HOW TO KNOW THE DIVISIBILITY TEST OF A NUMBER || DIVISIBILITY RULES FOR NUMBERS

How to know whether a given number, however large is divisible by 2, divisible by 3,4,5,6 and 10 .There are fixed divisibility test and divisibility rules for checking the divisibility of any given numbers. For different numbers there are  different rules to divide with . So in this post  we are going to discuss these divisibility rules with examples  one by one. Although there are divisibility test  for fraction also ,but this post will be restricted to natural numbers.

Divisibility rules for 2

To check whether the given number, however large  is divisible by 2 ,we have to check its right most digit/Unit place digit, if it is even number or zero.Then the given number is definitely divisible by 2

For Example

12 is divisible by 2 as its right most digit is 2 which is Even .

3548 is divisible by 2 as its right most digit is 8 which is Even 

999998 is divisible by 2 as its right most digit is 8 which is Even .

65989564 is divisible by 2 as its right most digit is 4 which is Even. 

22222229 is not divisible by 2 as its right most digit is 9 which is not A Even number.

589423100780      is divisible by 2 as its right most digit is 0 .

357913579571536   is divisible by 2 as its right most digit is 6 which is Even although all the remaining digits are odd.

Divisibility rules for 3

To check whether the given number, however large  is divisible by 3,we have to check the SUM of all its digits, if its sum is divisible by 3 then the  given number is divisible by 3, If sum of  all the digits of given number is again a large number then add the result so obtained and apply the rule again which is said earlier.

For Example

15 is divisible by 3 as sum of all its digits 1 + 5 = 6 is divisible by 3 .

833 is not divisible by 3 as sum of all its digits 8 + 3 + 3 = 14 = 1 + 4 = 5 is not divisible by 3 .

2678 is not divisible by 3 as sum of all its digits 2 + 6 + 7 + 8 = 23 = 3 + 3 = 5 is not divisible by3 .

98784552 is not divisible by 3 as sum of all its digits 9 + 8+ 7+ 8+ 4 + 5 + 5 + 2 = 48 = 4 + 8 = 12 = 1 + 2 = 3 is not divisible by 3.

359875269 is divisible by 3 as sum of all its digits 3 + 5 + 9 +8 +7 + 5 + 2 + 6 + 9 = 54 = 5 + 4 = 9 is divisible by 3.

3597841137 is divisible by 3 as sum of all its digits 3 + 5 + 9 + 7 + 8 + 4 + 1 + 1 + 3 + 7 = 48 = 12   is divisible by 3 .

Divisibility rules for 4

If the last two digits of a number is divisible by 4 ,Then the number is divisible by 4 . The number having two or more zeros at the end is also divisible by 4.

For Example

568928 : Here in this number last two digits are 28 ,which are divisible by 4,Hence the given number is divisible by 4.

134826900 : As ther are two zeros at the end,so the given number is divisible by 4.

13444255452 : As the last two digits number (52) is divisible by 4 ,the given number is divisible by 4.

35888875698549 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4

97971349999567776 : As the last two digits number (76) is divisible by 4 ,the given number is divisible by 4.

44444444444444444449 : As the last two digits number (49) is not divisible by 4 ,the given number is not divisible by 4.

Divisibility rules for 5

To check whether the given number, however large is divisible by 5 ,we have to check its right most digit/Unit place digit,if it is 5 or zero. Then the given number is definitely divisible by 5.I.e if the given number ends with 0 or 5 then it is divisible by by 5.

For Example

35 is divisible by 5 as its right most digit is 5.

97835 is divisible by 5 as its right most digit is 5.

6854940 is divisible by 5 as its right most digit is 0.

35000000355 is divisible by 5 as its right most digit is 5.

3579515465855 is divisible by 5 as its right most digit is 5.

12345678888880 is divisible by 5 as its right most digit is 0.

568954975311525 is divisible by 5 as its right most digit is 5.

55555555555556 is not divisible by 5 as its right most digit is 6.

