Least common multiplier, Greatest common Divider

Least common multiplier [LCM]

           The Least common multiplier [LCM] calculated for two or more numbers for small applications like food bags in particular specified area, arrangement of the students  in the playground area in the form of columns and rows etc.

            The LCM of any Two numbers can be found by dividing the numbers with least prime numbers and taking one time the common prime factors and product all prime factors of the numbers. in this case divisibility Technics are useful.

Eg: LCM of 3 and 4

For 3 and four there are no common prime factors

Hence the LCM=product of the numbers=3x4=12.

Conclusion: If there is no common factor ,the LCM will be product of the numbers

E.g 2] Find the LCM of 2, 4

2	2,        4
2	1,        2
	1,        1

Take 1^st prime number 2

LCM=2x2=4

The least common multiplier=4

E.g 2] Find the LCM of  3, 9, 18

2	3,        9,    18
3	3,        9,     9
3	1,        3.      3.
	1,        1,      1

Take 1^st prime number 2

Not divisible 2 so take 3

LCM of the numbers=2x3x3=18

Conclusion: If all numbers given are factor of the one big number among the numbers, the bigger number will be LCM.

E.g3: Find the LCM of 6,9 and 15

2	6,       9,       15
3	3,       9,        15
3	1,       3,         5
5	1,       1,          5
	1,       1,          1

Take the 1^st prime number

2 is not disable hence take next prime number 3

Do the same process until we get 1 in last row.

LCM=2x3x3x5=90

The least common multiplier=90

I.e the numbers 6, 9, 15 are divides the LCM 90 commonly with least quotients.

Check it   by 6       90/6=15

                By 9       90/9=10

                By 15     90/15=6

  Conclusion : The quotients are not having common divider.

• If there is no common factor among the given numbers, the LCM will be the product of the numbers.
• If one big number is divisible by remaining numbers given, the bigger number will be LCM.

Will be continue with some practical problems and solutions

Greatest Common Divider[GCD]

Greatest Common Divider [GCD] is calculated some practical solution like easy measurement of largest quantities with the greatest common measurement unit.

Eg. 1] GCD of two numbers    15, 6

This can be calculated by dividing the bigger number with smaller number and further dividing the smaller number with the remainder by dividing bigger number dropdown continuously.

 Take the 6 as divider                   6] 15 [2

                                                           12      

Drop down 6                                       3] 6 [2

 Divide with previous reminder 3            6

                                                                0

The reminder 0, hence the GCD=3

Calculate the LCM of 6,15 we get 30

The product of the numbers=6x15=90

The product of the LCM and GCD=30x3=90

Conclusion: The product of the given numbers =The product of the LCM and GCD of the numbers.

Examples
1] How many minimum number of the students can arrange in the form of 26 rows or 91 rows?

Sol: the number of rows of soldiers= 26 or 91

The minimum number of students can be arranged in 26 or 91 rows=LCM of 26, 91
                                                                           

2	26,    91
7	13,    91
13	13,   13
	1,       1

The LCM of the 26 and 91=2x7x13=182

The minimum students required for arrange as 26 or 91 rows in the play ground=182.

Example2] Find the least number that can be divisible by 12, 18, 28

            The numbers are 12,18, and 28

the least number that can be divisible by 12, 18, 28

2	12,     18,       28
2	6,       9,        14
3	3,       9,         7
3	1,       3,          7
7	1,       1,          7
	1,       1,           1

=The LCM of the numbers

The LCM of the numbers=2x2x3x3x7

                                        =252

The least number that can be exactly divisible by

12, 18 and 28=252

Example 3] Find the minimum number if the number divided by 7, 11 and 13, the reminder will be 5.

The numbers are 7, 11, and 13

The numbers are prime numbers i.e no common factor except 1

So LCM=Product of the numbers

          = 7x11x13=1001

For obtaining reminder as 5, add 5 to LCM.

Therefore the number=1001+5=1006.

