky equations with the same method.

1. Solving equations with undefined solutions

*Example 3:*

4x + 6 = 4x + 16

x = (16 - 6) / (4 - 4)

x = Undefined (This equation involved division by 0, hence the solution to this equation is undefined as it should be)

2. Solving equations expressed as ratio to one another

While solving equation expressed as ratio to one another, the solution as per vedic math trick would change a little bit as follows:

(ax + b) / (px + q) = y / z

x = (yq - zb) / (za - yp)

*Example 4:*

(3x + 5) / (6x + 4) = 7 / 8

x = (7*4 - 8*5) / (8*3 - 7*6)

x = (28 - 40) / (24 - 42)

x = (-12) / (-18)

x = 2 /3 (This solution can be checked by trying the traditional method)

Similarly, if (6x + 4) / (3x + 2) = 2, then the same vedic math trick can be used to solve this formula just by replacing 2 with (2/1) on the right hand side of the equation.

When faced with equations such as (2x + 3) / x = 13 / 5, the same vedic math trick can be used and we just need to rewrite the given equation in the format we have the solution, refer the below example.

*Example 5:*

(2x + 3) / x = 13 / 5

(2x + 3) / (1x + 0) = 13 / 5 (Rewritten the equation to match with the vedic math trick)

x = (13*0 - 5*3) / (5*2 - 13*1)

x = (0 - 15) / (10 - 13)

x = (-15) / (-3)

x = 5 (This solution can be checked by trying the traditional method)

Thus, any equation with single power of the unknown can be easily solved using this vedic math tricks. We just have to make sure to rewrite the equation in the structure provided by the vedic math trick adjust after transposing.

Solving equations is so easy ans fast with the help of vedic math tricks!

## Monday, 30 April 2012

### Trick to identify exact squares

While vedic math tricks would make squaring numbers very easy, these few tricks help identify perfect squares merely by observing the number.:

First trick is to understand what would be the structure of the square root just by observation.

**Structure of the square root**:First trick is to understand what would be the structure of the square root just by observation.

- The number of digits before decimal point would reduce to half when we take the square root of the number.(i.e. the square root of 1446.78 would have 2 digits before decimal number). This math trick would be applicable directly when the numbers of digits before decimal point are even.
- In case of odd number of digits before decimal point, the square root would still have half of the numbers but that half needs to be rounded up to the next integer. (i.e. square root of 144.678 would have two digits before the decimal point.)
- The number of digits after the decimal point would reduce to half when we take the square root of the number.(i.e. the square root of 1446.78 would have 1 digits after decimal number). This math trick would be applicable directly when the numbers of digits before decimal point are even.
- For numbers with odd digits after decimal point, the square root would be an irrational number, because any square of a whole rational number would have even digits only after the decimal point.

**Nature of the square root:**Next trick is to know the type of answer to expect while trying to derive square root of a number, the rule is:- For the numbers not exact squares of another whole number, their square root is always irrational. (i.e. the square root would have endless digits after decimal point and would not have any repeating patters in those digits.)
- Also, a number is odd number of zero's at the end, would never have a rational square root
- For the rest of the numbers the square root would be rational.

**The last digit of the answer:**This is a trick to identity an exact square and also to estimate the last digit of the square root.

- Number with last digits as 2, 3, 7 or 8 are not exact squares and inturn would result in an irrational number when their square root is calculated
- Also, there is a direct relation between last digit of the number and last digit of its square root, as follows:
- number 1, square root 1 or 9
- number 4, square root 2 or 8
- number 5, square root 5
- number 6, square root 4 or 6
- number 9, square root 3 or 7
- When the second last digit (from right) is even and the last digit is 6, the number is not a perfect square (i.e. 346 is not a perfect square)
- When the second last digit (from right) is odd, the last digit has to be 6 for the number to be a perfect square. (i.e. numbers ending with 34, 59, 11 are not perfect squares)
- When a number is even, and its last two digits taken together are not divisible by 4, that number is not a perfect square. (i.e. numbers ending with 42, 86, etc. are not perfect squares)

**Odd one out:**This trick will tell us whether the square root would be even or odd.

- Square root is odd when a perfect square is odd
- Square root is even when a perfect square is even

## Friday, 27 April 2012

### Solve simultaneous linear equations easily

**The vedic maths sutra to be used: "If one is in ratio, the other one is zero"**

While trying to solve simultaneous equations, especially with bigger numbers, you can use this vedic math sutra to simplify the process.

