Uncategorized

# Squares, Square roots, Cubes and Cube roots

1.2.1 Introduction

Look at the numbers of the form 1, 4, 9, 16, 25 and so on. We can recognise it as 4 = 2 x 2, 9 = 3 x 3 and 16 = 4 x 4. We see that each number is product of two equal numbers. Similarly, we see that, 8 = 2 x 2 x 2, 64 = 4 x 4 x 4, each number is product of three equal numbers.

1.2.2 Perfect squares:

Observe that, 1 = 1, 4 = 2 x 2, 9 = 3 x 3, 100 = 10 x 10.

If a is an integer and b = a x a, we say b is a perfect square.

Hence 1, 4, 9, 16, 25 are all perfect squares. Since 0 = 0 x 0, we see that 0 is a perfect square.

If a is an integer, we denote a x a = a2. We read it as square of a or simply a square. Thus 36 = 62 and 81 = 92. Thus a perfect square is of the m2, where m is an integer.

For example, 4 = 2 x 2 and 4 = (-2) x (-2); in the second representation, we again have equal integers, but negative this time.

Thus a perfect square is either equal to 0 or must be a positive integer. It can be a negative integer.

Look at the following table

 a 1 2 3 8 -7 -12 20 -15 a2 1 4 9 64 49 144 400 225

We see that squares of 2, 8, -12, 20 are even numbers and the squares of 1, 3, -7, -15 are odd numbers.

Statement 1: The Square of an even integer is even and the square of an odd integer is odd.

Consider first 10 perfect squares,

 12 22 32 42 52 62 72 82 92 102 1 4 9 16 25 36 49 64 81 100

If we observe that the units place in these squares are 1, 4, 9, 6, 5, 6, 9, 4, 1 and 0 in that order. Thus only the digits which can occupy digit’s place in perfect squares are 1, 4, 5, 6 and 9.

Statement 2: A perfect square always ends in one of the digits 0, 1, 4, 5, 6 and 9. If the last digit of a number is 2, 3, 7 or 8, it cannot be a perfect square.

Exercise 1.2.2

1. Express the following statements mathematically:

i) square of 4 is 16

Solution:

42 = 16

ii) square of 8 is 64

Solution:

82 = 64

iii) square of 15 is 225

Solution:

152 = 225

1. Identify the perfect squares among the following numbers:

1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000

Solution:

Perfect squares among the above are:

1, 36, 49, 81, 169, 625, 125, 900, 100

1. Make a list of all perfect squares from 1 to 500

Solution:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484

1. Write the 3 digit numbers ending with 0, 1, 4, 5, 6 , 9 one for each digit but one of them is a perfect square.

Solution:

We can take 200, 201, 204, 205, 206 and 209. None of these is perfect square lies between 142 = 196 and 152 = 225.

We can also take 300, 301, 304, 305, 306 and 309. None of these is a perfect square lies between 172 = 289 and 182 = 324.

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

Solution:

102 = 100, 112 = 121, 122 = 144, 132 = 169, 142 = 196, 152 = 225, 162 = 256, 172 = 289, 182 = 324, 192 = 361, 202 = 400

1.2.3 Some facts related to perfect squares

There are some nice properties about perfect squares. Let us study them here:

1. Look at following table:
 a 4 10 20 25 100 300 1000 a2 16 100 400 625 10000 90000 1000000 The number of zeros at the end of a2 0 2 2 0 4 4 6

We observe that, the number of zeros at end of a square is always an even number.

STATEMENT 3: /if a number has k zeros at the end, then its square ends in 2k zeros

Thus, if a number has odd number of zeros, it cannot be perfect square.

1. Look at the adjoining table:
 a a2 The remainder of a2 when divided by 3 The remainder of a2 when divided by 4 1 1 1 1 2 4 1 0 3 9 0 1 5 25 1 1 8 64 1 0 11 121 1 1 -6 36 0 0

We can see that, the remainder of a perfect square is either 1 or 0. we can also note that remainder of a perfect square when divided by 4 are either 0 or 1.

Since, when a number is divided by 3, the reminder used to be 0, 1 or 2. Similarly, when a number is divided by 4 the remainder used to be 0, 1, 2 or 3.

STATEMENT 4: The reminder of a perfect square when divided by 3 is either 0 or 1, but never be 2. The reminder of a perfect square when divided by 4 is either 0 or 1, but never be 2 and 3.