66666666666666665 is divisible by 5 as its right most digit is 5.

Divisibility rule for 6

For any number to be divisible by 6 ,it must be divisible by both 2 and 3 ,then the given number is divisible by 6, Therefore
1) The number should ends up with an even digits or zero and
2) The sum of its digit should be divisible by 3.


For Example :

56898 is divisible by 6 as sum of its digits is 5 + 6 + 8 + 9 + 8 = 36 = 3 + 6 = 9 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

3578952  As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 +7 + 8 + 9 + 5 + 2 = 39 = 4 + 2 = 6  which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

25689879798 is divisible by 6 as sum of its digits is 2 + 5+ 6 + 8 + 9 + 8 + 7 + 9 + 7 + 9 + 8 =78 = 15 = 1 + 5 = 6 so it is divisible by 3 and last digit is even ,as the given number is divisible by both 2 and 3 ,so it is divisible by 6.

35658999962 is not divisible by 6 as sum of its digits is 3 + 5 + 6 +  5 + 8 + 9 + 9 + 9 + 9 + 6 + 2 = 71 = 7 + 1 = 8 so it is not divisible by 3 and last digit is not even ,so it is not divisible by 6.

35789248956 As the last digit is even so the given number is divisible by 2 and sum of all its digits is 3 + 5 + 7+ 8+ 9+ 2+ 4 + 8+ 9 + 5 + 6 = 66 =  6 + 6 = 12 = 1 + 2 = 3 which is also divisible by 3 ,which implies the given number is divisible by both 2 and 3. Therefore the given number is divisible by 6 as well.

Divisibility rule for 10

It is the most easiest number to identify whether it is divisible by 10 , if the given number however large ends up with 0 then it is divisible by 0.

For Example

4546546546540 ,44545454560, 445474456110 5555598959550 are divisible by 10 as all the given numbers ends with 0.

and 445645489,454545,456445555 ,454545577 and 545454 are not divisible by 10 as all these numbers do not ends up with 0.

Conclusion

Thanks for giving your valuable time to read this post of the divisibility rules and divisibility test rules for  divisibility rule for 2 , divisibility rule for 3,    divisibility rule of 4 etc . If  you liked this post  , Then share it with your friends and family members . You can also read my others articles on Mathematics Learning and understanding Maths in easy ways.

Let us learn Multiplication , division , arithmatic and simplification short cut ,tips and  tricks after buying this Book of Magical Mathematics .

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HOW TO MULTIPLY TWO DIFFERENT NUMBERS IN FASTEST AND QUICKEST WAYS

        
  Whenever we are to multiply two numbers then most of the time we undergo long calculations , and if we have to do easy calculation then we were lucky.

But what to do when we have to multiply two numbers in very short time. Suppose we have to multiply 32 with 11 then it is easy, because just put right most digit as it is and then increase every digit by one to left and at last put the left most digit as it is and put 352 as answer.

Example

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COMPLEX NUMBERS || MULTIPLICATION || MODULUS || INVERSE || SQUARE ROOTS

Let us discuss concept of number, complex numbers, rational,numbers, real numbers, prime numbers ,complex imaginary numbers, factors of a number, complex number , introduction to complex numbers , operations with complex numbers such as addition of complex numbers , subtraction, multiplying complex numbers, conjugate, modulus  polar form and their Square roots of the complex numbers and complex numbers questions and answers .

Complex Numbers

Introduction to complex numbers

Any number of the type a+ib where a,b are Real numbers and i =√ (-1) is called complex number. Where "a" is called real part and "b" is called imaginary  part . The set of complex number is denoted by  C . These numbers are also called  non real numbers .

Complex Numbers Examples  

3,7 + 3i , -2 + 9i , 3i - 2 , 9i,  3 -7i , 2 , 3i, 6 + √5 i , 6 - √5 i ,√5
Here the numbers 2, 3, √5 are called purely Real and the numbers 3i, 9i are called Imaginary numbers .