Fun: Product of any three digit number and 7x11x13 gives six digit number that repeats the same number in the next three places. E.g 547x7x11x13=547547

Example 4 ] Find the minimum number if the number divided by 7, 11 and 13, the reminders will be 4, 8 and 10 respectively.

The numbers are 7, 11, and 13

The numbers are prime numbers i.e no common factor except 1

So  LCM of 7, 11, 13 [prime numbers]=Product of the numbers

          = 7x11x13=1001
The common difference=7-4=11-8=13-10=3

Therefore the number=1001-3=998

Check: 7]998[142

                 7

                 29

                 28

                  018

                    14

Reminder : 04

          

            11]998[90

                 99

                 08

                   0

Reminder:08

           13]998[76

                91

                 88

                 78

Reminder:10

The reminders are 4,8 and 10,so the solution justified.

Mathematical magic squares

Mathematical magic squares

The Mathematical magic squares can be easily made with any sequence of the arithmetic progression like numbers forms squares.

e.g the numbers 1 to 9 can be arranged as square which gives the sum of any row or column or cross line gives same.

i.e the total numbers[n]=9

The sum n number of natural numbers=n[n+1]

                                                                     2

                                                            =9[9+1]

                                                                   2

                                                             =9x10

                                                                  2

                                                            =9x5=45

The sum of the each row/column/ cross line=45/3=15

Please observe the process as below

 Finally check the each row or column or cross line total

Rows

         4+9+2=15

         3+5+7=15

         8+1+6=15

Columns

         4+3+8=15

         9+5+1=15

         2+7+6=15

Cross lines

         4+5+6=15

         2+5+8=15

The magic squares gives fun and creates interest on mathematics to children and improve the knowledge about numbers.

As above any arithmetic progression numbers can written as magic squares

e.g: arithmetic progression: 1, 3, 5, 7,…………17

Sum of the ‘n’numbers [i.e sum of odd numbers]=n²=9²=81

The sum of the each row or column or cross line=81/3=27

Check the sum of each line i.e row or column or cross line

Sum of 1^st row numbers=7+17+3=27
      Sum of 2^nd row numbers=5+9+13=27

            Sum of 3^rd row numbers=15+1+11=27

             Sum of 1^st column numbers=7+5+15=27

     Sum of 2^nd column numbers=17+9+1=27

Sum of 3^rd column numbers=3+13+11=27

Sum of 1^st cross line numbers=7+9+11=27

          Sum of 2nd cross line numbers=3+9+15=27

Easy Solution of  quadratic equation ax² +bx+c=0

                                 ax² +bx+c=0
First we should understand that the equation is not solvable, as the
 unknown terms are two i.e  x and  x^2. so we can't solve the 

equation until the equation can be written as single unknown 

variable like [kx+l]² =m

^{So let us try.}
                                                  ax² +bx+c  =0

                            =>ax² +bx       =-c

                            =>[√ax]² +bx =-c              let A=√ax

Try to write as A²+2AB on RHS without violating  the equation

 let 2AB =bx =>2[√ax]B=bx 

 =>B=bx/[2√ax]

:.B=b/[2√a]

                =>[√ax]² +2√ax [b/2√a]=-c i.e in the form of A²+2AB

                    For getting A²+2AB+B² add B² on both sides

               =>[√ax]² +2√ax [b/2√a]+[b/2√a]²=-c+[b/2√a]²

As A²+2AB+B²=[A+B]²

                   =>[√ax+b/2√a]²=-c+[b/2√a]²     [b/2√a]²=b²/4a

               =>[√ax+b/2√a]²=-c+b²/4a         -c+b²/4a=[b²-4ac]/4a

            

              => √ax+b/2√a= +√[[b²-4ac]/4a]

                 
                    =>√ax             =-b/2√a+√[[b²-4ac]/4a]
           
             =>x              =[-b/2√a+√[[b²-4ac]/4a]]/√a

            =>x              =[-b/2a+√[[b²-4ac]/4a²]

                           :.  x=[-b+√[[b²-4ac]]/2a 

  Solution of  quadratic equation ax² +bx+c=0 is

 x=[-b+√[[b²-4ac]]/2a the values of the x are two i.e 

 x=[-b+√[[b²-4ac]]/2a or  x=[-b-√[[b²-4ac]]/2a
                             

^{Properties of the roots of the quadratic equation} 

1] If   b²-4ac<0, ^{the roots of the quadratic equation are imaginary or complex numbers and unequal.}

2] If   b²-4ac=0, ^{the roots of the quadratic equation are real and      equal.}

3]If   b²-4ac>0, <sup>the roots of the quadratic equation are real and unequal.