STATEMENT 5: When the product of 4 consecutive integers is added to 1, the resulting number is perfect square.

Ex:

(1 x 2 x 3 x 4) + 1 = 24 + 1 = 25 = 52

(5 x 6 x 7 x 8) + 1 = 1680 + 1 = 412

STATEMENT 6: The sum of the first n odd natural numbers is equal to n2, for every natural number n.

Ex:

1 = 1 = 12

1 + 3 = 4 = 22

1 + 3 + 5 = 9 = 32

1 + 3 + 5 + 7 = 16 = 42

1 + 3 + 5 + 7 + 9 = 25 = 52

STATEMENT 7: consider the number N = 1000….01, where zeros appear k times. (For example, for k = 6, you get N = 10000001; there are 6 zeros in the middle.) Then N2 = 1000…02000…01, where the number of zeros on both the sides of 2 is k.

Ex:

112 = 121

1012 = 10201

10012 = 1002001

100012 = 100020001 and so on.

STATEMENT 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Ex: The dots are now arranged in shapes. Now count the number of dots in each triangle. (Single dot is considered as the generate triangle.) They are 1, 3, 6, 10, 15, 21, 28, 36 and so on. These are called triangular numbers.

For nth triangular number, we form a triangle of dots with n- rows and each row contains as many points as index of that row.  If you want find the 8th triangular number, the number of points in the 8th triangle is

1 + 2 + 3 + 4+ 5 + 6 + 7 + 8 = 36

The first few triangular numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91.

Take any two consecutive triangular number and find their sum. For example : 10 + 15 = 25 = 52; 21 + 28 = 49 = 72. Which holds the statement 8.

Exercise 1.2.3

1. Find the sum 1 + 3 + 5 + ……. +51 (the sum of all odd numbers from 1 to 51) without actually adding them.

Solution:

[We have 5 odd numbers from 1 to 10, therefore there are 5×5 = 25 odd numbers are there from 1 to 50, we know 51 is also an odd number…. So, there are 25+1 = 26 ]

There are 26 odd numbers are there from 1 to 51.

Thus, 1 + 3 + 5 + …….. + 51 = 262 = 676.

1. Express 144 as a sum of 12 odd numbers.

Solution:

We know that, the sum of the first n odd natural numbers is equal to n2, for every natural number n.

Therefore, the sum of first 12 odd natural numbers is equal to 122 = 144.

Thus,

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 22 = 122 = 144.

1. Find the 14th and 15th triangular numbers, and find their sum. Verify the statement 8 for this sum.

Solution:

We know, statement 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Sum of 14th triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105

Sum of 15th triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120

Now, let us verify, statement 8: The sum of nth and (n+1)th triangular number is (n+1)2.

i.e., Sum of 14th triangular number(n) + sum of 15th triangular number(n+1) = 105 + 120

= 225 = 152 = (n+1)2, which holds the statement8.

1. What are the remainders of a perfect square when divided by 5?

Solution:

 a a2 The remainders of a2 when divided by 5 1 1 1 2 4 4 5 25 0 6 36 1 -7 49 4 11 121 1

The remainders of perfect squares when divided by 5 are 0, 1 and 4.

1.2.4 Methods of squaring a number

Many times it is easy to find the square of a number without actually multiplying the number to itself.

Consider 42 = 40 + 2

Thus,

422 = (40+2)2

= 402 + 2(40)(2) + 22

Now it is easy to recognise 402 = 1600; 2x40x2 = 160; 22 = 4.

Therefore, 422 = 1600 + 160 + 4 = 1764.

Example: Find 892

Solution: 892 = (80 + 9)2

= 802 + 2 x 80 x 9 + 92

= 6400 + 1440 + 81

= 7921

Exercise 1.2.4

1. Find the squares of:

i) 31

Solution:

31 = 30 + 1

312 = (30 + 1)2

= 302 + 2 x 30 x 1 +1

= 900 + 60 + 1

= 961

ii) 72

Solution:

72 = 70 + 2

722 = (70 + 2)2

= 702 + 2 x 70 x 2 + 22

= 4900 + 280 + 4

= 5184

iii) 37

Solution:

37 = 30 + 7

372 = (30 + 7)2

= 302 + 2 x 30 x 7 + 72

= 900 + 420 + 49

= 1369

iii) 166

Solution:

