Some multiplication [product] short cuts or tips:

Some multiplication [product] short cuts or tips:

1] multiplication of the numbers with 5 in ones place;

                                                            1000's and 100's places    10's 1's places

 A] Squares: eg: 1]      35X35=1225   [by 3x[3+1]                             5x5]]

                            2]     75X75=5625   [by 7x[7+1]                              5x5]]

*For n5xn5=n[n+1]25 [n[n+1]is in 1000’s and 100’s place +25 is in 10’s and one’s place]

B] the numbers with 5 in one’s place and differ by 10                                         

Eg:1]  35X45=1575

       2] 65x55=3575

*For [n-1]5xn5=[n²-1]   75

2]the numbers multiplied by 5

   E.g1] 256x5=1280 [i.e [256/2]x10

   E.g2] 445x5=2225 [i.e [445/2]x10+5]]

If an even number n multiplied by 5 the product is [n/2]x10

If odd number n multiplied by 5 the product is [[n-1]/2]x10+5

3] square of the numbers near to 10,100,100…etc.

                                      10,s   one’s

            E.g.1] 9x9=81 [9-1]   1     =81

                                     100’s  one’s 

            Eg. 2] 98x98= [[98-2] [100-98]x[100-98]=9604

i.e 98-[100-98] is in 1000’s and 100’s place further [100-98][100-98] is in 10’s place and one’s place.

 Same way 96x94=[96-[100-94] [100-94]x[100-96]]=9024.

3] product of the numbers near to 10,100,100…etc.

E.g1]                           [two places left to 10’s]          two places left from start

1000’s, 100’s                           10’s 1’s 

Eg. 98x97=9506         [98-[100-97]]                    [100-98]x[100-97] as the number of digits in every number is ‘’2’’

E.g2] Can you imagine the product of the two numbers 99994x99997

5places                                                5 places          

Simple: 99994-[1000000-99997]    [1000000-99994]x[100000-99997]=9999100018.

The number of digits in product=2xthe number of digits in each number.

So we have obtained the product by multiplying the each number differences  to 100000 is 6x3=18 it should be in first 5 digits so that is 00018. Next, the difference between the one of the numbers to Nearest Round number 100000-another number]

Product of any two numbers which are near to 50, 500, 5000 etc

Two places       Two places

E.g1: 47X48=[47-[50-48]]/2      [50-47]x[50-48] here the left side number should be multiplied by 50/100 i.e ½

                    =[47-2]/2                3x2

                    =45/2                       06

                    =22.5                       06

                                 =2256

The fraction 0.5 in 100,s place= direct number 5 in 10’s place [0.5x10=5]

Two places       Two places

Eg2] 44x46= [44-[50-46]]/2         [50-44]x[50-46]

                    =[44-4]/2                   6x4

                   = 40/2                          24

                    =20                             24

            The product=2024

                            Three places                three places

E.g3] 498x493=[498-[500-493]]/2      [500-498]x[500-493]

                        =[498-7]/2                   2x7

                        = 491/2                        014

                        =245.5                         014

                        =245514

The fraction 0.5 in 1000’s place =direct integer in 100’s place.

Product of any two numbers which are near to 25, 250, 2500 etc

Two places       Two places

Eg1] 23x24= [23-[25-24]]/4    [25-23]x[25-24] ,  the left side number should be multiplied by 50/100 i.e ¼

e.g2]           = [23-1]/4             2x1

                    =22/4                   02 as two places

                    =5.5                     02

                                  =552

The fraction 0.5 in 100’s place = direct integer 5 in 10’s place.

The all remaining product types are to be practiced as the tips applied for some special cases to easy to solve.

The above particular cases only, the multiplication tips applied.

For remaining, the normal method of practice is sufficient.

 3] The 9 is a magic number can remember 9^th table easily.

A]  9^th Table:                          Ten’s place      one’s place

                        9x1=9              0                                  [9-0]

                        9x2=18            1                                  [9-1]

                        9x3=27            2                                  [9-2]

                        9x4=36            3                                  [9-4]

                        9x5=45            4                                  [9-4]

                        9x6=54            5                                  [9-5]

                        9x7=63            6                                  [9-6]

                        9x8=72            7                                  [9-7]

                        9x9=81            8                                  [9-8]

                        9x10=90          9                                  [9-9]

                                                 10's place                    1's place

                        9xn=                [n-1]                            [10-n]

If we/ remind the 9 table up to 9x10=90 we can solve the many multiplications by 9

The remainder when a number divided by 9 with out division

1] 95÷9 quotient is 10 and remainder 5

2] 782÷9 quotient is 86 and remainder 8

In each event we find out quotient for finding out remainder.