Algebra of complex numbers

Addition of complex numbers

If   Z1 = a1+ ib1 and  Z2 = a2+ib2  are two complex numbers then their addition can be calculated by adding their real parts and imaginary parts separately. 

 Z1 + Z2= (a1 + ib1 ) + (a2  + ib2 )
Z1 + Z2= (a1 + a2  ) + (ib1 + ib2 )

Z1 + Z2= (a1 + a2  ) + i(b1 + b2 )

Here   (a1 + a2  ) is called Real Part and   (b1 + b2 ) is called Imaginary Part of resultant .

Let Z1 = 3+5i  and Z2  = 9 - 6i then

Z1 + Z2  = (3 + 5i) + (9 - 6i)

Z1 + Z2 = (3 + 9) + (5i - 6i)  

Z1 + Z2  = 12 - i

Here  12  is its Real Part   -1 is its Imaginary Part

More Examples 

 $\displaystyle{\left({6}+{7}{j}\right)}+{\left({3}-{5}{j}\right)}=$(6+7i)
$Z1=6+7iand$ Z2 =3−5i
Z1  +  Z2  = (3

Z1  +  Z2   = (6+3)+(7−5)i

Z1  +  Z2 = 9+2i

If   Z1  +  Z2  = (12+6i) + (-4 - 5i) ,
Z1  +  Z2 = (12−4) + (6i−5i),
Z1  +  Z2= (12−4) + (6−5)i  ,
Z1  +  Z2 = 8+i

If   Z1 =  6+ √5 i and   Z2  = 7 - √4 i

then  Z1  +  Z2  =  ( 6 + √5 i ) + ( 7 - √4 i) 

          Z1  +  Z2 = (6 + 7) + (√5 i - √4 i) 

           Z1  +  Z2  = 13 - (√5 - √4)i

Subtracting Complex Numbers

If   Z1 = a1 + ib1 and  Z2 = a2 + ib2  are two complex numbers then their subtraction can be calculated by subtracting their real parts and imaginary parts from each  separately. 

 Z1 - Z2= (a1 +ib1 ) - (a2  + ib2 )

Z1 - Z2= (a1 - a2  ) + (ib1 - ib2 )

Z1 - Z2  = (a1 - a2  ) + i(b1 - b2 )

Here   (a1 - a2  ) is its Real Part  and   (b1 - b2 )  is its Imaginary Part.

Let Z1 = 8+5i  and Z2  = 4-6i then

Z1  - Z2  = (8+5i) - (4-6i) = (8-4) - (5i-6i)  = 4+i
Z1 - Z2  = (3+5i) - (9-6i) = (3+9) - (5i-6i)  = 12+i

Examples

 $\displaystyle{\left({6}+{7}{j}\right)}+{\left({3}-{5}{j}\right)}=$(6+7i)

$Z1=6+7iand$ Z2 = 3−5i

Z1  -  Z2  = (6+7i) - (3-5i)

Z1  -  Z2 = (6-3)+(7+5)i

Z1  -  Z2  =3+ 12i

If  Z1  = 12+6i ,  Z2  = -4-5i

   Z1  -  Z2 = (12+6i)-(-4−5i)
  Z1  -  Z2 = (12+4)+(6i+5i) =12+11i 

  Z1 -  Z2  = (12+4)+(6+5)i
 Z1  -  Z2 = 18+9i

Multiplying Complex Numbers

If   Z1 = a1 + ib1 and  Z2 = a2 + ib2  are two complex numbers then their Multiplication can be done as  follows

 Z1 Z2 = (a1 +ib1 ) (a2  + ib2 )

Z1  Z2 = a1  (a2  + ib2 )+ib1(a2  + ib2 )

Z1  Z2  = a1  a2  + i a1 b2 +ib1a2  + ib2 ib1  (because  i²  = −1)

Z1  Z2 = a1  a2  + i a1 b2 +ib1a2  +  i² b2 b1     

Z1  Z2 = (a1  a2  - b1b2)  +i(a1 b2 + a2b1 )