166 = 160 + 6

1662 = (160 + 6)2

= 1602 + 2 x 160 x 6 + 62

= 25600 + 1920 + 36

= 27556

2. Find the squares of:

i) 85

Solution:

852 = (80 + 5)2

= 802 + 2 x 80 x 5 + 52

= 1600 + 800 + 25

=2425

ii) 115

Solution:

1152 = (100 + 15)2

= 1002 + 2 x 100 x 15 + 152

= 10000 + 3000 + 225

= 13225

iii) 165

1652 = (160 + 5)2

= 1602 + 2 x 160 x 5 + 52

= 25600 + 1600 + 25

= 27225

1. Find the squares of 1468 by writing this as 1465+3

14682 = (1465 + 3)2

= (1465)2 + 2 x 1465 x 3 + 32

= 2146225 + 8790 + 9

= 2155024

1.2.5 Square roots

Consider the following perfect squares:

1 = 12, 4 = 22, 9 = 32, 16 = 42, 49 = 72, 196 = 142.

In each case the number is obtained by the product of two equal numbers. Here we say 1 is the square root of 1; 2 is the square root of 4; 3 is the square root of 9 and so on.

Suppose N is a natural number such that N= M2. The number M is called a square root of N.

We have seen earlier m2 = m x m = (-m) x (-m). Thus m2 has 2 roots m and –m. For example, 16 = 42 = (-4)2, thus both 4 and -4 are the roots of 16. Mathematically both 4 and -4 are accepted as the square root of 16. Thus,

Whenever the word square root is used, it is always meant to be the positive square root. The square root on n is denoted by √N.

Square root of a perfect square by factorization:

Example 1: Find the square root of 5929.

Solution: Thus, 7 x 7 x 11 x 11 = 5929

Therefore, [group it]

(7 x 11) x (7 x 11) = 77 x 77 = 772 = 5929.

Example 2: Find the square root of 6724.

Solution: Thus, 6724 = 2 x 2 x 41 x 41

6724 = (2 x 41) x (2 x 41)

= 82 x 82

Example 3: Find the smallest positive integer with which one has to divide 336 to get a perfect square.

Solution:

We observe that 336 = 2 x 2 x 2 x 2 x 3 x 7. Here both 3 and 7 occur only once. Hence we have to remove them to get a perfect square.

We divide 336 by 3 and 7

42.

The required least number is 21.

Exercise 1.2.5

1. Find the square roots of the following numbers by factorization:

i) 196

Solution:

196 = 2 x 2 x 7 x 7

= (2 x 7) x (2 x 7)

= 14 x 14

196 = 142

ii) 256

Solution:

256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2)

= 16 x 16

= 162

iii) 10404

Solution:

10404 = 2 x 2 x 3 x 3 x 17 x 17

= (2 x 3 x 17) x (2 x 3 x 17)

= (102) x (102)

= 1022

iv) 1156

Solution:

1156 = 2 x 2 x 17 x 17

= (2 x 17) x (2 x 17)

= 34 x 34

= 342

v) 13225

Solution:

13225 = 5 x 5 x 23 x 23

= (5 x 23) x (5 x 23)

= 115 x 115

= 1152

2. Simplify:

i) √100 + √36

Solution:

√100 + √36 = 10 + 6

= 16

ii) √(1360 + 9)

Solution:

√(1360 + 9) = √1369 = √(37 x 37) = 37

iii) √2704 + √144 + √289

Solution:

√2704 + √144 + √289 = √(52 x 52) + √(12 x 12) + √(17 x 17)

= 52 + 12 + 17

= 81

iv) √225 – √25

Solution:

√225 – √25 = √(15 x 15) – √(5 x 5)

= 15 – 5

= 10.

v) √1764 – √1444

Solution:

√1764 – √1444 = √(42 x 42) – √(38 x 38)

= 42 – 38

= 4

vi) √169 x √361

√169 x √361 = √(13 x 13) x √(19 x 19)

= 13 x 19

= 247

1. A square yard has area 1764m2. From a corner of this yard, other square part of area 784m2 is taken out for public utility. The remaining portion is divided in to 5 equal parts. What is the perimeter of each of these equal parts?

