Identities

Let us consider this green square.

Diagonal products are:

2×8 = 16

7×3 = 21

Difference = 21-16 = 5

Now let us consider this green square.

Diagonal products: 9× 15 = 135

14× 10 = 140

Difference = 140 – 135 = 5

Now consider this green square.

Diagonal products: 11× 17 = 187

16× 12 = 192

Difference = 192 – 187 = 5

So, we observe that in all the above cases the difference of diagonal products is coming out to be same and is equal to 5.

Let us consider this green square.

Diagonal products are:

2×8 = 16

7×3 = 21

Difference = 21-16 = 5

Now let us consider this green square.

Diagonal products: 9× 15 = 135

14× 10 = 140

Difference = 140 – 135 = 5

Now consider this green square.

Diagonal products: 11× 17 = 187

16× 12 = 192

Difference = 192 – 187 = 5

So, we observe that in all the above cases the difference of diagonal products is coming out to be same and is equal to 5.

Let us consider this green square.

Diagonal products are:

2×8 = 16

7×3 = 21

Difference = 21-16 = 5

Now let us consider this green square.

Diagonal products: 9× 15 = 135

14× 10 = 140

Difference = 140 – 135 = 5

Now consider this green square.

Diagonal products: 11× 17 = 187

16× 12 = 192

Difference = 192 – 187 = 5

So, we observe that in all the above cases the difference of diagonal products is coming out to be same and is equal to 5.

Let the first number in the square be x, the others can be filled in as below,

Diagonal products: x (x+6) = x² + 6x

(x+5)(x+1) = x(x+1) + 5(x+1) = (x² + x) + (5x +5) = x² + 6x + 5

Clearly, the difference between the two diagonal products is always 5.

Let the first number in the square be x, the others can be filled in as below,

Diagonal products: x (x+6) = x² + 6x

(x+5)(x+1) = x(x+1) + 5(x+1) = (x² + x) + (5x +5) = x² + 6x + 5

Clearly, the difference between the two diagonal products is always 5.

Consider the green square and consider the four corner elements ( red circles)

Diagonal products: 1× 13 = 13

3× 11 = 33

Difference between diagonal products = 33 - 13 = 20

Now, consider this green square and consider the corner elements (red circles)

Diagonal products: 8× 20 = 160

18× 10 = 180

Difference between diagonal products = 180 – 160 = 20

So, we observe that the difference between diagonal products in the above cases is also coming out to be same and is equal to 20.

Explanation using Algebra,

Let the first number in the square be x, the others can be filled in as below,

Diagonal products: x (x+12) = x² + 12x

(x+10)(x+2) = x(x+2) + 10(x+2) = (x² + 2x) + (10x+ 20)

= x² + 12x + 20

Clearly, we can observe that the difference between diagonal products is always equal to 20.

Consider the green square and consider the four corner elements ( red circles)

Diagonal products: 1× 13 = 13

3× 11 = 33

Difference between diagonal products = 33 - 13 = 20

Now, consider this green square and consider the corner elements (red circles)

Diagonal products: 8× 20 = 160

18× 10 = 180

Difference between diagonal products = 180 – 160 = 20

So, we observe that the difference between diagonal products in the above cases is also coming out to be same and is equal to 20.

Explanation using Algebra,

Let the first number in the square be x, the others can be filled in as below,

Diagonal products: x (x+12) = x² + 12x

(x+10)(x+2) = x(x+2) + 10(x+2) = (x² + 2x) + (10x+ 20)

= x² + 12x + 20

Clearly, we can observe that the difference between diagonal products is always equal to 20.

MULTIPLICATION TABLE

In this part, we are considering this red square of a multiplication table.

Diagonal sums: 6+20 = 26

10+12 = 22

Difference between diagonal sums = 26 – 22 = 4

MULTIPLICATION TABLE

In this part, we are considering this red square of a multiplication table.

Diagonal sums: 6+20 = 26

10+12 = 22

Difference between diagonal sums = 26 – 22 = 4

we can write a row of multiplication table as

x 2x 3x 4x 5x 6x 7x 8x 9x

Next row will be:

(x+1) 2(x+1) 3(x+1) 4(x+1) 5(x+1) 6(x+1) 7(x+1) 8(x+1) 9(x+1)

If we take a general number from the first row as yx then the next number in this row will be (y+1)x.

