Squares, Square-roots, Cubes And Cube-roots

(i) square of 4 is 16;

4² = 16

(ii) square of 8 is 64;

8² = 64

(iii) square of 15 is 225

15² = 225

Since,

6² = 36

7² = 49

9² = 81

13² = 169

25² = 625

30² = 900

10² = 100

15² = 225

Hence, 36, 49, 81, 169, 625, 900 and 100 are perfect squares.

1² = 1

2² = 49

3² = 9

4² = 16

5² = 25

6² = 36

7² = 49

8² = 64

9² = 81

10² = 100

11² = 121

12² = 144

13² = 169

14² = 196

15² = 225

16² = 256

17² = 289

18² = 324

19² = 361

20² = 400

21² = 441

22² = 484

So, perfect squares from 1 to 500 are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400,441, 484

3-digit numbers ending with 0, 1, 4, 5, 6, 9, are 110, 111, 114, 115, 116, 119.

There are many other numbers too, satisfying the conditions mentioned.

numbers from 100 to 400 that end with 0, 1, 4, 5, 6 or 9, which are perfect squares are:

10² = 100

11² = 121

12² = 144

13² = 169

14² = 196

15² = 225

16² = 256

17² = 289

18² = 324

19² = 361

20² = 400

Total terms =

= 26

As we know, that sum of first n odd natural numbers is n²

⇒ Sum of above digits = 26²

= 676

As we know, that sum of first n odd natural numbers is n²

Sum of above digits = 144

⇒ Number of digits = √144

= 12

So, digits are 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

14^th triangular number = 1 + 2 + 3 + 4 + … + 14 = 105

15^th triangular number = 1 + 2 + 3 + 4 + … + 14 + 15 = 120

According to statement 8, the sum of n^th and (n + 1)^th triangular number is (n + 1)²

Here, 120 + 105 = 225

And (14 + 1)² = 15²

= 225

Hence Statement 8 is verified

Since all perfect squares terminate with digits 0, 1, 4, 5, 6, 9 , i.e., digits at unit place.

So, remainder when the perfect square is divided by 5 is either of 1, 4, and 0.

Using the identity (a + b)² = a² + b² + 2ab

Here, 31² = (30 + 1)²

= 30² + 1² + 2×30×1

= 900 + 1 + 60

= 961

Using the identity (a + b)² = a² + b² + 2ab

Here, 72² = (70 + 2)²

= 70² + 2² + 2×70×2

= 4900 + 4 + 280

= 5184

Using the identity (a-b)² = a² + b²- 2ab

Here, 37² = (40 - 3)²

= 40² + 3² - 2×40×3

= 1600 + 9 - 240

= 1369

Using the identity (a-b)² = a² + b²- 2ab

Here, 72² = (170 - 4)²

= 170² + 4² - 2×170×4

= 28900 + 16 - 1360

= 27556

(i) 85

Since 85 terminates with 5.

⇒ 8 × (8 + 1)

8 × 9 = 72

So, 85² = 7225

(ii) 115

Since 115 terminates with 5.

⇒ 11 × (11 + 1)

11 × 12 = 132

So, 115² = 13225

(iii) 165

Since 165 terminates with 5.

⇒ 16 × (16 + 1)

16 × 17 = 272

So, 165² = 27225

Using the identity (a + b)² = a² + b² + 2ab

Here, 72² = (1465 + 3)²

= 1465² + 3² + 2×1465×3 ..(i)

Since 1465 terminates with 5,

1465²⇒ 146 × (146 + 1)

= 146 × 147

= 21462

1465² = 2146225

Substituting 1465² in equation (i)

= 1465² + 3² + 2×1465×3

= 2146225 + 9 + 8790

= 2155024

196 = 7 × 7 × 2 × 2

196 = 7² × 2²

∴ square root of 196 is 14

256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

256 = 2⁸

∴ square root of 256 is 16

10404 = 2 × 2 × 3 × 3 × 17× 17

10404 = 2² × 3² × 17²

∴ square root of 10404 is 102

1156 = 2 × 2 × 17× 17

1156 = 2² × 17²

∴ square root of 1156 is 34

13225 = 5 × 5 × 23 × 23

13225 = 5² × 23²

∴ square root of 13225 is 115

√100 + √36

√100 = 10

√36 = 6

⇒ √100 + √36 = 10 + 6

= 16

∴ √100 + √36 = 16

√(1360 + 9)