Solution:

Area of square = a2

Area of given square yard = 1764 m2

Area used for public utility = 784 m2

∴ Area of remaining portion = 1764 – 784 = 980 m2

If the area is divided into 5 parts

Then, the area of each Square = 980/5 = 196 m2

Length of each side = √ 196, a = 14m

Perimeter of square = 4a = 4 x 14 = 56m

4. Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square:

i) 847

Solution:

847 = 11 x 11 x 7

Here 7 occur only once. Hence we have to multiply by 7 them to get a perfect square.

ii) 450

Solution:

450 = 5 x 5 x 3 x 3 x 2

Here 2 occur only once. So we have multiply by 2 to get a perfect square.

iii) 1445

Solution:

1445 = 17 x 17 x 5

Here 5 occur only once. So we have multiply by 5 to get a perfect square.

iv) 1352

Solution:

1352 = 2 x 2 x 13 x 13 x 2

Here 2 occur only three times. So we have multiply by 2 to get a perfect square.

5. Find the largest perfect square factor of each of the following numbers:

i) 48

Solution:

48 = 2 x 2 x 2 x 2 x 3 = (2 x 2) x (2 x 2) x 3

= 4 x 4 x 3 = 16 x 3 = 42 + 3

Therefore, 16 is the largest perfect square factor of 48.

ii) 11280

Solution:

11280 = 2 x 2 x 2 x 2 x 3 x 47

Therefore, 16 is the largest perfect square factor of 11280

iii) 729

Solution:

729 = 3 x 3 x 3 x 3 x 3 x 3

= (27) x (27)

= 272

Therefore, 729 is the largest perfect square factor of 729.

iv) 1352

Solution:

1352 = 2 x 2 x 2 x 13 x 13

= (2 x 13) x (2 x 13) x 2

= 26 x 26 x 2

Therefore, 676 is the largest perfect square of 1352.

1. Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs?

Solution:

The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Multiples of 48 are = 48, 96 , 144, 192 , 240, 280 ….

1) 48 + 1 = 49 = 72

2 ) 192 + 4 = 196 = 142

3)288 + 1 = 289 = 172

4 ) 96 + 4 = 100 = 102

5)96 + 48 = 144 = 122

6) 48 + 16 = 64 = 82

7)240 + 16 = 256 = 162

Yes, we can prove that there are infinitely many such fairs.

1.2.6 Perfect squares near to a given number

Example 5: If the area of a square is 90 cm2, what is its side-length rounded to the nearest integer?

Solution:

Since, area A = a2, we have a2 = 90.

But 81 < 90 < 100 and 81 is the nearer to 90 than 100. Hence, the nearest integer to √90 is √81 = 9.

Example 6: A square piece of land has area 112m2. What is the closest integer which approximates the perimeter of the land?

Solution:

If a is the side length of a square, its perimeter is 4a. We know that a2 = 112 m2.

Hence, (4a)2 = 16a2 = 16 x 112 = 1792

But 422 = 1764 < 1792 < 1849 = 432 .

!764 is nearer to 1792 than 1849. Therefore, the integer approximation for √1792 is 42. The approximate value of perimeter is 42cm.

Exercise 1.2.6

1. Find the nearest integer to the square root of the following numbers:

i) 232

Solution:

We have, 152 = 225 and 162 = 256.

We know, 225 < 232 < 256

152 < 232 < 162

∴ Square root of 232 is nearest to 15

ii) 600

Solution:

We have, 242 = 576 and 252 = 625.

We know, 576 < 600 < 625

242 < 600 < 252

∴ Square root of 600 s nearest to 24.

iii) 728

Solution:

We have, 262 = 676 and 272 = 729

676 < 728 < 729

262 < 728 < 272

∴ Square root of 728 is nearest to 27.

iv) 824

Solution:

We have, 282 = 784 and 292 = 841

784 < 824 < 841

282 < 824 < 292

∴Square root of 824 is nearest to 29

v) 1729

Solution:

We have, 412 = 1681 and 422 = 1764

1681 < 1729 < 1764

412 < 1729 < 422

∴Square root of 1729 is nearest to 42

1. A piece of land is in the shape of a square and its area is 1000m2 . This has to be fenced using barbed wire. The barbed wire is available only in integral lengths. What is the minimum length of the barbed wire that has to be bought for this purpose.

Solution:

Area of the land = 1000m2

Area = a2 = 1000m2

Perimeter = 4a

Squaring , (Perimeter)2 = ( 4a)2 = 16a2 = 16 x 1000

(perimeter)2 = 16,000

We have, square of the perimeter i.e., 16000. now we have to find the square root of 16000. As 16000 does not have a perfect square root, let us find the square root nearest to square root of 16000.

We know, 1262  = 15876 and 1272 = 16129.