Examples

  If    Z1  = 2 + 6i ,  Z2  = 4 - 5i

 Z1Z2 = (2 + 6i)(4−5i)

Z1Z2 =2(4−5i)+ 6i(4−5i)

Z1Z2 = 8-10i+24i−30i²

Z1Z2 = 8-10i + 24i + 30  

Z1Z2 = (8 + 30)+ (-10i + 24i)

 Z1Z2 = (38)+ (14i)

Z1Z2 = 38 +14i

 If    Z1  = -2-i ,  Z2  = 4+5i

 Z1Z2 = (-2-i)(4+5i)
Z1Z2 = -2(4 + 5i) -i(4 + )

 Z1Z2  =  -8-10i-4i−5i²

 Z1Z2 =   -8-10i-4i+5

 Z1Z2 = (-8+5)+ (-10i-4i)

 Z1Z2 = (-3)+ (-14i)

Z1Z2 = -3 -14i

Now=(3+3i)(1-7i)
= 3×1 + 3×(-7i) + 3i×1+ 3i×(-7i)

 = 3 - 21i + 3i -21i²

 = 3 - 21i + 3i +21 (because i² = −1)

 = 24 - 18i

 How  to Simplify Huge Power  of Iota  

In order to reduce huge/big power of iota to smallest power . First divide the power if iota with 4 and then write first term as power of (4 × quotient) of iota and second term as power of remainder of iota as follows . we know that i⁴ = 1 , implies the contribution of 1st term reduces to "1" .

Note : In order to get quotient after division by 4  , it is sufficient to divide the last two digits of the given power by 4.

Therefore it is the second term which will contribute to answer.

Before we proceed further Remember these result   i² = -1 ,          i³ = -i ,       i⁴ = 1.

So these are some Simplification 

i⁴⁵⁰ =   i^{4×112 .} i²  .= 1× (-1) = -1

i⁴⁵¹ =   i^{4×112 .} i³  .= 1× (-i) = -i

 i⁴⁵² =   i^{4×113 .} i⁰  .= 1× 1 = 1

i⁴⁵³ =   i^{4×113 .} i¹  .= 1× i = i

 ^{i3523 =   i4×855} i³ = 1×(-i)= -i

 ^{i9998 =   i4×2449} i² = 1×(-1) = -1

 ^{i9997 =   i4×2449} i¹ = 1×(i) = i

^{i2221 =   i4×555} i¹ = 1×(i) =  i

^{i2222 =   i4×555} i² = 1×(-1) = -1

^{i2223 =   i4×555} i³ = 1×(-i) = -i

Complex Conjugate of complex Number 

If  z = + iy be any complex number then  x - iy is called its Conjugate.  Simply change the sign of   i   to -i.

Examples

Let Z= 5-7i then Conjugate of this complex number is 5+7i

Let  5 2 - 7 3i  be any complex number  then

 Conjugate of this complex number is 
5 2 + 7 3i  and
  Z= 2+ i be the complex number then Conjugate of this complex number is 2 - i

Modulus of a Complex Number

If  z = + iy be any complex number then  Positive square root of sum of the squares of its Real and Imaginary Parts is called its Conjugate.It is denoted by |Z|

Then 

e. g 

Let Z= -3+2i

then | Z | = √(${-3)}^2+({2)}^2$ )  = √ (13)

If Z = -2-i

Then 

     |Z|   $=$ √ ( (${-2)}^2+({-1)}^2$ )

     |Z|   = √5

If Z= 6 - 8i
Then |Z| = √( 36+64   )  
          |Z|= √ (100) 

          |Z|= 10

Division of two Complex Numbers

To divide Z1 With Z2 Multiply and divide the numerator and Denominator by the Conjugate of  Denominator and then write in a + ib form   after simplification.