There is a precondition for using this formula though, and it is that there has to be a linear relationship with ratio of coefficient of one of the unknown in the equation with the constant term. For all the linear equations satisfying this condition, can be solved by simply putting the other unknown (other than the one with ratio of coefficients in proportion of the constant) as zero.

**Example:**

5x + 8y = 24

23x + 16y = 48

In these equations, we have two unknown (x & y), so let us test if any of them satisfy the required condition:

*Testing for x*

Ratio of coefficients of x = 23 / 5 = 4.6

Ratio of constant number = 48 / 24 = 2

Not satisfied for x.

*Testing for y*

Ratio of coefficients of y = 16 / 8 = 2

Ratio of constant number = 48 / 24 = 2

Satisfied for y.

Therefore, the solutions to this equation is x = 0 and therefore y = 3.

however, if both the unknown satisfy the condition, then the equations can not be solved.

**The Logic:**

The equations consist of two sides LHS and RHS.

Now, RHS is a constant number the ratio of increase is RHS is obvious and can be understood just by observation, which in this case is 2.

This reflects that the LHS of second equation has to deliver a value exactly double as compared to what LHS to first equation delivered.

Now, LHS comprises of two terms with unknown x & y and having different coefficients. For LHS to second equation to become exactly double as compared to the first, both the coefficients should double. But in this case, y is exactly doubling and x is increasing randomly, which clearly means that whatever be the coefficient of x, it has to be zero so as not to disturb the proportion.

Now enjoy solving simultaneous liner equations just by looking at them with the use of this vedic math sutra.

### Solve quadratic equations easily with vedic maths

The vedic math sutra used for simplifying equations is : "It is zero if the Samuccaya is Same"

This formula aims to simplify the solution of quadratic equations by setting some simple ground rules. But, before we understand the application of this vedic math sutra with quadratic equation, let us start with basic equations and understand the applicability of this vedic mathss sutra.

This formula aims to simplify the solution of quadratic equations by setting some simple ground rules. But, before we understand the application of this vedic math sutra with quadratic equation, let us start with basic equations and understand the applicability of this vedic mathss sutra.

This vedic math sutra (It is zero if the Samuccaya is same) has applicability in four different situations:

1. When a common term is there on all the terms of left hand side as well as right hand side of an equation, then that equation can be solved by assuming that common term to be zero.

Example:

11x + 5x = 6x + 7x

Since "x" occurs as a common factor in all the terms (of both LHS & RHS), therefore, x = 0 is a solution.

2. When multiplication of all the numbers appearing on each side of the equation is equal, the solution to that equation can be derived by assuming the unknown to be zero.

Example:

(x + 6) (x + 5) = (x + 3) (x + 10)

6 * 5 = 3 * 10 (multiplication of each side's numbers is equal = 30)

Solution is x = 0

3. The sum of the fractions with same numerical numerator can be derived by adding the denominators and equating it to zero.

Example:

3/ (4x − 1) + 3/ (6x − 1) = 0

In this case the numerator of both the fractions is the same number 3, therefore the addition can be re-written as:

10x-2 = 0 (LHS simplified by adding the denominators)

x = 1/5

4. For a quadratic equation expressed as N1/D1 = N2/D2 and if N1 + N2 = D1 + D2, then this sum (i.e. either N1+N2 or D1+D2) is zero. This would help solve the quadratic equation easily.

4. For a quadratic equation expressed as N1/D1 = N2/D2 and if N1 + N2 = D1 + D2, then this sum (i.e. either N1+N2 or D1+D2) is zero. This would help solve the quadratic equation easily.

Example:

(3x + 8) / (3x + 4) = (3x + 4) / (3x + 8)

In this case, the condition can be assessed as:

N1 + N2 = 3x + 8 + 3x + 4 = 6x + 12

D1 + D2 = 3x + 4 + 3x + 8 = 6x + 12

N1 + N2 = D1 + D2

Therefore,

N1 + N2 = D1 + D2 = 0

6x + 12 = 0

x = -2 (Solution to quadratic equation)

Now with the knowledge of this vedic math sutra solving quadratic equation faster is very easy.

### Subtracting from a large power of ten

The vedic math sutra - "All from nine and the last from ten"

Subtraction is not a very complex mathematics calculation but it would get difficult when you are dealing with large numbers and even more difficult when it results in carrying over from the left hand site number.