So, we can obtain a general square of 9 terms from this multiplication table as:

Diagonal sums: (yx)+(y+2)(x+2) = yx+y(x+2)+2(x+2) = yx + yx + 2y +2x +4

= 2yx +2y +2x +4

Y(x+2)+ (y+2)x = (yx +2y)+(yx + 2x) = 2yx +2y +2x

Clearly, the difference between diagonal sums is equal to 4.

we can write a row of multiplication table as

x 2x 3x 4x 5x 6x 7x 8x 9x

Next row will be:

(x+1) 2(x+1) 3(x+1) 4(x+1) 5(x+1) 6(x+1) 7(x+1) 8(x+1) 9(x+1)

If we take a general number from the first row as yx then the next number in this row will be (y+1)x.

So, we can obtain a general square of 9 terms from this multiplication table as:

Diagonal sums: (yx)+(y+2)(x+2) = yx+y(x+2)+2(x+2) = yx + yx + 2y +2x +4

= 2yx +2y +2x +4

Y(x+2)+ (y+2)x = (yx +2y)+(yx + 2x) = 2yx +2y +2x

Clearly, the difference between diagonal sums is equal to 4.

Now, if we take a square of 16 numbers, then the general terms will be:

Diagonal sums: yx + (y+3)(x+3) = yx + y(x+3) + 3(x+3) = yx +yx +3y +3x +9 = 2yx +3y+3x+9

Y(x+3) + (y+3)x = yx +3y + yx +3x = 2yx + 3y + 3x

Clearly, the difference between the diagonal sums is equal to 9.

Now, if we take a square of 16 numbers, then the general terms will be:

Diagonal sums: yx + (y+3)(x+3) = yx + y(x+3) + 3(x+3) = yx +yx +3y +3x +9 = 2yx +3y+3x+9

Y(x+3) + (y+3)x = yx +3y + yx +3x = 2yx + 3y + 3x

Clearly, the difference between the diagonal sums is equal to 9.

i) 5× 8 = (6× 7) – 2

6× 9 = (7× 8) – 2

ii) The above pattern is based on this statement only.

Let the four consecutive numbers be 1,2,3,4.

Product of first and last = (1× 4) = 4

Product of middle two terms = (2× 3) = 6

and 4 = 6 - 2

Let the four consecutive numbers be 4,5,6,7.

Product of first and last = (4× 7) = 28

Product of middle two terms = (5× 6) = 30

And 28 = 30 - 2

iii) General principle,

Let the four consecutive natural numbers be x , (x+1) , (x+2) , (x+3).

Product of first and last = x(x+3) = x² + 3x

Product of middle two terms = (x+1)(x+2) = x(x+2) + 1(x+2)

= x² + 2x + x + 2

= x² + 3x + 2

Clearly, we can see that the difference between the product of middle two terms and the product of first and last term is equal to 2.

i) 5× 8 = (6× 7) – 2

6× 9 = (7× 8) – 2

ii) The above pattern is based on this statement only.

Let the four consecutive numbers be 1,2,3,4.

Product of first and last = (1× 4) = 4

Product of middle two terms = (2× 3) = 6

and 4 = 6 - 2

Let the four consecutive numbers be 4,5,6,7.

Product of first and last = (4× 7) = 28

Product of middle two terms = (5× 6) = 30

And 28 = 30 - 2

iii) General principle,

Let the four consecutive natural numbers be x , (x+1) , (x+2) , (x+3).

Product of first and last = x(x+3) = x² + 3x

Product of middle two terms = (x+1)(x+2) = x(x+2) + 1(x+2)

= x² + 2x + x + 2

= x² + 3x + 2

Clearly, we can see that the difference between the product of middle two terms and the product of first and last term is equal to 2.

i)

ii) We know that we can express any 2-digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

So, let (10m+n) and (10x+y) be the two numbers to be multiplied.