Example

Divide  2+3i with 4+7i  

4 + 7i

Multiply Numerator And Denominator  by the conjugate of 4+7i 

2 +3i4 +7i×4 - 7i4 - 7i  =  8 -14i + 12i - 21i²16 + 49

=  29 - 2i65

Now change  into a +ib form

=  29 65 - i265

Example

Divide  3+4i with 2+3i

  3+4i2+3i

Multiply Numerator And Denominator by the conjugate of2+3i 

3+4i2+3i × 2-3i2-3i  =  6 -9i +8i -12i²4+9

=  18- i13

Now change  into a +i b form:

=  18 13 - i113

Multiplicative Inverse of complex Number

If Z = x + iy be any complex number then 1/Z is called it Multiplicative Inverse.

Example

If Z = 1 - ithen Z -1= 1/(1-i)
If  Z = 4+3i 
then Z -1= 1/(4+3i)




If  Z = 1-√ 2i
then Z -1= 1/(1-√ 2 i)

Polar Form of a Complex Number

To Convert  Z = x + iy complex number into Polar Form 

Put $x=r\cos \theta$  ....................(1)

$\backslash displaystyle\{y\}=\{r\}\backslash \; \backslash sin\{\backslash theta\}$

 Then Squaring and adding  (1) and (2)
$we\; get$


 
 
 $r^2=x^2+y^2$ and 

upon dividing  (2)equation  by (1) we get

$\displaystyle \tan{\ }\theta=\frac{y}{{x}}$$\displaystyle{x}={r}\ \cos{\theta}$

Then Z = r( cos θ + i sinθ ) is in Polar Form

Example

Convert the complex number 

7 − 5i    into Polar Form , We need to find r and θ.

Put $x=r\cos \theta$  

$\backslash displaystyle\{y\}=\{r\}\backslash \; \backslash sin\{\backslash theta\}$

\displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}
​

$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

$\displaystyle=\sqrt{{{49}+{25}}}$

$\displaystyle=\sqrt{{{74}}}\approx{8.6}$
8

To find θ, we first find the acute angle α 

which is equal to

$\displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

As the Points (7,-5) lies in fourth Quadrant in Argand Plane ,So   $\displaystyle{7}-{5}{j}$ will be  in the fourth quadrant, 

  θ= 360° - α  = 324.5°

So, expressing $\displaystyle{7}-{5}{j}$ in polar form

7-5i = 8.59 ( cos 324.5° + i sin 324.5° )

Example

Convert the complex number $\backslash displaystyle\{7\}-\{5\}\{j\}$

We need to find r and θ.

$\displaystyle{r}=\sqrt{{{}}}$

$\displaystyle=\sqrt{{{7}^{2}+{\left(-{5}\right)}^{2}}}$

       =    1+2=3

   r = √2

 

To find θ, we first find the acute angle α 

which is equal to

$\displaystyle={{\tan}^{{-{1}}}{\left(\frac{5}{{7}}\right)}}$

α = 45°

As the Points (1,-√2) lies in fourth Quadrant in Argand Plane ,So  1−√2i will be  in the fourth quadrant, 

So θ=360 - 45°

θ= 315°

So, expressing 1$-$√2i  in polar form


Z =  √2(cos 315° + i sin 315°) 

Also Watch   HOW TO CONVERT ANY COMPLEX NUMBER TO IT POLAR FORM 

Square root of complex number

Example 1

Find the square root of 8 + 6i

 Let        z² = (x + yi)²  = 8+6i   Squaring  Both Sides

          \    (x² – y²) + (2xy)i     = 8+6i

          Compare real parts and imaginary parts,we have 

       x² – y² = 8   .............................    (1)

        2xy = 6          .............. ...............  (2)

         

 Using   Identity

  ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

    ( x² + y² )²  = 8² + 6²

   ( x² + y² )²  =100

     x² + y² = √(8² + 6²) = 10                  

   x² + y² = 10    .............................. (3)

To find the values of  x and y 

 Adding   (1) and (3) 

 we get  x² – y² + x² + y² = 10 +8   

           2 x² = 18 

          x² = 9 

           x = ±3  

Put   x = 3  and x = -3 and    in equation (2)    

we get     y = 1 when x=3  

and          y = -1 when x= -3

From (2) we concluded that  x and y are of same  sign,

 (x = 3 and y = 1)    or   (x = -3 or y =-1)