~~ 1~~00000

What if there is a vedic math sutra which shows how to do this calculation without even having to write to down numbers and also without the hustle of having to carry over from the left. Let us lean about the vedic math sutra:

The sutra: All from nine and the last from ten

Method: As the name suggests, you just need to subtract all the numbers from nine and the last one from 10

So in this case, last number is 9 which is to be subtracted from 10 = 1

and the rest of the numbers 8754 to be subtracted from 9 = 1245

So the answer is 12451

Subtraction is not a very complex mathematics calculation but it would get difficult when you are dealing with large numbers and even more difficult when it results in carrying over from the left hand site number.

Let us understand where to effectively use this vedic maths sutra:

100000 - 87459

The normal method of subtraction would require us to write down these numbers and do the calculations as follows:

100000

__87549__

9 9 9 910

__87549__(carrying over from the left hand side number)

12451

What if there is a vedic math sutra which shows how to do this calculation without even having to write to down numbers and also without the hustle of having to carry over from the left. Let us lean about the vedic math sutra:

The sutra: All from nine and the last from ten

Method: As the name suggests, you just need to subtract all the numbers from nine and the last one from 10

So in this case, last number is 9 which is to be subtracted from 10 = 1

and the rest of the numbers 8754 to be subtracted from 9 = 1245

So the answer is 12451

Thus, though subtraction is easy, this vedic math sutra makes it even easier!

## Thursday, 26 April 2012

### What is vedic maths

While wonders that vedic mathematis can do in simplifying maths have been discussed at length in Vedic Mathematics, this post is intended for background about vedic maths. Knowing about vedic maths (history, origination) would make it more interesting and appealing.

Vedic Mathematics was rediscovered from the Vedas (Indian Spiritual and knowledge books) between 1911 and 1918 by Sri Bharati Krsna Tirthaji (1884-1960). According to his research all of mathematics is based on sixteen Sutras (formulas). Each of these vedic maths sutras (formulae) describe easier method of how to carry out mathematical calculations mentally and offer best value to students in cracking through exams with high scores.

The most interesting thing about vedic maths is that this method is that it is fully integrated and offers both way proofs of all concepts. meaning multiplication methods can be reversed to use for one-line divisions and the simple squaring method can be reversed to get square roots.

So complex mathematics problems often be solved immediately is you know about vedic maths. These striking and beautiful methods are just a part of a complete system of mathematics which is far more systematic than the modern 'system'. Vedic Mathematics manifests the coherent and unified structure of mathematics and the methods are complementary, direct and easy.

The simplicity of Vedic Mathematics implies easier and mental calculations. There are many advantages in using a flexible, mental system. Students can invent their own methods, they are not limited to the one 'correct' method. This leads to more creative, interested and intelligent learning and would make mathematical concepts easier to grasp for students.

With increasing focus on competitive exams with requirement of faster calculation speeds and, focus on Vedic maths system is growing in education. Lot of new research is being carried out to augment established vedic maths sutras and also enhance their usability in many fields not only for students.

Now that we have understood a lot about vedic maths, let us get going with vedic maths sutras and explore the immense potential of this method to make mathematics easy. After studying this method, the question "mathematics how to" would never arise!

### Vedic maths sutras

**Primary Sutras (formulas)**

These vedic maths sutras deal with individual mathematical operations (i.e. multiplication, division, squares, factorization, etc.)

These vedic maths sutras help augment the basic calculations while using the primary sutras or while doing normal mathematical calculations.

- Last from 10 and the rest from 9
- Cross-wise and Vertical
- By one more than the one before.
- Apply after transposing
- It is zero if the Samuccaya is Same
- If One is in Ratio the Other is Zero
- By Subtraction and by Addition
- By the Completion or Non-Completion
- Differential Calculus
- Use Deficiency
- Specific and General
- The Ultimate and Twice the Penultimate
- The Remainders by the Last Digit
- By One Less than the One Before
- All the Multipliers.
- The Product of the Sum

**Secondary Sutras (Formula)**These vedic maths sutras help augment the basic calculations while using the primary sutras or while doing normal mathematical calculations.

- Proportionately
- The Remainder Remains Constant
- For 7 the Multiplicand is 143
- The First by the First and the Last by the Last
- By Osculation
- Lessen by the Deficiency
- The Product of the Sum is the Sum of the Products
- Set up the Square of the Deficiency
- Whatever the Deficiency lessen by that amount and
- Last Totalling 10
- The Sum of the Products
- Only the Last Terms
- On the Flag
*(Not mentioned in vedic maths original books)* - By Alternative Elimination and Retention
- By Mere Observation

## Tuesday, 17 April 2012

### Mental Multiplication

Here we are talking about the sutra "Nikhilam" which would help us solve the mathematical multiplication in our mind without the use of calculator and even without having to write anything down! Let us understand the method with the help of examples.