(10m+n)(10x+y) = 10m(10x+y)+n(10x+y) = 100 mx +10my +10nx + ny

= (100× mx) + ((m×y)+(n×x))× 10 +(n×y)

Now, carefully observe the above method, you will see that we were finding these terms only and then addition of these terms gives us the required product.

i)

ii) We know that we can express any 2-digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

So, let (10m+n) and (10x+y) be the two numbers to be multiplied.

(10m+n)(10x+y) = 10m(10x+y)+n(10x+y) = 100 mx +10my +10nx + ny

= (100× mx) + ((m×y)+(n×x))× 10 +(n×y)

Now, carefully observe the above method, you will see that we were finding these terms only and then addition of these terms gives us the required product.

we can write generally as .

So, to find the square we have to calculate

using, (x+y)² = x² + 2xy + y²

we can write generally as .

So, to find the square we have to calculate

using, (x+y)² = x² + 2xy + y²

Now, let’s calculate 72²

Now, let’s calculate 72²

Now lets try to understand this method using algebra.

We know that we can express any 2 digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

Now we have to find its square, i.e. (10m + n)²

(10m + n)² = (10m)² + 2× 10m× n + n² ……….using, (x+y)² = x² + 2xy + y²

=( m² × 100) + ( 2mn× 10 )+ (n² )

Now, carefully observe the above method, you will see that we were finding these terms only and then addition of these terms gives us the required square.

Now lets try to understand this method using algebra.

We know that we can express any 2 digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

Now we have to find its square, i.e. (10m + n)²

(10m + n)² = (10m)² + 2× 10m× n + n² ……….using, (x+y)² = x² + 2xy + y²

=( m² × 100) + ( 2mn× 10 )+ (n² )

Now, carefully observe the above method, you will see that we were finding these terms only and then addition of these terms gives us the required square.

Method to find the square of number ending with 5.

Suppose we have to find 45².

Let the number at tens place be x.

So, we first have to find x(x+1).

For example, here we have x=4.

So, x(x+1) = 4(4+1) = 4×5 = 20

Now, we just need to write this 20 before 25, i.e. 2025 and it is the required square.

Let us now find the square of 85.

So, 8(8+1) = 8×9 = 72

∴ 85² = 7225

Method to find the square of number ending with 5.

Suppose we have to find 45².

Let the number at tens place be x.

So, we first have to find x(x+1).

For example, here we have x=4.

So, x(x+1) = 4(4+1) = 4×5 = 20

Now, we just need to write this 20 before 25, i.e. 2025 and it is the required square.

Let us now find the square of 85.

So, 8(8+1) = 8×9 = 72

∴ 85² = 7225

i) 4² + (4×5) = 6²

5² + (4×6) = 7²

ii) General principle,

By observing the pattern we can write a general line as:

x² + (4× (x+1)) = ( x+2 )²

Now, let us prove this using algebra.

x² + ( 4× (x+1)) = x² + 4x + 4

(x+2)² = x² + 4x + 4 ……using , (x+y)² = x² + 2xy + y²

Cleary, both these terms are equal.

Hence, proved.

i) 4² + (4×5) = 6²

5² + (4×6) = 7²

ii) General principle,

By observing the pattern we can write a general line as:

x² + (4× (x+1)) = ( x+2 )²

Now, let us prove this using algebra.

x² + ( 4× (x+1)) = x² + 4x + 4

(x+2)² = x² + 4x + 4 ……using , (x+y)² = x² + 2xy + y²

Cleary, both these terms are equal.

Hence, proved.

We know that, We can express a number which is not a multiple of 3 as ( 3m + 1) or ( 3m +2).

( 3m+1)² = (3m)² + 2×3m×1 + 1² = 9m² + 6m + 1 = 3 ( 3m² + 2m) + 1 ………..(1)

…………..using, (x+y)² = x² + 2xy + y²

( 3m+2)² = (3m)² + 2×3m×2 + 2² = 9m² + 12m + 4 = 9m² + 12m + 3 +1

= 3( 3m ² + 4m +1) +1 ………..(2)

…………..using, (x+y)² = x² + 2xy + y²

Now, observe (1) and (2) carefully, clearly on division by 3 both these terms will leave remainder 1.