Therefore required Square roots of   8+6i are

Z = 3+i and Z = -3-i

Example 2

Find the square root of -7+24i

 Let        z² = (x + yi)²  = -7+24i   Squaring  Both Sides

  (x² – y²) + (2xy)i     = -7+24i

  Compare real parts and imaginary parts,we have 

 x² – y² = -7  .............    (1)

  2xy = 24      ..............  (2)

         

 Using   Identity

   ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

   ( x² + y² )²  = (-7)² + (24)²

   ( x² + y² )²  = 49 + 576=625

    x² + y² = √(25)                  

   x² + y² = 25    ...................   (3)           

 To find the values of  x and y 

 Adding   (1) and (3) 

 we get         x² – y² + x² + y² = -7 +24   

 2 x² = 18 

  x² = 9 

 x = ±3  

Put   x = 3  and x = -3 and    in equation (2)    

we get     y = 4 when x=3  

and          y = -4when x= -3

  From (2) we concluded that  x and y are of same  sign,

   (x = 3 and y = 4)    or   (x = -3 or y = -4)

Therefore required Square roots of   8 + 6i are

Z= 3+4i and Z = -3-4i

Example 3

Find the square root of -15 - 8i

 Let        z² = (x + yi)²  = -15 - 8i

  Squaring  Both Sides

 (x² – y²) + (2xy)i     = -15 - 8i

 Compare real parts and imaginary parts,we have 

 x² – y² = -15   .............................    (1)

  2xy = -8         .............. ...............  (2)

         

 Using   Identity

  ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

  ( x² + y² )²  = -15² + -8²

  ( x² + y² )²  =225+64

   x² + y² = √289                   

 x² + y² = 17    ..............................   (3)           

 To find the values of  x and y 

 Adding   (1) and (3) 

 we get         x² – y² + x² + y² = -15 +17   

 2 x² = 2 

   x² = 1 

  x = ±1 

Put   x = 1  and x = -1 and    in equation (2)    

we get     y =  -4  when x = 1  

and          y = 4 when x = -1

  From (2) we concluded that  x and y are of same  sign,

  (x = 1 and y = -4)    or   (x = -1 or y = 4)

Therefore required Square roots of   -15 - 8i are

Z=1 - 4i  and Z  = -1+ 4i

Example 4

Find the square root of -3 + 4i

 Let        z² = (x + yi)²  =  -3 + 4i

  Squaring  Both Sides
  (x² – y²) + (2xy)i     = -3 + 4i

          Compare real parts and imaginary parts,we have 

                        x² – y² = -3   .............................    (1)

                        2xy =   4         .............. ...............  (2)

         

 Using   Identity

    ( x² + y² )²  = (   x² - y² )²  - 4 x²  y² 

      ( x² + y² )²  = -3² + 4²

       ( x² + y² )²  =9+4

        x² + y² = √25                  

    x² + y² = 5    ..............................   (3)           

 To find the values of  x and y 

  Adding   (1) and (3) 

 we get  x² – y² + x² + y² = -3 +5   

          2 x² = 2 

          x² = 1 

         x = ±1 

Put   x = 1  and x = -1 and    in equation (2)    

we get     y =  2  when x = 1  

and          y = -2 when x = -1

          From (2) we concluded that  x and y are of same  sign,

                  \ (x = 1 and y = 2)    or   (x = -1 or y =-2)

Therefore required Square roots of   -15 - 8i are

Z=1 +2i  and Z  = -1- 2i

HOW TO FIND THE SQUARE ROOTS OF ANY COMPLEX NUMBER IN HINDI

Conclusion

In this post I have discussed Complex number, rational numbers,
real numbers ,prime numbers, modulus ,inverse, polar form,square roots of complex numbers. If this post helped you little bit, then share it with your friends and like this post to boost me to do better, and follow this Blog .We shall meet in next post till then Bye .

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