But we can find out remainder individually just by adding the digits of the number up to 1digit.

E.g. 78932185÷9 then remainder 7+8+9+3+2+1+8+5=43 then 4+3=7 so the remainder is 7.

By eliminating the 9,s from sum obtained ≥ 9 while calculating sum,   we can get remainder easily

[7+8]-9+9-9+[3+2+1+8-9]+5=6+5+5 again-9=7

Add upto 9 and eliminate

6+[1+8]+9+[3+2+[1+8]]+5 after eliminating the red colour sums we get  =6+5+5=[6+3]+2+5 again after eliminating the red colour sums we get =7

2] By The same way if  a number divided by 3, we can find the remainder, without knowing quotient. 

E.g 57÷9, the remainder is 5+7=12, further 1+2=3 but the 3 is divisible by 3 so remainder is 0

E.g 43÷9 , the remainder is 4+3=7,  the 7 is not divisible by 3, so remainder is 7-3x2=1

All the best wishes for competitive examination and aware of mathematics. [ a well known subject means ‘’can be expressed the subject in mathematics’’]

[Next we approach to provide best understanding on divisibility and number systems i.e types of numbers] 

1] multiplication of the numbers with 5 in ones place;

                                                            1000's and 100's places    10's 1's places

 A] Squares: eg: 1]      35X35=1225   [by 3x[3+1]                             5x5]]

                            2]     75X75=5625   [by 7x[7+1]                              5x5]]

*For n5xn5=n[n+1]25 [n[n+1]is in 1000’s and 100’s place +25 is in 10’s and one’s place]

B] the numbers with 5 in one’s place and differ by 10                                         

Eg:1]  35X45=1575

       2] 65x55=3575

*For [n-1]5xn5=[n²-1]   75

2]the numbers multiplied by 5

   E.g1] 256x5=1280 [i.e [256/2]x10

   E.g2] 445x5=2225 [i.e [445/2]x10+5]]

If an even number n multiplied by 5 the product is [n/2]x10

If odd number n multiplied by 5 the product is [[n-1]/2]x10+5

3] square of the numbers near to 10,100,100…etc.

                                      10,s   one’s

            E.g.1] 9x9=81 [9-1]   1     =81

                                     100’s  one’s 

            Eg. 2] 98x98= [[98-2] [100-98]x[100-98]=9604

i.e 98-[100-98] is in 1000’s and 100’s place further [100-98][100-98] is in 10’s place and one’s place.

 Same way 96x94=[96-[100-94] [100-94]x[100-96]]=9024.

3] product of the numbers near to 10,100,100…etc.

E.g1]                           [two places left to 10’s]          two places left from start

1000’s, 100’s                           10’s 1’s 

Eg. 98x97=9506         [98-[100-97]]                    [100-98]x[100-97] as the number of digits in every number is ‘’2’’

E.g2] Can you imagine the product of the two numbers 99994x99997

5places                                                5 places          

Simple: 99994-[1000000-99997]    [1000000-99994]x[100000-99997]=9999100018.

The number of digits in product=2xthe number of digits in each number.

So we have obtained the product by multiplying the each number differences  to 100000 is 6x3=18 it should be in first 5 digits so that is 00018. Next, the difference between the one of the numbers to Nearest Round number 100000-another number]

Product of any two numbers which are near to 50, 500, 5000 etc

Two places       Two places

E.g1: 47X48=[47-[50-48]]/2      [50-47]x[50-48] here the left side number should be multiplied by 50/100 i.e ½

                    =[47-2]/2                3x2

                    =45/2                       06

                    =22.5                       06

                                 =2256

The fraction 0.5 in 100,s place= direct number 5 in 10’s place [0.5x10=5]

Two places       Two places

Eg2] 44x46= [44-[50-46]]/2         [50-44]x[50-46]

                    =[44-4]/2                   6x4

                   = 40/2                          24

                    =20                             24

            The product=2024

                            Three places                three places

E.g3] 498x493=[498-[500-493]]/2      [500-498]x[500-493]

                        =[498-7]/2                   2x7

                        = 491/2                        014

                        =245.5                         014

                        =245514

The fraction 0.5 in 1000’s place =direct integer in 100’s place.