We know that, We can express a number which is not a multiple of 3 as ( 3m + 1) or ( 3m +2).

( 3m+1)² = (3m)² + 2×3m×1 + 1² = 9m² + 6m + 1 = 3 ( 3m² + 2m) + 1 ………..(1)

…………..using, (x+y)² = x² + 2xy + y²

( 3m+2)² = (3m)² + 2×3m×2 + 2² = 9m² + 12m + 4 = 9m² + 12m + 3 +1

= 3( 3m ² + 4m +1) +1 ………..(2)

…………..using, (x+y)² = x² + 2xy + y²

Now, observe (1) and (2) carefully, clearly on division by 3 both these terms will leave remainder 1.

We know that we can express any 2 digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

∴ the number ending with 3 can be expressed as 10m+3.

(10m + 3)² = (10m)² + 2×10m×3 + 3² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 10× 6m + 9

Clearly, we can observe that the square ends in 9.

The number ending with 5 can be expressed as 10m+5.

(10m + 5)² = (10m)² + 2×10m×5 + 5² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 100m + (25)

= 100 (m² + m) + (10×2 + 5 )

Clearly, we can observe that the square ends in 5.

The number ending with 4 can be expressed as 10m+4.

(10m + 4)² = (10m)² + 2×10m×4 + 4² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 10× 8m + (16)

= 100m² + 10× 8m + (10 + 6)

= 100 m² + 10× (8m+1) + 6

Clearly, we can observe that the square ends in 6.

We know that we can express any 2 digit number as 10m+n , where m is the digit at tens place and n is the digit at ones place.

∴ the number ending with 3 can be expressed as 10m+3.

(10m + 3)² = (10m)² + 2×10m×3 + 3² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 10× 6m + 9

Clearly, we can observe that the square ends in 9.

The number ending with 5 can be expressed as 10m+5.

(10m + 5)² = (10m)² + 2×10m×5 + 5² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 100m + (25)

= 100 (m² + m) + (10×2 + 5 )

Clearly, we can observe that the square ends in 5.

The number ending with 4 can be expressed as 10m+4.

(10m + 4)² = (10m)² + 2×10m×4 + 4² …………using, (x+y)² = x² + 2xy + y²

= 100m² + 10× 8m + (16)

= 100m² + 10× 8m + (10 + 6)

= 100 m² + 10× (8m+1) + 6

Clearly, we can observe that the square ends in 6.

=

Using identity = + - 2

So =

= + - 2

= 2500 + 1- 100

= 2401

=

Using identity = + - 2

So =

= + - 2

= 2500 + 1- 100

= 2401

98 can be written as (100-2)

=

Using identity = + - 2

So =

= +- 2

= 10000+4 - 400

= 9604

98 can be written as (100-2)

=

Using identity = + - 2

So =

= +- 2

= 10000+4 - 400

= 9604

can be written as

Using identity= + + 2

so ₌

₌₊

_{= 49 ++}

_{= 49+ ()}

_{= 49 +}

_{= 49}

can be written as

Using identity= + + 2

so ₌

₌₊

_{= 49 ++}

_{= 49+ ()}

_{= 49 +}

_{= 49}

9.25²can be written as(9 + 0.25)²

Using identity = + + 2

(9 + 0.25)² = (9)² + (0.25)² + 290.25

= 81 + 0.0625 + 4.5

= 85.5625

9.25²can be written as(9 + 0.25)²

Using identity = + + 2

(9 + 0.25)² = (9)² + (0.25)² + 290.25

= 81 + 0.0625 + 4.5

= 85.5625

we can write it as ₊

Using identity = + - 2

and = + + 2

= 1+_{- 1 + 1++1}

_{= 2 +}

₌

Similarly, we can do it for other numbers.

we can write it as ₊

Using identity = + - 2

and = + + 2

= 1+_{- 1 + 1++1}

_{= 2 +}

₌

Similarly, we can do it for other numbers.

Let’s use algebra. starting with x,y, the square of the difference is is = + - 2

the square of the sum is

= + + 2

what if we subtract the square of the difference from the square of the sum.