605 x 504 = ?

We first need to select a base number (preferably a multiplier of 100 in case of three digit numbers to make it easier). In this case the base would be 500.

Now, define the distance of multipliers from the base as given below

605 -500 = +105 (the sign is very important)

504 -500 = +004

Now lets reproduce these numbers as:

605 + 105

504 + 004

From this, we derive two numbers and their names would be as follows:

LHS (is the result of multiplying the deficiencies) = +105 x +104 = +420

RHS (by adding diagonally on any one side) = 105 + 504 = 609 (remember, you can select any diagonal side, the answer will remain the same)

Now because our base is not a 10 series number (i.e. 10, 100, 1000, 10000, etc.) we need to establish the relation of the base with next highest 10 series number and the relation is 1000 / 2 = 500

Therefore, we have to express the LHS and RHS as follows:

(609 /2) | (+420)

304.5 + 420

304.5 | 420

Now see that the L.H.S has a decimal point ,so carry out that half to the R.H.S side from L.H.S.

Now R.H.S side becomes 420 + 500 = 920

304 | 920

Therefore ,605 x 504 = 304920

485 x 475 = ?

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above both numbers are less than the base.

How much deficient are are the above numbers are from base ?

485 -500 = -015

475 -500 = -025

485 - 015

475- 025

--------------

460 | ( +375) -->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

230 | 375 ---- (Half of 460 is 230)

230 | 375

Therefore 485 x 475 = 230375

614 x 495 =?.

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above numbers .One is greater than the base and one is less than the base.

How much deficient are are the above numbers are from base ?

614 - 500 = +114

495 - 500 = -005

614 + 114

495 - 005

------------

609 | (-570)--->-->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

(609 /2) | (-570)

304.5 | (-570) ----- > the minus sign is removed by 1000 -570 =430 and remove one from the L.H.S side

303.5 | 430

303| (430+500) ----> the 0.5 when carried to R.H.S, becomes 500 and add it to R.H.S

303| 930

Therefore ,614 x 495= 303930

These are general rules. There would not be issues when multiplications are above or higher, but when one is greater and one is lesser, follow the rules in the last example. All in all, the rules are:

The (0.5) on L.H.S is carried as 500 and added to the R.H.S every time you fall into a decimal (1/2).

The Minus signs on R.H.S is removed by adding 1000 to it.

It would seem very easy once you get in this mode of calculations.

*Example 1:*605 x 504 = ?

We first need to select a base number (preferably a multiplier of 100 in case of three digit numbers to make it easier). In this case the base would be 500.

Now, define the distance of multipliers from the base as given below

605 -500 = +105 (the sign is very important)

504 -500 = +004

Now lets reproduce these numbers as:

605 + 105

504 + 004

From this, we derive two numbers and their names would be as follows:

LHS (is the result of multiplying the deficiencies) = +105 x +104 = +420

RHS (by adding diagonally on any one side) = 105 + 504 = 609 (remember, you can select any diagonal side, the answer will remain the same)

Now because our base is not a 10 series number (i.e. 10, 100, 1000, 10000, etc.) we need to establish the relation of the base with next highest 10 series number and the relation is 1000 / 2 = 500

Therefore, we have to express the LHS and RHS as follows:

(609 /2) | (+420)

304.5 + 420

304.5 | 420

Now see that the L.H.S has a decimal point ,so carry out that half to the R.H.S side from L.H.S.

Now R.H.S side becomes 420 + 500 = 920

304 | 920

Therefore ,605 x 504 = 304920

*Example 2*

485 x 475 = ?

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above both numbers are less than the base.

How much deficient are are the above numbers are from base ?

485 -500 = -015

475 -500 = -025

485 - 015

475- 025

--------------

460 | ( +375) -->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

230 | 375 ---- (Half of 460 is 230)

230 | 375

Therefore 485 x 475 = 230375

*Example 3*

614 x 495 =?.

The nearest base for the above two numbers are 500.So we fix this 500 as our base.Now, see that the above numbers .One is greater than the base and one is less than the base.

How much deficient are are the above numbers are from base ?