√1360 + 9 = √1369

√1369 = 37

∴

√2704 + √144 + √289

√2704 = 52

√144 = 12

√289 = 17

⇒ √2704 + √144 + √289 = 52 + 12 + 17

= 81

∴ √2704 + √144 + √289 = 81

√225 – √25

√225 = 15

√25 = 5

⇒ √225 - √25 = 15 - 5

= 10

∴ √225 - √25 = 10

√1764 – √1444

√1764 = 42

√1444 = 38

⇒ √1764 - √1444 = 42 - 38

= 4

∴ √1764 - √1444 = 4

√169 × √361

√169 = 13

√361 = 19

⇒ √163 × √361 = 13 × 19

= 247

∴ √169 × √361 = 247

Area of yard = 1764 m²

Area of part taken out = 784 m²

Area of remaining part = 1764 – 784

= 980 m²

∴ Area of one part out of 5 equal parts

= 196 m²

Side of the square of area 196 m² = √196

= 14 m

∴Perimeter of each small square part = 4 × 14

= 56 m

(i) 847

847 = 7 × 11 × 11

847 = 7× 11 × 11 × 7

Hence required integer is 7

(ii) 450

450 = 5 × 5 × 6 × 3

450 = 5 × 5 × 6 × 3 × 2

Hence required integer is 2

(iii) 1445

1445 = 5 × 17 × 17

1445 = 5 × 17 × 17 × 5

Hence required integer is 5

(iv) 1352

1352 = 2 ×2 × 2 ×13 × 13

1352 = 2 ×2 × 2 ×13 × 13 × 2

Hence required integer is 2

(i) 48

48 = 2 ×2 × 2 × 2 × 3

48 = 16 × 3

Hence largest perfect square factor is 16

(ii) 11280

11280 = 2 ×2 × 2 × 2 × 5 × 3 × 47

11280 = 16 × 705

Hence largest perfect square factor is 16

(iii) 729

729 = 3 ×3 × 3 × 3 × 3 × 3

729 = 729

Hence largest perfect square factor is 729

(iv) 1352

1352 = 2 ×2 × 2 × 13 × 13

1352 = 2 × 676

Hence largest perfect square factor is 676

(i) 232

Nearest perfect square < 232 = 225

Nearest perfect square > 232 = 256

⇒ Nearest integer to the square root of 232 = 15

(ii) 600

Nearest perfect square < 600 = 576

Nearest perfect square > 600 = 625

⇒ Nearest integer to the square root of 600 = 24

(iii) 728

Nearest perfect square < 728 = 676

Nearest perfect square > 728 = 729

⇒ Nearest integer to the square root of 232 = 27

(iv) 824

Nearest perfect square < 824 = 784

Nearest perfect square > 824 = 841

⇒ Nearest integer to the square root of 824 = 29

(v) 1729

Nearest perfect square < 1729 = 1681

Nearest perfect square > 1729 = 1764

⇒ Nearest integer to the square root of 1729 = 42

Area of land = 1000 m²

Length of side = √1000 m

⇒ Perimeter of land = 4 × √1000

= 127 m (approx.)

∴ Length of wire required is 127m

√961 = 31

26² = 676 < 691 < 27 = 729

⇒ 691 is nearer to 676.

Thus, the nearest integer to is 26.

Difference = 31 − 26 = 5

∴ His number was smaller than the correct answer by 5.

The cube of even number is always even and that of odd be always odd.

1³ = 1

2³ = 8

3³ = 27

4³ = 64

⇒ There are 4 cubes from 1 to 100.

0 = 0

(−1)³ = −1

(−2)³ = −8

(−3)³ = −27

(−4)³ = −64

∴ There are 9 cubes from −100 to 100.

1³ = 1

2³ = 8

3³ = 27

4³ = 64

5³ = 125

6³ = 216

7³ = 343

8³ = 512

Thus, there are 7 perfect cubes from 1 to 500 and 64 i.e., (8²) is the only perfect square.

10³ = 1000

30³ = 27000

100³ = 10,00,000

1000³ = 1,00,00,00,000

The number of zeroes is always a multiple of 3

1³ = 1

2³ = 8

3³ = 27

4³ = 64

5³ = 125

6³ = 216

7³ = 343

8³ = 512

9³ = 729

10³ = 1000

The digits at units place are 1, 8, 7, 4, 5, 6, 3, 2 and 9.

Since, all digits are at end of some or other cube.

So, it is not possible to say that a number is not a perfect cube by looking at the digit in unit’s place of the given number.

1728 = 2 × 2 ×2 × 2 × 2 × 2 × 3 × 3 × 3

1728 = 2³ × 2³ × 3³

∴ cube root of 1728 is 12

3375 = 5 × 5 ×5 × 3 × 3 × 3

3375 = 5³ × 3³

∴ cube root of 3375 is 15

10648 = 2 × 2 ×2 × 11 × 11 × 11 10648 = 2³ ×11³

∴ cube root of 10648 is 22

46656 = 2 × 2 ×2 × 2 × 2 ×2 × 9× 9 ×9 46656 = 2⁶ ×9³

∴ cube root of 46656 is 36

Let n³ = 91125

Here, the unit’s place digit is 5;

∴ The units digit of n must be 5.