Then, 15876 < 16000 < 16129

1262 < 16000< 1272

∴ Nearest number is 126 But this is not enough to cover the hard.

∴Length of barbed wire required = 127 m.

1. A student was asked to find √961. He read it wrongly and found √691 to the nearest integer. How much small was his number forms the correct answer?

Solution:

We know, √ 961 = 31

It is given that student read it wrongly and found the result for √691. So, we have to find out the difference between  √961 and √691.

√676 < √691 < √729

262 < √691 < 272

26 is nearest number to √ 691

∴ difference = 31 – 26 = 5

1.2.7 Perfect cubes

Let,

1 = 1 x 1 x 1;

8 = 2 x 2 x 2;

27 = 3 x 3 x 3;

125 = 5 x 5 x 5;

We observe that, each number is written as a product of 3 equal integers.

We say that an integer N is a perfect cube if N can be written as a product of three equal integers. If N = m x m x m. we say N is the cube of m and write N = m3 (read as cube of m or simply m-cube)

Consider a few more examples:

(-4) x (-4) x (-4) = -64 = (-4)3

(-5) x (-5) x (-5) = -125 = (-5)3

We see that the negative numbers are also perfect cubes.

Example 7: Find the cube of 6.

Solution:

63 = 6 x 6 x 6 = 216

Example 8: What is the cube of 20?

Solution:

203 = 20 x 20 x 20 = 400 x 20 = 8000.

Example 9: If a cube has side length 10cm, what is its volume?

Solution:

Volume, V = 10 x 10 x 10 = 1000m3.

Exercise 1.2.7

1. Looking at the pattern, fill in the gaps in the following:
 2 3 4 -5 —— 8 —— 23 = 8 33 = —- —- = 64 —- = —- 63= —– —- = —- — = -729

Solution:

 2 3 4 -5 6 8 -9 23 = 8 33 = 27 43 = 64 (-5)3 = -125 63= 216 83 = 512 (-9)3 = -729

1. Find the cubed of the first five odd natural numbers and the cubes of the first five even natural numbers. What can you say about parity of the odd cubes and even cubes?

Solution:

 Odd cubes Even cubes 13 = 1 23 = 8 33 = 27 43 = 64 53 = 125 63 = 216 73 = 343 83 = 512 93 = 729 103 = 1000

Parity:

Cubes of odd numbers is always odd.

Cubes of even numbers is always even.

1. How many perfect cubes you can find from 1 to 100 ? How many from – 100 to 100?

Solution:

We have 4 cubes from 1 to 100, those are,

13 = 1, 23 = 8, 33 = 27, 43 = 64. The next cube will be 53 = 125, which is larger than 100. so we have only 4 cubes from 1 to 100.

We have 8 cubes from -100 to 100. Those are, (-1)3 = 1, (-2)3 = 8, (-3)3 = 27, (-4)3 = 64 and 13 = 1, 23 = 8, 33 = 27, 43 = 64.

1. How many perfect cubes are there from 1 to 500? How many are perfect square among cubes?

Solution:

We have 7 cubes from 1 to 500, i.e., 13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343.

1. Find the cubes of 10, 30 , 100, 1000. What can you say about the zeros at the end?

Solution:

103 = 1000

303 = 27000

1003 = 1000000

10003 = 10000000000

The number of zero of a cube are 3 times, the no. of zero of numbers

1. What are the digits in the unit’s place of the cubes 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10? Is it possible to say that a number is not a perfect cube by looking at the digit in unit’s place of the given number, just like you did for squares?

Solution:

 a a3 digit in the units place 1 1 1 2 8 8 3 27 7 4 64 4 5 125 5 6 216 6 7 343 3 8 512 2 9 729 9 10 1000 0

No, its not possible to tell a number is not a perfect cube by looking at the digit in unit’s place of the given number.

1.2.8 Cube root

If N is number and n is another number such that N = n3, we say n is the cube root of N and write n =   .

Example 12: Find the cube root of 216 by factorization.

Solution:

216 = 2 x (108) = 2 x 2 x 54 = 2 x 2 x 2 x 27 = 2 x 2 x 2 x 3 x 9 = 2 x 2 x 2 x 3 x 3 x 3

216 = (2 x 3) x (2 x 3) x (2 x 3)

216  = 6 x 6 x 6 = 63

Example 13: find the cube root of -17576 using factorization.