Product of any two numbers which are near to 25, 250, 2500 etc

Two places       Two places

Eg1] 23x24= [23-[25-24]]/4    [25-23]x[25-24] ,  the left side number should be multiplied by 50/100 i.e ¼

e.g2]           = [23-1]/4             2x1

                    =22/4                   02 as two places

                    =5.5                     02

                                  =552

The fraction 0.5 in 100’s place = direct integer 5 in 10’s place.

The all remaining product types are to be practiced as the tips applied for some special cases to easy to solve.

The above particular cases only, the multiplication tips applied.

For remaining, the normal method of practice is sufficient.

 3] The 9 is a magic number can remember 9^th table easily.

A]  9^th Table:                          Ten’s place      one’s place

                        9x1=9              0                                  [9-0]

                        9x2=18            1                                  [9-1]

                        9x3=27            2                                  [9-2]

                        9x4=36            3                                  [9-4]

                        9x5=45            4                                  [9-4]

                        9x6=54            5                                  [9-5]

                        9x7=63            6                                  [9-6]

                        9x8=72            7                                  [9-7]

                        9x9=81            8                                  [9-8]

                        9x10=90          9                                  [9-9]

                                                 10's place                    1's place

                        9xn=                [n-1]                            [10-n]

If we/ remind the 9 table up to 9x10=90 we can solve the many multiplications by 9

The remainder when a number divided by 9 with out division

1] 95÷9 quotient is 10 and remainder 5

2] 782÷9 quotient is 86 and remainder 8

In each event we find out quotient for finding out remainder.

But we can find out remainder individually just by adding the digits of the number up to 1digit.

E.g. 78932185÷9 then remainder 7+8+9+3+2+1+8+5=43 then 4+3=7 so the remainder is 7.

By eliminating the 9,s from sum obtained ≥ 9 while calculating sum,   we can get remainder easily

[7+8]-9+9-9+[3+2+1+8-9]+5=6+5+5 again-9=7

Add upto 9 and eliminate

6+[1+8]+9+[3+2+[1+8]]+5 after eliminating the red colour sums we get  =6+5+5=[6+3]+2+5 again after eliminating the red colour sums we get =7

2] By The same way if  a number divided by 3, we can find the remainder, without knowing quotient. 

E.g 57÷9, the remainder is 5+7=12, further 1+2=3 but the 3 is divisible by 3 so remainder is 0

E.g 43÷9 , the remainder is 4+3=7,  the 7 is not divisible by 3, so remainder is 7-3x2=1

All the best wishes for competitive examination and aware of mathematics. [ a well known subject means ‘’can be expressed the subject in mathematics’’]

[Next we approach to provide best understanding on divisibility and number systems i.e types of numbers] 

Small difference between mathematic operator and sign

            Most of the students could not find the deference between mathematical operator like subtraction, addition, multiplication and division and sign +ve and –ve and fails in competitive mathematics with mistake. To avoid the mistake, please read the following examples.

Find out operator and sign

E.g 1:   2+3=5 in this equation + is operator i.e addition and the sign of two numbers is +ve [the numbers without sign means +ve numbers] simply addition of two +ve numbers.

E.g2:    7-3=4 here – is an  operator of subtraction and the two numbers 7 and 3 are +ve

            The same can be written as 7+[-3] here operator is + i.e addition but sign of the 3 is –ve i.e addition of one +ve number 7 and one –ve number -3.

E.g3:    8-[-3]=11 here operator is 1^{st “} –“ and sign of 8 is +ve and sign of 3 is –ve

E.g4:    -8/[-16]=1/2 here operator of the two numbers is / sign of the numbers is –ve.

Finding of unknown number ‘X’

E.g1:    if X/[-2]=8 then the X=8x[-2] the sign of the denominator cannot be changed from RHS to LHS to find out value of the X only operator  / can be changed i.e the number -2 is dividing X on LHS and to find  X valve the -2 multiply 8 one RHS without changing of it’s sign.

E.g2:    if X-2=8, then X=8+2 here 2is subtracted from X on LHS and to find out X, 2 is to be added on RHT without changing it’s sign +ve. [as X-[+2]=8].