Let us split the number 91125 as 91 and 125

Now, 4³ < 91 < 5³ = 125

∴ 40³ = 64000 < 91000 < 50³ = 125000

Since, the unit’s place digit of n is 5, the only possible number is 45.

Also, 45³ = 91125

Let n³ = 166375

Here, the unit’s place digit is 5;

∴ Unit’s place digit of n must be 5.

Let us split the number 166375 as 166 and 375

Now, 5³ < 166 < 6³

∴ 50³ = 125000 < 166000 < 60³ = 216000

Since, the unit’s place digit of n is 5, the only possible number is 55.

Also, 55³ = 166375

Let n³ = 704959

Here, the unit’s place digit is 9;

∴ Unit’s place digit of n must be 9.

Let us split the number 704959 as 704 and 959

Now, 8³ < 704 < 9³

∴ 80³ = 512000 < 166000 < 90³ = 729000

Since, the unit’s place digit of n is 9, the only possible number is 89.

Also, 89³ = 704969

(i) 331776

Since, 6³ = 216 < 331 < 7³ = 343

∴ 60³ = 216000 < 331000 < 70³ = 343000

⇒ 68³ = 314432, 69³ = 328509, 70³ = 343000

So nearest integer to the cube root of 331776 is 69

(ii) 46656

Since, 3³ = 27 < 46 < 4³ = 64

∴ 30³ = 27000 < 46000 < 40³ = 64000

⇒ 31³ = 29791, 32³ = 32768, 35³ = 42875, 36³ = 46656

So nearest integer to the cube root of 46656 is 36

(iii) 373248

Since, 7³ = 343 < 373 < 8³ = 512

∴ 70³ = 343000 < 373000 < 80³ = 512000

⇒ 71³ = 357311, 72³ = 357911, 73³ = 389017

So nearest integer to the cube root of 737248 is 72

(1) 5² = 5 × 5 = 25

(2) 8² = 8 × 8 = 64

(3) 2² = 2 × 2 = 4

(4) (-6)² = -6 × -6 = 36

(5) (-22)² = -22 × -22 = 484

(6) 12² = 12 × 12 = 144

Let us write the square of numbers starting with the square of 1:

1, 4, 9,16,25,36,49,64,81,100

121,144,169,196,225,256,289,324,361,400, 441,484, 529

We see that 484 is the last square that appears from 1 to 500. Also, (22)² = 22 × 22 = 484.

Hence, the number of perfect squares from 1 to 500 is 22.

A. 1² = 1, 11² = 121, and so on. The last digit of these perfect squares is 1.

C. 5² = 25, 15² = 225, and so on. The last digit of these perfect squares is 5.

D. 3² = 9, 13² = 169, and so on. The last digit of these perfect squares is 9.

Hence, the last digit of a perfect square can never be 3.

The number of zeroes of a number gets doubled in its perfect square. For example:

10² = 100, 100² = 10000, and so on.

So, if a number ends in 5 zeroes, its square ends in 10 zeroes.

On dividing a perfect square by 8, we get 0, 1 and 4 as the remainder.

So, the remainder among when a per feet square is divided by 8 could be 1.

The 6^th triangular number = 1 + 2 + 3 + 4 + 5 + 6 = 21

Squares of all integers from -10 to 5:

(-10)² = -10 × -10 = 100

(-9)² = -9 × -9 = 81

(-8)² = -8 × -8 = 64

(-7)² = -7 × -7 = 49

(-6)² = -6 × -6 = 36

(-5)² = -5 × -5 = 25

(-4)² = -4 × -4 = 16

(-3)² = -3 × -3 = 9

(-2)² = -2 × -2 = 4

(-1)² = -1 × -1 = 1

0² = 0 × 0 = 0

1² = 1 × 1 = 1

2² = 2 × 2 = 4

3² = 3 × 3 = 9

4² = 4 × 4 = 16

5² = 5 × 5 = 25

There are 11 distinct numbers.

We know that if two numbers have same unit place then the unit place of their squares is also same.

4² = 16, unit place of 4² = 6

5² = 25, unit place of 5² = 5

9² = 81, unit place of 9² = 1

24: 4² = 16, unit place of 24² = 6

17: 7² = 49, unit place of 17² = 9

76: 6² = 36, unit place of 76² = 6

34: 4² = 16, unit place of 34² = 6

52: 2² = 4, unit place of 52² = 4

33: 3² = 9, unit place of 33² = 9

2319: 9² = 81, unit place of 2319² = 1

18: 8² = 64, unit place of 18² = 4

3458: 8² = 64, unit place of 3458² = 4

3453: 3² = 9, unit place of 3453² = 9

All the numbers from 400 to 425 which end in 2, 3, 7 or 8 are

402, 403, 407, 408, 412, 413, 417, 418, 422, 423

We know that every square has one of these as its unit place digit 1, 4, 9, 6, 5, 0. So, none of these is a perfect square.