Solution:

-17576 = 2 x (-8788) = 2 x 2 x (-4394) = 2 x 2 x 2 x (-2197) = 2 x 2 x 2 x (-13) x (-13) x (-13)

= [2 x (-13)] x [2 x (-13)] x [2 x (-13)]

= (-26) x (-26) x (-26)

= (-26)3

Example 15: Find the cube root of 103823.

Solution:

Here unit in the digits place is 3. If n3 = 103823. Then, the unit in the digits place of n must be 7.

Let us split this as 103 and 823.

We observe that, 43 = 64 < 103 < 125 = 53.

Hence 403 = 64000 < 103823 < 125000 = 503. Hence n must lie between 40 and 50. Since the unit in the digits place is 7, therefore n must be 47.

473 = 103823.

Exercise 1.2.8

1. Find the cube root by prime factorization.

i) 10648

Solution:

10648 =  2 x 5324

= 2 x 2 x 2662

= 2 x 2 x 2 x 1331

= 2 x 2 x 2 x 11 x 121

= 2 x 2 x 2 x 11 x 11 x 11

= (2 x 11) x (2 x 11) x (2 x 11)

= 22 x 22 x 22

= 223.

ii) 46656

Solution:

46656 = 2 x (23328)

= 2 x 2x (11664)

= 2 x 2 x 2 x (5832)

= 2x2x2x2x(2916)

= 2x2x2x2x2x(1458)

= 2x2x2x2x2x2x(729)

= 2x2x2x2x2x2x9x(81) = 2x2x2x2x2x2x9x9x9 = (2x2x9)x(2x2x9)x(2x2x9) = 36 x 36 x36

= 363.

iii)      15625

Solution:

15625 = 5 x (3125)

= 5 x 5 x (625)

= 5 x 5 x 5 x (125)

= 5 x 5 x 5 x 5 x (25)

= 5 x 5 x 5 x 5 x 5 x 5

= (5×5)x(5×5)x(5×5)

= (25) x (25) x (25)

= 253

1. Find the cube root of the following by looking at the last digit and using estimation.

i) 91125

Solution:

Here unit in the digits place is 5. If n3 = 91125. Then, the unit in the digits place of n must be 5.

Let us split this as 91 and 125.

We observe that, 43 = 64 < 91 < 125 = 53.

Hence 403 = 64000 < 91125 < 125000 = 503. Hence n must lie between 40 and 50. Since the unit in the digits place is 5, therefore n must be 45.

ii) 166375

Solution:

Here unit in the digits place is 5. If n3 = 166375. Then, the unit in the digits place of n must be 5.

Let us split this as 166 and 375.

We observe that, 53 = 125 < 166 < 216 = 63.

Hence 503 = 12500 < 166375 < 216000 = 603. Hence n must lie between 50 and 60. Since the unit in the digits place is 5, therefore n must be 55.

iii)      704969

Solution:

Here unit in the digits place is 9. If n3 = 704969. Then, the unit in the digits place of n must be 9.

Let us split this as 704 and 969.

We observe that, 83 = 512 < 704 < 729 = 93.

Hence 803 = 512000 < 704969 < 729000 = 903. Hence n must lie between 80 and 90. Since the unit in the digits place is 9, therefore n must be 89.

1. Find the nearest integer to the cube root of each of the following.

i) 331776

Solution:

For easy simplification let us split 331776 as 331 and 776,

63 = 216 < 331 < 343 = 73

603 = 216000 < 331776 < 343000 = 703.

Now it is closer to 703 than 603.

Let us go for more accurate, 693 = 328509 < 331776 < 34300 = 703.

Therefore , 331776 is closest to 693.

ii46656

Solution:

For easy simplification let us split 46656 as 46 and 656,

33 = 27 < 46 < 64 = 43

303 = 27000 < 46656 < 64000 = 403.

Now it is closer to 403 than 303.

The number closer to 40 than 30 are 36, 37, 38, 39. Let us go through one by one.

363 = 46656, satisfies the condition.

iii.      373248

Solution:

For easy simplification let us split 372248 as 373 and 248,

73 = 343 < 373 < 512 = 83

703 = 343000 < 373248 < 512000 = 803.

Now, it is closer to 703 than 803.

The numbers which are closer to 70 than 80 are : 71, 72, 73, 74, 75

Let us go for more accurate, 713 = 357911 < 373248 < 373248 = 723.

Therefore, 373248 is closest to 723.