E.g 3:   If Xx[-2]=8 then X=8/ -2  here -2 is multiplying the X on LHS and to find out the X the number -2 is dividing the 8 on RHS without changing it,s sign –ve.

E.g4:    If X/[-2]=8 then X=8x[ -2]  here -2 is dividing the X on LHS and to find out the X the number -2 is multiply the 8 on RHS without changing its sign –ve.

Conclusion:

The operator is different from sign. The operation between two numbers changes from LHS to RHS while finding out of unknown values, without changing its sign of the number other than the number to be findout on LHS.

[Note : Aim of the blog is to make the students bright in mathematics and make the parents as teachers.]

Addition of the numbers

Addition of the two +ve numbers gives a +ve number.

+ve number+ +ve number=+ve number

e.g : 3+5=7

Addition of the two –ve numbers gives a –ve number

-ve number+ -ve number=-ve number

-7+[-3]=-10

Addition of one –ve number and one +ve number vice versa gives the difference between the numbers with the bigger numbers sign.

-ve number+ +ve number= Sign of big number [difference of the numbers]

e.g: -3+2=-1 here the sign of the big number is –ve and difference between the numbers 3 and 2 is 1.

e.g: -7+10=3 here the sign of the big number is +ve and difference between the numbers 10 and 7 is 3.  

Conclusion

*Addition of the similar sign numbers gives the some of the numbers with same  sign.

*Addition of the different sing numbers give the difference of the numbers with bigger number sign.

Multiplication

Multiplication or division between two +ve numbers give a +ve number

e.g: 3x5=15

Multiplication or division of between two –ve numbers gives +ve number

E.g: -3x[-5]=15

Multiplication or division between one –ve number and one +ve number vice versa gives –ve number.

E.g: -3x5=-15

E.g: -15/3=-5

E.g: 3x[-5]=-15

E.g: 15/[-5]=-3

 Conclusion

*Multiplication or division between the similar sign numbers gives always +ve number.

*Multiplication or division between the different sign numbers gives always –ve number.

Square roots-General mistakes

        1-3=4-6 is it correct? yes it is.

             1^2_2x[3/2]x1=2²-2x[3/2]x2    add [3/2]² both side on both side

                   1^2_2x[3/2]x1+[3/2]²=2²-2x[3/2]x2+[3/2]²   identified as a²+2ab+b²=[a+b]²

                         [1-3/2]²=[2-3/2]² cancel the squares on both sides

                              1-3/2=2-3/2

^{:        .}1=2     is it correct ? Actually where is the mistake?

simply we  cancelled the squares on both sides

But it is not correct.

We don’t forget about + on right hand side or left hand side of the equation.

Don't commit mistakes by forgetting fundamentals.

As above, many examples can be taken as a^2_2a[a+b]/2=b^2_2b[a+b]/2 both sides we can get -ab on both sides so satisfied the equation.

Take another example
take a=3 and b=4
then a^2_2a[a+b]/2=b^2_2b[a+b]/2
       3^2_2x3[[3+4]/2]=4^2_2x4[3+4]/2
       9-21=16-28
        -12=-12
Equation satisfied

3^2_2x3[3+4]/2=4^2_2x4[3+4]/2 add {[3+4]/2}² on both sides.
3^2_2x3[3+4]/2+{[3+4]/2}² =4^2_2x4[3+4]/2+{[3+4]/2}² it is in the form of  a²+2ab+b²=[a+b]²
 [3-7/2]²=[4-7/2]²
if cancel the squares on both sides, equation becomes 3=4,which is practically wrong.
3-7/2=-1/2, 4-7/2=1/2, -1/2 is not equal to 1/2 but [-1/2]²=[1/2]²

Square roots Simple method
Any real +ve number in the form of [a+b]²=a²+2ab+b²
+ve numbers are mainly perfect squares and imperfect squares.
step1 identify a number's possibility of perfect square.
All perfect squares have either of the perfect square digit in 1s place as below
i.e 0,1,4,9,6.
256 mayor may not  be perfect square
252 is not perfect square.
456789 mayor may not  be perfect square
789235 is not a perfect square
by this we can estimate a number either chances of perfect square and not.

Finding of Square root of a number

I.e Find the square root of 7225