We know that:

11² = 121

111² = 12321

1111² = 1234321

We see a pattern according to which

(111111111)² = 12345678987654321

whose digit sum is 81.

(i) Given that x² + y² = z²

where x = 4 and y = 3

⇒ z²= 4² + 3²

⇒ z² = 16 + 9

⇒ z² = 25

⇒ z = ±5

(ii) Given that x² + y² = z²

where x = 5 and z = 13

⇒ y²= 13² - 5²

⇒ y² = 169 - 25

⇒ y² = 144

⇒ y = ±12

(iii) Given that x² + y² = z²

where y = 15 and z = 17

⇒x²= 17² - 15²

⇒x² = 289 - 225

⇒x² = 64

⇒ x = ±8

Let the number of persons be x.

According to the question,

Money given to each person = Rs x

⇒ No. of persons × money given to each person = Total sum distributed

⇒x × x = 2304

⇒x² = 2304

∵ 48² = 2304

⇒ x = 48

Each person gets Rs 48.

Consider m*n = m² + n².
(i)
Yes, ℕ is closed under *.
This is because, for any natural number m, m² is also a natural number.
Further, on adding two natural numbers, we get a natural number only.
So, if m and n belong to ℕ then m² + n² also belongs to ℕ.

(ii) Commutative means x*y = y*x
where x and y belongs to ℕ

⇒m*n = m² + n²

And n*m = n² + m²

⇒ m² + n² = n² + m²

⇒m*n = n*m

Hence, * is commutative on ℕ

(iii) Associative means (x*y)*z = x*(y*z)
where x, y and z belongs to ℕ

m*n = m² + n²

Further,

(m*n)*o = (m² + n²)*o = (m² + n²)² + o²

Similarly, n*o = n² + o²

Further,

m*(n*o) = m*(n² + o²) = m² + (n²+ o²)²

⇒ (m² + n²)² + o² ≠ m² + (n² + o²)²

⇒ (m*n)*o ≠ m*(n*o)

Hence, * is not associative.

(iv) An identity element is a special type of element of a set with respect to a binary operation on that set,
which leaves other elements unchanged when combined with them.

For *, let k be the identity element
then m*k = m² + k² = k² + m² = m²

This is possible only when k = 0 but k needs to be a natural number.

Hence, * does not have an identity element.

We know that the integers that satisfy

A² + B² = C²

are called Pythagorean triplets.

So, all perfect squares from 1 to 500, each of which is a sum of two perfect squares are –

25 = 9 + 16

⇒ 5² =3²+ 4²

100 = 36 + 64

⇒10² =6²+ 8²

169 = 25 + 144

⇒13² =5²+ 12²

289 = 64 + 225

⇒17² =8²+ 15²

Given area of a square = 7396 m²

We know that the area of a square with side s = s²

⇒ s² = 7396

⇒ s = √7396

⇒ s = 86 m

Also, the perimeter of a square = 4 × side

⇒ Perimeter = 4 × 86 m

⇒ Perimeter = 344 m

We are required to find two perfect squares such that 1010 can be written as a difference of two perfect squares.

This means 1010 = A²- B²

∵1010 is even number

∴ Either A and B are even numbers or odd numbers

So, A² - B² is divisible by 4 but 1010 is not divisible by 4 because 1010 = 10 × 101 = 2 × 5 × 101

Hence, 1010 cannot be expressed as a difference of two perfect squares.

Let us find the remainder by dividing each cube by 7.

Hence, the possible remainders are 0, 1, and 6.

To find the least perfect square which leaves the remainder 1 when divided by 7 as well as by 11:

L.C.M. of 7 and 11 = 77

Then, required number is of the form = 77x + 1 where x = 1, 2, 3, 4, and so on.

34² = 1156

Hence, 1156 is the least perfect square which leaves the remainder 1 when divided by 7 as well as by 11.

Hence, 4 and 16 are the two smallest perfect squares whose product is a perfect cube.

Proper Factor is a factor of a number other 1 and itself.

Proper factors of 48 = 2, 3, 4, 6, 8, 12, 16, 24

Proper Multiple is a multiple other than itself.

Proper Multiples of 48 = 96, 144, 192, 240, 288, 336, 384 ….

A proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square are:

4 + 96 = 100 = 10²

4 + 192 = 196 = 14²

16 + 240 = 256 = 16²

16 + 384 = 400 = 20²

Hence, there are infinitely many such pairs