Some of the solved examples.</sup>

^{1] Find the roots of}  2x² +5x+6 =0

Assume the equation  2x² +5x+6 =0 as  ax² +bx+c  =0

=>a=2, b=5, c=6

Solution of  quadratic equation ax² +bx+c=0 is

                           x=[-b+√[[b²-4ac]]/2a 

                           x=[-5+√[[5²-4x2x6]]/2x2

                           x=[-5+√[[25-48]]/4

                           x=[-5+√[[-23]]/4

                           x=-5 + √-23
                                 4      4

^{2] Find the roots of}  x² -8x+16 =0

Assume the equation   x² -8x+16 =0 as  ax² +bx+c  =0

=>a=1, b=-8, c=16

Solution of  quadratic equation ax² +bx+c=0 is

                           x=[-b+√[[b²-4ac]]/2a 

                           x=[-[-8]+√[[[-8]²-4x1x16]]/2x1

                           x=[8+√[[64-64]]/2

                           x=[8+√0]/2

                           x= 8
                                2

                           x=4
                                 
       Here the determinant b²-4ac=0, so the roots are real and equal

3]^{Find the roots of}  x²-5x+6 =0

Assume the equation x²-5x+6 =0 as  ax² +bx+c  =0

=>a=1, b=-5, c=6

Solution of  quadratic equation ax² +bx+c=0 is

                           x=[-b+√[[b²-4ac]]/2a 

                           x=[-[-5]+√[[[-5]²-4x1x6]]/2x1

                           x=[5+√[[25-24]]/2

                           x=[5+√1]/2

                            x=[5+1]/2

                           x=[5+1]/2 or [5-1]/2

                           x=6/2 or 4/2

                           x=3 or 2

 Here the determinant b²-4ac>0, so the roots are real and unequal [b²-4ac=1].

Simple tips for divisibility Technics

                                                  Checking of the Divisibility of number by the smallest prime numbers is facilitate the prime factoring any number and further uses to find out square roots and cube roots of the numbers easily.These are use full for upper class mathematics in geometry, mensuration etc.

1. Divisible by 2

                               A number can be divisible by 2, if the number is even number or the number has            even integer in it,s 1's place.
         e.g: 8546794 is divisible by 2 as 4 in the 1's place is divisible by 2,hence the entire number is              divisible by 2 exactly.

           Divisible by4
                                 a number can be divisible by 4,if the number has minimum of  first two digits                forming a number can be divisible by 4 [4, 8 are also divisible by 4], the entire number can be            divisible by 4 exactly [i.e with reminder 0] This Technic is used for checking the number                    having the number of digits>2
         E.g: let us check the number 5236 is divisible by 4 or not
        simply check the first 2 digit number 36
        36/4=9 remainder=0 hence the total number can be divisible by 4
         Justification 5236/4=1309 with reminder=0

1. Divisible by8
2.                        a number can be divisible by 4,if the number has minimum of  first three digits forming a number can be divisible by 8 [the numbers having number of places<3 can be checked by direct method], the entire number can be divisible by 8 exactly [i.e with reminder 0] This Technic is used for checking the number having the number of places>3 only.
3. Eg: let us check the number 37952128 is divisible by 8 or not
           simply check the first 3 digit number 128
   128/8=16 remainder=0 hence the total number can be divisible by 8
    Justification 37953128/8=4744141 with reminder=0

1. Divisible by3

                             A number can be divisible 3, if the total digits of the number is divisible by 3. If the           sum of the digits is more than 1 places can be added the digits further to single digit such that             the number can check easily. The remainder become the same as the reminder of the main                   number divided by 3.