- = (+ + 2 - (+ - 2)

- =

writing this in reverse

= -

for example, 24= 4

= here x=6 and y=1

= -

= -

= 7² – 5²

similarly 24 can also be written as

24 = 432

= here x=3 and y=2

= -

= -

= 5² – 1²

so 24 =7² – 5²= 5² – 1²

other multiple of 8,

32 = 9² – 7² = 6² – 2²

32 = 481

= here x=8 and y=1

= -

= -

= 9² – 7²

similarly 32 can also be written as

32 = 442

= here x=4 and y=2

= -

= -

32 = 9² – 7² = 6² – 2²

Let’s use algebra. starting with x,y, the square of the difference is is = + - 2

the square of the sum is

= + + 2

what if we subtract the square of the difference from the square of the sum.

- = (+ + 2 - (+ - 2)

- =

writing this in reverse

= -

for example, 24= 4

= here x=6 and y=1

= -

= -

= 7² – 5²

similarly 24 can also be written as

24 = 432

= here x=3 and y=2

= -

= -

= 5² – 1²

so 24 =7² – 5²= 5² – 1²

other multiple of 8,

32 = 9² – 7² = 6² – 2²

32 = 481

= here x=8 and y=1

= -

= -

= 9² – 7²

similarly 32 can also be written as

32 = 442

= here x=4 and y=2

= -

= -

32 = 9² – 7² = 6² – 2²

Let’s use algebra. starting with x,y, the square of the difference is = + - 2

the square of the sum is

= + + 2

what if we subtract the square of the difference from the square of the sum.

- = (+ + 2 - (+ - 2)

- =

writing this in reverse

= -

multiples of 16 are 16,32,48,64,80…..starting with 48

48 = 4121

= here x=12 and y=1

= -

= -

= -

similarly 48 can also be written as

48 = 443

= here x=4 and y=3

= -

= -

= -

48 can also be written as

48 = 462

= here x=6 and y=2

= -

= -

= -

Similarly for 64 = 4161 = -

64 = 482 = -

64 = 444 = -

Similarly we can do it for other multiples of 16.

Let’s use algebra. starting with x,y, the square of the difference is = + - 2

the square of the sum is

= + + 2

what if we subtract the square of the difference from the square of the sum.

- = (+ + 2 - (+ - 2)

- =

writing this in reverse

= -

multiples of 16 are 16,32,48,64,80…..starting with 48

48 = 4121

= here x=12 and y=1

= -

= -

= -

similarly 48 can also be written as

48 = 443

= here x=4 and y=3

= -

= -

= -

48 can also be written as

48 = 462

= here x=6 and y=2

= -

= -

= -

Similarly for 64 = 4161 = -

64 = 482 = -

64 = 444 = -

Similarly we can do it for other multiples of 16.

using identity - =

68² – 32² = (68+32)(68-32)

= (100) (36)

= 3600

using identity - =

68² – 32² = (68+32)(68-32)

= (100) (36)

= 3600

using identity - =

- =

= (

= 6

= 6

using identity - =

- =

= (

= 6

= 6

using identity - =

(3.6)² – (1.4)² = (3.6+1.4) (3.6 - 1.4)

= (5) (2.2)

= 11

using identity - =

(3.6)² – (1.4)² = (3.6+1.4) (3.6 - 1.4)

= (5) (2.2)

= 11

it can be written as (200+1)(200 - 1)

using identity -

(200+1)(200 - 1)= -

= 40000-1

= 39999

it can be written as (200+1)(200 - 1)

using identity -

(200+1)(200 - 1)= -

= 40000-1

= 39999

(

using identity (x+y) (u+v) = xu+xv+yu+yv

=(2 +() + (

= 2+++

= 2++

=

(

using identity (x+y) (u+v) = xu+xv+yu+yv

=(2 +() + (

= 2+++

= 2++

=

it can be written as (10+0.7) (10-0.7)

using identity -

(10+0.7) (10-0.7) = -

= 100 - 0.49

= 99.51

it can be written as (10+0.7) (10-0.7)

using identity -

(10+0.7) (10-0.7) = -

= 100 - 0.49

= 99.51

we can write it as _-

Using identity = + - 2

and = + + 2

= 1+_{+ 1 -( 1+-1)}

_{= 1 ++1 - 1 -+1}

_{= 2}

Similarly it can be done for other numbers.

we can write it as _-

Using identity = + - 2

and = + + 2

= 1+_{+ 1 -( 1+-1)}

_{= 1 ++1 - 1 -+1}

_{= 2}

Similarly it can be done for other numbers.

