Forming Greatest and Smallest Number with Repetitive Digits.

FORMING GREATEST AND SMALLEST NUMBER WITH REPETITIVE DIGITS:

DIGITS REPETITING  TWICE IN THE NUMBER :

(a) Forming Greatest Number:

In Forming a greatest number with the Repetitive digits, we need to check the given digits: we need to follow three steps.

(i) The number of digits given, by which the greatest place value occupied by the greatest digit can be found.

(ii)  The digits are written in Descending order.

(iii) The greater digit is written twice ( two times ) : first in the greatest  place value and the same greater number is written in the next greater place value.

The number other than greater number is written in next corresponding place values.By writing the greater number twice, one of the digit is left out from the given digits.we can simply leave out that remained number.

The greater digit in the given digits occupies greatest place value twice ,the next greater digit occupies the next place value and so on.

1) Digits: 1 , 3, 7 , 8 .

Answer:

Step - 1: The  number of digits = 4 ;

 So the greatest place value is THOUSANDS (Th).

Thousands     (Th)

Hundreds     (H)

Tens   (T)

Ones   (O)

 

 Step-2 : The Descending order of the given digits: 8 , 7 , 3, 1.

                    8 > 7 > 3 > 1 .

 Step -3: The greater number in the given digit is : 8 ; which is repeated twice in the Place value System.

i.e; digit  8 is written in Thousands (Th) and the next greater place is Hundreds (H).The next greater digits are 7 and 3 , which are put in Tens(T) and Ones(O) place value. 

The remaining digit '1 '  can leaved out.No need to represent in place value.

Thousands     (Th)	Hundreds     (H)	Tens   (T)	Ones   (O)
8	8	7	3

The greatest number formed with the given digits ,repeating twice.

 8873.

 

2) Digits : 6 , 4 , 3 , 9 , 5 , 1.

Answer:

Step – 1:  The number of digits = 6   

So , the greatest place value is Lakhs(L).

Lakhs    (L)

TenThousands       (TTh)

Thousands     (Th)

Hundreds     (H)

Tens   (T)

Ones   (O)

 Step – 2 : The Descending order of the digits : 9 > 6 > 5 > 4 > 3 > 1.

Step – 3: The digit 9, which is greater need to put in two place values. i.e; Lakhs(L) , TenThousands(TTh) ,  the next greater numbers are placed in next place values.

We leave out the remained digit '1' from the given digits.

Lakhs    (L)	TenThousands       (TTh)	Thousands     (Th)	Hundreds     (H)	Tens   (T)	Ones   (O)
9	9	6	5	4	3

 Thus the greatest number formed is :  9,96,543.

 

Note: If the greater digit  to be repeated Thrice, then the greater is digit is to put in First three greater place values ,followed by the next greater numbers in the corresponding Place values.

  

(b) Forming Smallest Number:

In Forming a smallest number with the given digits, we need to check the given digits:

(i) The number of digits given, by which the greatest place value occupied by the smallest digit can be found.

(ii)  The digits are written in Ascending order.

(iii) The Smallest  number is to put in the greatest place value.The same greater value is to be put in the next greater place value.The next smaller digits is to be put in the next corresponding place values.

If  digit Zero ‘0’ is found in the given digits, then it should not put in the greatest place value,as the digit ‘0 ‘ in the left most of the number,the number  has no value.

So, we have to find the next greater digit than Zero i.e; next smallest number and is to be put first in the greatest place value . Then digit Zero is to put in the next place values two times .The next greater digits occupies the next corresponding place values.     

 

1) Digits: 5 , 8 ,3 , 0

Answer:

Step – 1 :  The number of digits = 4 .

 So, the greatest place value is Thousands(Th).

 

Thousands     (Th)

Hundreds     (H)

Tens   (T)

Ones   (O)

Step-2: The Ascending order of the digits: 0 < 3 < 5 < 8 .