        e.g number=825

       the sum of the digits=8+2+5=15, further 1+5=6,6 is divisible by 3 so the number 825 is divisible       by 3

         i.g number=1735, 

         the sum of the digits=1+7+3+5=16 further 1+6=7 is can't be divisible by 3, as the remainder=1

         the reminder is same as the reminder by main divided by 3 

• Divisible by9    
• A number can be divisible 9, if the total digits of the number is divisible by 9. If the sum of the digits is more than 1 places can be added the digits further to single digit such that the number can check easily. The remainder become the same as the reminder of the main number divided by 9.

1. e.g number=825

   the sum of the digits=8+2+5=15, further 1+5=6, 6 is not divisible by 9 so the number 825 is not divisible by 9. the reminder is 6.
   e.g number=173511 
   the sum of the digits=1+7+3+5+1+1=18 further 1+8=9, 9 can be divisible by 9,

           Divisible by 5 

          A number cab be divisible by 5, if the number has 5 or 0 in it,s 1,s place.

        e.g 5, 15, 254589731525, 125462189510 are divisible by 5.

           Divisible by 10

                                  A number cab be divisible by 10, if the number has  0 in it,s 1,s place.

        e.g 10, 150, 2545897315250, 125462189510 are divisible by 10

           Divisible by 7

                              A number cab be divisible by 7. If a number has the digits it,s places, such that the            difference between twice of the 1's place or 1st digit to the number formed remaining digits                should be divisible by 7.

           e.g is 63 divisible by 7 or not

           difference between twice the 1's place number to remaining number=2x3-6=0,0 can be                        divisible 7 hence the number can be divisible by 7

           e.g: is the number 51513 divisible by 7

           step1: 5151-2x3=5145

           step2; 514-2x5=504

           Step3: 50-2x4=42

           step4:  4-2x2=0, 0 can be divisible by 7, so the entire  number can be divisible by 7.

           If the number has more places the same process can be repeated until getting single digit.

           Divisible by 11

                                 A number can be divisible by 11, if the number has digits such that the                           difference between sums of the alternate digits is divisible by 11.

          E.g: 1059531

          1st. sum of Alternate numbers=1+5+5+1=12

          2nd sum of Alternate numbers=0+9+3=12

          Difference= 12-12=0, 0 is divisible by 11 so the entire number divisible by 11

Some example problems for competitive examinations: 

1]  Find the missing digit in the number 1?1 if the number is exactly divisible by 3
     let ?=x
     The sum of the digits in the number should be divisible by 3
     Sum of the digits=1+x+1
                           =x+2
The least number >2 can be divisible by 3 is 3
                           =>=x+2=3
                              x=3-2=1
So the missing number=1

2] Find the missing digit in the number 53?2 if the number is exactly divisible by 3
    let ?=x
    The sum of the digits in the number should be divisible by 3
     Sum of the digits=5+3+x+2
                           =x+10
   The least number >10 can be divisible by 3 is 12.
                           =>x+10=12
                                x=12-10=2
   So the missing number=2

3] Find the missing digit in the number 53?2 if the number is exactly divisible by 9
    let ?=x
    The sum of the digits in the number should be divisible by 3
     Sum of the digits=5+3+x+2
                           =x+10
  The least number >10 can be divisible by 9 is 18.
                           =>x+10=18
                                x=18-10=8
   So the missing number=8
 or again add the digits obtained in the total 10=1+0=1
  The least number >10 can be divisible by 9 is 9.
so                      x+1=9
                      => x=9-1=8.
the missing number=8

4] Find the missing digit in the number 53?2 if the number is exactly divisible by 11
    let ?=x
The difference among sum of  alternate  digits must be divisible by 11
1st alternate sum of digits=5+x
2nd alternate sum of digits=3+2=5
difference=5+x-5=x
The least  number can be divisible by 11=0
Hence the difference may equal to zero
hence x=0
So the missing number=0
Finally the number=5302.