Using identity -

25 × 75 = (50-25)(50+25) = -

26 × 74 = (50- 24)(50+24)= -

clearly 26 × 74 is the larger product

since in this we are subtracting from whereas in 25 × 75 we are subtracting from .

example- (10-5) and (10-4)

(10-4) is larger since we are subtracting 4 from 10 whereas we are subtracting 5 from 10 in (10-5).

Using identity -

25 × 75 = (50-25)(50+25) = -

26 × 74 = (50- 24)(50+24)= -

clearly 26 × 74 is the larger product

since in this we are subtracting from whereas in 25 × 75 we are subtracting from .

example- (10-5) and (10-4)

(10-4) is larger since we are subtracting 4 from 10 whereas we are subtracting 5 from 10 in (10-5).

Using identity -

76 × 24 = (50+26)(50-26) = -

74 × 26 = (50+24)(50-24) = -

clearly 74 × 26 is the larger product

since in this we are subtracting from whereas in 76 × 24 we are subtracting from . since we are subtracting smaller number from same number.

Using identity -

76 × 24 = (50+26)(50-26) = -

74 × 26 = (50+24)(50-24) = -

clearly 74 × 26 is the larger product

since in this we are subtracting from whereas in 76 × 24 we are subtracting from . since we are subtracting smaller number from same number.

Using identity-

10.6 × 9.4 = (10+0.6)(10-0.6)= -

10.4 × 9.6 = (10+0.4)(10-0.4)= -

clearly 10.4 × 9.6 is the larger product

since in this we are subtracting from whereas in 10.6×9.4 we are subtracting from

Using identity-

10.6 × 9.4 = (10+0.6)(10-0.6)= -

10.4 × 9.6 = (10+0.4)(10-0.4)= -

clearly 10.4 × 9.6 is the larger product

since in this we are subtracting from whereas in 10.6×9.4 we are subtracting from

(125 × 75) can be written as (100+25)(100-25)

using identity -

(100+25)(100-25) = -

= 10000 - 625

= 9375

now (126 × 74) can be written as (100+26)(100-26)

using identity -

(100+26)(100-26) = -

= 10000 - 676

= 9324

now putting values in question

(125 × 75) – (126 × 74) = 9375-9324

= 51

(125 × 75) can be written as (100+25)(100-25)

using identity -

(100+25)(100-25) = -

= 10000 - 625

= 9375

now (126 × 74) can be written as (100+26)(100-26)

using identity -

(100+26)(100-26) = -

= 10000 - 676

= 9324

now putting values in question

(125 × 75) – (126 × 74) = 9375-9324

= 51

(124 × 76) can be written as (100+24)(100-24)

using identity -

(100+24)(100-24) = -

= 10000 - 576

= 9424

now (126 × 74) can be written as (100+26)(100-26)

using identity -

(100+26)(100-26) = -

= 10000 - 676

= 9324

now putting values in question

(124 × 76) – (126 × 74) = 9424-9324

= 100

(124 × 76) can be written as (100+24)(100-24)

using identity -

(100+24)(100-24) = -

= 10000 - 576

= 9424

now (126 × 74) can be written as (100+26)(100-26)

using identity -

(100+26)(100-26) = -

= 10000 - 676

= 9324

now putting values in question

(124 × 76) – (126 × 74) = 9424-9324

= 100

(224 × 176) can be written as (200+24)(200-24)

using identity -

(200+24)(200-24) = -

= 40000 - 576

= 39424

now (226 × 174) can be written as (200+26)(200-26)

using identity -

(200+26)(200-26) = (200)²-

= 40000 - 676

= 39324

now putting values in question

(224 × 176) – (226 × 174) = 39424 - 39324

= 100

(224 × 176) can be written as (200+24)(200-24)