614 - 500 = +114

495 - 500 = -005

614 + 114

495 - 005

------------

609 | (-570)--->-->the RHS is got by adding diagonally on any one side ,the LHS is the result of multiplying the deficiencies .

(609 /2) | (-570)

304.5 | (-570) ----- > the minus sign is removed by 1000 -570 =430 and remove one from the L.H.S side

303.5 | 430

303| (430+500) ----> the 0.5 when carried to R.H.S, becomes 500 and add it to R.H.S

303| 930

Therefore ,614 x 495= 303930

These are general rules. There would not be issues when multiplications are above or higher, but when one is greater and one is lesser, follow the rules in the last example. All in all, the rules are:

The (0.5) on L.H.S is carried as 500 and added to the R.H.S every time you fall into a decimal (1/2).

The Minus signs on R.H.S is removed by adding 1000 to it.

It would seem very easy once you get in this mode of calculations.

### Divisibility Testing

We all know divisibility tests for simple numbers like 2,3,4.etc but here I wish to demonstrate how to test for divisibility by 7, 11 and by higher prime numbers such as 13, 17, 19 and so on.

The first glance may force you to think that this method is complicated, but little time getting familiar with it, would make your life very simple.

For testing that a number is divisible by 11, add up two separate numbers. The first sum comprises the sum of the digits of the odd numbered columns of the number (1st, 3rd, 5th and so on). The second comprises the sum of the digits of the even numbered columns of the number (2nd, 4th, 6th and so on).

If the difference between these two numbers is zero, 11 or a multiple of 11, the number is divisible by 11.

Let us understand this with the example of number: 142,857.

Here, we now test 142,857 for divisibility by 11.

1 + 2 + 5 = 8

4 + 8 + 7 = 19

19 - 8 = 11

142,857 is divisible by 11.

So now we come to the divisibility test for the only number remaining, 7. When using this test, which can be applied to any other prime number, you only need to know the multiples of the candidate factor from zero and the factor itself through to five times the factor.

For example, in working out how to divide a number by 7, you only need to know 0, 7, 14, 21, 28 and 35. If you are testing for divisibility by 13, you only need to know 0, 13, 26, 39, 52 and 65.

The working behind this method is outlined below.

First of all, if a number a is a multiple of x:-

a = n1x

then any multiple b of a is also going to be a multiple of x:-

b = n2a ==> b = n2n1x

Next, any integer above 1 can be shown to be the sum of two smaller integers:-

c = d + e (d < c; e < c; d > 0; e > 0).

If we know that one of these two right hand terms is a known integer multiple of the candidate factor, we can eliminate it and test for divisibility with the smaller term, For example, substitute b for d:-

c = n2a + e

e = c - n2a

If, when we test e, we discover that it too is a multiple of x:-

e = n3x

we can deduce that the whole number c must be divisible by x.

This is the test: by eliminating known multiples of the candidate factor, if we obtain a final residue of 0 or the candidate factor, the number is divisible by the candidate factor.

To test this, we once again turn back to 142,857 and test it to see if it is divisible by 7. We already know that 142,857 is divisible by 11.

We can begin by eliminating the easiest multiples of 7. When eliminating a multiple, do not be afraid to take big bites.

142,857 - 140,000 = 2,857

2,857 - 2,800 = 57

Since the nearest multiple of 7 is now 56, eliminating that yields the final residue, 1. So 142,857 is not divisible by 7.

Let us test for divisibility by another prime factor, namely 17.

142,857 - 17 = 142,840

14,284 - 34 (2x17) = 14,250

1,425 + 85 (5x17) = 1,510

151 - 51 (3x17) = 100

100 - 85 = 15

Since the residue is neither 0 nor 17, 142,857 is not divisible by 17.

This technique can be applied to divisibility tests by any prime factor, and all it requires is the knowledge of the multiples of the candidate factor from 0 to 5x the factor.

Before wrapping up this article, I will conclude my business with 142,857. In my analysis of the candidate factors which go into 142,857 I determined that this number has the following factors, and only these factors:=

142,857 = 3 x 3 x 3 x 11 x 481

142,857 is an interesting number with connections to the number 7, which we shall discuss in a later post. However, 142,857 itself is not divisible by 7.

This rules are based on simple mathematical rules extrapolated to come up with a solution easier and practical for day to day issues. I hope you enjoyed knowing this!

**The Rule of 11***Let us first try out divisibility by 11 rule.*The first glance may force you to think that this method is complicated, but little time getting familiar with it, would make your life very simple.