Step – 3: Here, digit 0 is their, so we need to check the next smaller number other than digit ‘0’. i.e;  3.

The smaller digit 3 is to put in the greatest place value i.e : in Thousands place.Then Digit ‘ 0 ‘ is written in  the next two place values.i.e; Hundreds(H) and Tens(T) place value. The next smaller digit is 5 ,is written in Ones(O) place value.

We leave out the remained digit '8'.

Thousands     (Th)	Hundreds     (H)	Tens   (T)	Ones   (O)
3	0	0	5

Thus the Smallest number formed : 3,005.

 

2) Digits: 6 , 3 , 8.

Answer:

Step – 1: The number of digits = 3.

So , the greatest place value is Hundreds(H) place.

Hundreds  (H)

Tens  (T)

Ones (O)

 Step – 2 : The Ascending order of the digits: 3 < 6 < 8 .

Step – 3 : Here, the smallest digit is ‘ 3 ‘ , which is repeated twice in place values.i.e; Hundreds(H) and Tens(T).we leave out the remained Digit ' 8' . 

Hundreds  (H)	Tens  (T)	Ones (O)
3	3	6

 Thus the smallest number is :  336.

The Table shows Greatest and Smallest number with digits repeating twice in the number formed:

S.No	DIGITS	SMALLEST NUMBER	GREATEST NUMBER
1	3,9,2,1	1,123	9,932
2	6,7,8,3	3,367	8,876
3	5,0,3,1,9	10,035	99,531
4	8,0,3,6,4	30,046	88,643

Factorize using Cubes of Binomial Identity

FACTORIZE USING CUBES OF BINOMIAL IDENTITY:

1. Cube of a Binomial:

 (a + b)³  = a³ + 3a²b + 3ab² + b³.

 ( a – b )³ = a³ – 3a²b + 3ab² – b^3.

2. Sum of Cubes:

a³ + b³  = (a + b) ( a²   - ab + b² ) 

              = (a + b)³ – 3ab( a + b )

3. Difference of Cubes:

a³ – b³ = ( a – b )(a² + ab + b² ).

            = ( a – b )³ + 3ab ( a – b )

 

Factorize the following:

 1)  (2x – 2/x)³

Answer:

 The given expression: (2x – 2/x)³ ; is in the form of : ( a – b )³  .

Here,  a = 2x ; b = 2/x .

Substituting a , b  values in the Cube of Binomial Identity :

 ( a – b )³ = a³ – 3a²b + 3ab² – b^3.

= (2x – 2/x)³

=  (2x)³ – 3 * (2x)² * (2/x) + 3 * (2x) * (2/x)² – (2/x)³

=  8x³ – 3 * 4x^{2 *} 2/x +  3* 2x * 4/x² – 8 / x³

=  8x³ – 24x + 24/x – 8/x³ .

 

2) 2x³ – 54

Answer:

In  the given expression: 2x³ – 54; if we take out number ‘2’ as common , the expression changes in to : 2 ( x³ – 27 ) = 2 ( x³ – 3³  ) as we know 27 = 3³  and the new expression is in the form of : Difference of Cubes.

 The Difference of Cubes Identity :  a³ – b³ = ( a – b )(a² + ab + b² ).

 Here, a = x ; b = 3 .

 a²  = x² ; b² = 3² = 9 ; ab = 3x .

Therefore, 2 ( x³ – 3³  )

= 2 ( x – 3 ) ( x² + 3x + 3² ).

= 2 (x-3)( x² + 3x + 9 ).

    ( OR )

Substituting a,b values in  :( a³ –  b³ ) =  ( a – b )³ + 3ab ( a – b )

Therefore , 2 ( x³ – 3³  ) becomes

= 2 (( x – 3 )³ + 3 * 3x ( x- 3 ) ).

3) If  x =1  + √ 2; then find ( x – 1/x )³ ?

Answer:

Here,  we need to find ( x – 1/x )³  ,  which is in the form of : ( a – b )³ .