using identity -

(200+24)(200-24) = -

= 40000 - 576

= 39424

now (226 × 174) can be written as (200+26)(200-26)

using identity -

(200+26)(200-26) = (200)²-

= 40000 - 676

= 39324

now putting values in question

(224 × 176) – (226 × 174) = 39424 - 39324

= 100

(10.3 × 9.7) can be written as (10+0.3)(10-0.3)

using identity -

(10+0.3)(10-0.3) = -

= 100 - 0.09

= 99.91

now (10.7 × 9.3) can be written as (10+0.7)(10-0.7)

using identity -

(10+0.7)(10-0.7) = -

= 100 - 0.49

= 99.51

now putting values in question

(10.3 × 9.7) – (10.7 × 9.3) = 99.91 - 99.51

= 0.40

(10.3 × 9.7) can be written as (10+0.3)(10-0.3)

using identity -

(10+0.3)(10-0.3) = -

= 100 - 0.09

= 99.91

now (10.7 × 9.3) can be written as (10+0.7)(10-0.7)

using identity -

(10+0.7)(10-0.7) = -

= 100 - 0.49

= 99.51

now putting values in question

(10.3 × 9.7) – (10.7 × 9.3) = 99.91 - 99.51

= 0.40

(11.3 × 10.7) can be written as (11+0.3)(11-0.3)

using identity -

(11+0.3)(11-0.3) = -

= 121 - 0.09

= 120.91

now (11.7 × 10.3) can be written as (11+0.7)(11-0.7)

using identity -

(11+0.7)(11-0.7) = -

= 121 - 0.49

= 120.51

now putting values in question

(11.3 × 10.7) – (11.7 × 10.3) = 120.91- 120.51

= 0.40

(11.3 × 10.7) can be written as (11+0.3)(11-0.3)

using identity -

(11+0.3)(11-0.3) = -

= 121 - 0.09

= 120.91

now (11.7 × 10.3) can be written as (11+0.7)(11-0.7)

using identity -

(11+0.7)(11-0.7) = -

= 121 - 0.49

= 120.51

now putting values in question

(11.3 × 10.7) – (11.7 × 10.3) = 120.91- 120.51

= 0.40

Let the numbers be 6,7,13 and 14

the squares of the diagonal are

6² and 14²

7² and 13²

Add the squares of the diagonal pair

6² + 14²

= 36+196

= 232

7² + 13²

= 49+ 169

= 218

find the difference of these sums:

232 - 218

=14

Let the numbers be 6,7,13 and 14

the squares of the diagonal are

6² and 14²

7² and 13²

Add the squares of the diagonal pair

6² + 14²

= 36+196

= 232

7² + 13²

= 49+ 169

= 218

find the difference of these sums:

232 - 218

=14

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

the squares of diagonal are

x² and (x+8)²

(x+7)² and (x+1)²

Add the squares of the diagonal pair

x² + (x+8)²

= x² + (x²+8²+2 [Using identity = + + 2]

⁼ 2x² + 64+16

Add the squares of the other diagonal pair

= (x+1)²+(x+7)²

⁼ (x² +1 +2 (x²+7²+2

= (x² +1 +2 (x² +49 +14)

= 2x² +16+ 50

find the difference of these sums:

= (2x² + 64+16- (2x² +16+ 50)

= 2x² + 64+16- 2x² -16- 50

= 64-50

= 14

Hence the difference is 14, we can take any number as x; which means this holds in any part of the calendar.

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

the squares of diagonal are

x² and (x+8)²

(x+7)² and (x+1)²

Add the squares of the diagonal pair

x² + (x+8)²

= x² + (x²+8²+2 [Using identity = + + 2]

⁼ 2x² + 64+16

Add the squares of the other diagonal pair

= (x+1)²+(x+7)²

⁼ (x² +1 +2 (x²+7²+2

= (x² +1 +2 (x² +49 +14)

= 2x² +16+ 50

find the difference of these sums:

= (2x² + 64+16- (2x² +16+ 50)

= 2x² + 64+16- 2x² -16- 50

= 64-50

= 14

Hence the difference is 14, we can take any number as x; which means this holds in any part of the calendar.