For testing that a number is divisible by 11, add up two separate numbers. The first sum comprises the sum of the digits of the odd numbered columns of the number (1st, 3rd, 5th and so on). The second comprises the sum of the digits of the even numbered columns of the number (2nd, 4th, 6th and so on).

If the difference between these two numbers is zero, 11 or a multiple of 11, the number is divisible by 11.

*Example:*Let us understand this with the example of number: 142,857.

Here, we now test 142,857 for divisibility by 11.

1 + 2 + 5 = 8

4 + 8 + 7 = 19

19 - 8 = 11

142,857 is divisible by 11.

**The Rule of 7**So now we come to the divisibility test for the only number remaining, 7. When using this test, which can be applied to any other prime number, you only need to know the multiples of the candidate factor from zero and the factor itself through to five times the factor.

For example, in working out how to divide a number by 7, you only need to know 0, 7, 14, 21, 28 and 35. If you are testing for divisibility by 13, you only need to know 0, 13, 26, 39, 52 and 65.

*Method:*The working behind this method is outlined below.

First of all, if a number a is a multiple of x:-

a = n1x

then any multiple b of a is also going to be a multiple of x:-

b = n2a ==> b = n2n1x

Next, any integer above 1 can be shown to be the sum of two smaller integers:-

c = d + e (d < c; e < c; d > 0; e > 0).

If we know that one of these two right hand terms is a known integer multiple of the candidate factor, we can eliminate it and test for divisibility with the smaller term, For example, substitute b for d:-

c = n2a + e

e = c - n2a

If, when we test e, we discover that it too is a multiple of x:-

e = n3x

we can deduce that the whole number c must be divisible by x.

This is the test: by eliminating known multiples of the candidate factor, if we obtain a final residue of 0 or the candidate factor, the number is divisible by the candidate factor.

To test this, we once again turn back to 142,857 and test it to see if it is divisible by 7. We already know that 142,857 is divisible by 11.

We can begin by eliminating the easiest multiples of 7. When eliminating a multiple, do not be afraid to take big bites.

142,857 - 140,000 = 2,857

2,857 - 2,800 = 57

Since the nearest multiple of 7 is now 56, eliminating that yields the final residue, 1. So 142,857 is not divisible by 7.

Let us test for divisibility by another prime factor, namely 17.

142,857 - 17 = 142,840

14,284 - 34 (2x17) = 14,250

1,425 + 85 (5x17) = 1,510

151 - 51 (3x17) = 100

100 - 85 = 15

Since the residue is neither 0 nor 17, 142,857 is not divisible by 17.

This technique can be applied to divisibility tests by any prime factor, and all it requires is the knowledge of the multiples of the candidate factor from 0 to 5x the factor.

Before wrapping up this article, I will conclude my business with 142,857. In my analysis of the candidate factors which go into 142,857 I determined that this number has the following factors, and only these factors:=

142,857 = 3 x 3 x 3 x 11 x 481

142,857 is an interesting number with connections to the number 7, which we shall discuss in a later post. However, 142,857 itself is not divisible by 7.

This rules are based on simple mathematical rules extrapolated to come up with a solution easier and practical for day to day issues. I hope you enjoyed knowing this!

### Finding Squares

You know the squares of 30, 40, 50, 60 etc.

but if you are required to calculate square of 31 or say 61 then you will scribble on paper and try to answer the question.

Can it be done mentally?

Some of you will say may be and some of you will say may not be.

But if I give you a formula then all of you will say, yes! it can be.

What is that formula….. The formula is simple and the application is simpler.

Say you know 60sq = 3600

Then 61sq will be given by the following

61sq = 60sq + (60 + 61) = 3600 + 121 = 3721

or Say you know 25sq = 625 then 26sq = 625 + (25 + 26) = 676

Like above, you can find out square of a number that is one less than the number whose square is known.

but if you are required to calculate square of 31 or say 61 then you will scribble on paper and try to answer the question.

Can it be done mentally?

Some of you will say may be and some of you will say may not be.

But if I give you a formula then all of you will say, yes! it can be.

What is that formula….. The formula is simple and the application is simpler.

Say you know 60sq = 3600

Then 61sq will be given by the following

61sq = 60sq + (60 + 61) = 3600 + 121 = 3721

or Say you know 25sq = 625 then 26sq = 625 + (25 + 26) = 676

Like above, you can find out square of a number that is one less than the number whose square is known.

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