First we have to solve the given expression :

 ( x – 1/x )³  = (  ( x-1) / x)³

Substituting the given ‘ x’value in the expression:

= ( ( 1 + √ 2 - 1)  / (1  + √ 2) )³

= ( √ 2 /(1  + √ 2)³ ) ; the denominator is in the form of  Cube of Binomial Identity: 

( a + b )³ = a³ + 3a²b + 3ab² + b³.

Here,(1  + √ 2)³ expression: a = 1 ; b= √ 2

= 1³ + 3* 1²* √ 2 + 3* 1 * (√ 2)² + (√ 2)³

= 1 + 3√ 2 + 6 + 2 √ 2 

= (5 √ 2 + 7 )  ; substituting in ( √ 2 /(1  + √ 2)³ ) , then

 ( √ 2 / (1  +√ 2 )³ ) = ( √ 2 /( 5√ 2 + 7); multiplying numerator and denominator with 

(5√ 2 -7 ),i.e: rationalizing. Hence :

= ( √ 2 * (5 √ 2 - 7) / (5 √ 2 + 7) *( 5 √ 2 -7) ); denominator is in the form of Difference of squares Identity : (a+b)(a-b) = a² – b² .

= ( (10 - 7√ 2 ) / ( 50 – 49)

=  ( (10 - 7√ 2) / 1 )

=  10 - 7√ 2.

Factorizing using Product of Two Binomials Identity

Product of Two Binomials Identity:

( x + a)( x + b ) = x² + (a + b) x + ab. 

Factorize the following:

 1) p² + 10pq + 21   

Answer:

 In the given expression : p² q² + 10pq + 21, the number of terms are three.

1^st term is a Perfect square, where as the 3^rd term is not a Perfect square. Here,we have to use Product of Two Binomials Identity.

x² + (a + b) x + ab = ( x + a)( x + b ) .

In this , we need to find  common factors : N_1,  N₂

p² q²  * 21 = 21 p² q² ,the multipliers of  21 p² q²  are 7pq and 3 pq .

Therefore, the factors  N₁ ,  N_{2 =}  7 pq , 3 pq.

The middle term is N₁  + N₂ =  7pq + 3pq = 10 pq.

So, the expression :  p² q² + 10pq + 21   is written as in terms of their product of factors.

 = p² q² + 7pq + 3pq + 21

 =  ( pq + 7 ) ( pq + 3).     

                            (OR)

 p² q² + 10pq + 21    

Answer:

The expression is not a perfect square.To make it into Perfect square expression ,we need to add 4  and subtract 4 to keep the expression unchanged .

Therefore,the expression changes into:

 p² q² + 10pq + 21 + 4 – 4

=  (p² q² + 10pq + 25 ) – 4

= (  (pq)² + 2 * pq * 5 + (5)² ) – 4    

( (pq)² + 2 * pq * 5 + (5)² ;it is in the form of  Squares of Binomials Identity: 

a² + 2 ab + b² = (a+b)²  ;  Therefore,

= ( pq + 5 )² – 2²     

It is in the form of  Differrence of Squares Identity :

 a²  - b² = ( a+b)(a - b)

Here, a = pq+5  and  b = 2.

Therefore; ( pq + 5 )² – 2²  =  ( pq +5 +2 ) ( pq + 5 – 2 )    

                                              = ( pq + 7 ) ( pq + 3 ).

 

         

2) u² + uw + uv + vw

Answer:

The expression: u² + uw + uv + vw  is given in terms of their factors.

By using the Product of Two Binomials Identity: 

x² + (a + b) x + ab = ( x + a)( x + b ) .

 we need to factorize this expression.

The common Factor is taken out from the expression: u is common in first two terms and v is common in last two terms.Therefore;

= u² + uw + uv + vw

= u ( u + w) + v( u + w )

= (u + w ) ( u + v) .