let the other numbers be

The squares of diagonal are

6² and 22²

20² and 8²

Add the squares of the diagonal pair

6² + 22²

= 36+ 484

= 520

20² + 8²

= 400+64

= 464

find the difference of these sums:

= 520 - 464

= 56

let the other numbers be

The squares of diagonal are

6² and 22²

20² and 8²

Add the squares of the diagonal pair

6² + 22²

= 36+ 484

= 520

20² + 8²

= 400+64

= 464

find the difference of these sums:

= 520 - 464

= 56

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

The squares of diagonal are

x² and (x+16)²

(x+14)² and (x+2)²

Add the squares of the diagonal pair

= x² + (x+16)²

= x² +(x²+256+ 2 [Using identity = + + 2]

=2 x² +256+ 32x

Add the squares of the other diagonal pair

= (x+14)²+(x+2)²

=(x² +14² +2 (x²+2²+2

= (x² +196 +28 (x² +4 +4)

= 2x² +32+ 200

find the difference of these sums:

(2 x² +256+ 32x) - ( 2x² +32+ 200)

= 2 x² +256+ 32x - 2x² - 32- 200

= 256-200

= 56

Hence the difference is 56, we can take any number as x; which means this hold in any part of the calendar.

Yes we can take x in the center but this will complicate the calculations.

since then the numbers will be

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

The squares of diagonal are

x² and (x+16)²

(x+14)² and (x+2)²

Add the squares of the diagonal pair

= x² + (x+16)²

= x² +(x²+256+ 2 [Using identity = + + 2]

=2 x² +256+ 32x

Add the squares of the other diagonal pair

= (x+14)²+(x+2)²

=(x² +14² +2 (x²+2²+2

= (x² +196 +28 (x² +4 +4)

= 2x² +32+ 200

find the difference of these sums:

(2 x² +256+ 32x) - ( 2x² +32+ 200)

= 2 x² +256+ 32x - 2x² - 32- 200

= 256-200

= 56

Hence the difference is 56, we can take any number as x; which means this hold in any part of the calendar.

Yes we can take x in the center but this will complicate the calculations.

since then the numbers will be

let the other numbers be

Multiply the diagonal pairs

6x22 = 132

20x8 = 160

Difference of these products

160-132

= 28

let the other numbers be

Multiply the diagonal pairs

6x22 = 132

20x8 = 160

Difference of these products

160-132

= 28

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

Multiply the diagonal pairs

(x)(x+16)

= x²+16x

Other diagonal product

(x+14)(x+2) [using identity (x+y)(u+v)= xu+xv+yu+yv]

= x²+2x+14x+28

= x²+16x+28

Difference of these products

Let’s use algebra to see this.

Taking the first number in the square as x, the others can be filled as below

Multiply the diagonal pairs

(x)(x+16)

= x²+16x

Other diagonal product

(x+14)(x+2) [using identity (x+y)(u+v)= xu+xv+yu+yv]

= x²+2x+14x+28

= x²+16x+28

Difference of these products

Using (x+y)² = x² + xy +yx +y²

(52)² = (50 + 2)²

=(50)² + 2×50×2 + (2)

= 2500 + 200 + 4

= 2704

Using (x+y)² = x² + xy +yx +y²

(52)² = (50 + 2)²

=(50)² + 2×50×2 + (2)

= 2500 + 200 + 4

= 2704

Using (x+y)² = x² + xy +yx +y²

(105)² = (100 + 5)²

=(100)² + 2×100×5 + (5)²

= 10000 + 1000 + 25

= 11025

Using (x+y)² = x² + xy +yx +y²

(105)² = (100 + 5)²

=(100)² + 2×100×5 + (5)²

= 10000 + 1000 + 25

= 11025

Using (x+y)² = x² + xy +yx +y²

Using (x+y)² = x² + xy +yx +y²

(10.2)² = (10 + 0.2)²

=(10)² + 2×10×0.2 + (0.2)²

= 100 + 4 + 0.04

= 104.04

(10.2)² = (10 + 0.2)²

=(10)² + 2×10×0.2 + (0.2)²

= 100 + 4 + 0.04

= 104.04