A2ZLY

Chapter 1:Β A Square and A Cube

by

In this chapter, we explore squares, cubes, and their properties. Our Class 8 Maths Ganita Prakash NCERT Solutions provide step-by-step guidance to calculate areas, volumes, and understand the relationship between numbers easily.

Textbook Page 1 – 3

Q. Before the process begins, Khoisnam realises that he already knows which lockers will be open at the end. How did he figure out the answer?
Hint: Determine the number of times each locker is switched (toggled).

Solution:

He observed that the number of times a locker is toggled corresponds to the number of factors of its number. A locker toggled an odd number of times will remain open, while one toggled an even number of times will be closed. Since only perfect squares have an odd number of factors, the lockers numbered with perfect squares (1, 4, 9, 16, …) are the ones left open.

Q. Does every number have an even number of factors?
Solution:

Not every number has an even count of factors. Only perfect squares have an odd count, since one of their factors is repeated β€” the square root, which multiplied by itself gives the number.

Q. Can you use this insight to find more numbers with an odd number of factors?
Solution:

Q. Write the locker numbers that remain open.
Solution:

The lockers numbered with perfect squares β€” 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 β€” will remain open.

Q. Which are these five lockers?
Solution:

Lockers that are toggled exactly twice are the prime-numbered ones, as each prime has only two factors: 1 and itself. Hence, the code is 2-3-5-7-11.

Textbook Page 4

Q. Find the squares of the first 30 natural numbers and fill in the table below.

Solution:

Q. What patterns do you notice? Share your observations and make conjectures.
Solution:

Every perfect square ends with one of the digits 0, 1, 4, 5, 6, or 9, and never with 2, 3, 7, or 8.

Q. If a number ends in 0, 1, 4, 5, 6, or 9, is it always a square?
Solution:

A number cannot be confirmed as a perfect square simply by its unit digit. However, the unit digit can help us rule out some cases. If a number ends in 2, 3, 7, or 8, it is definitely not a perfect square. For instance, 26 ends with 6, yet it is not a perfect square.

Q. Write 5 numbers such that you can determine by looking at their unit digit that they are not squares.
Solution:

Here are 5 such numbers whose unit digit shows they are not perfect squares:

  • 23 (ends with 3)
  • 47 (ends with 7)
  • 58 (ends with 8)
  • 92 (ends with 2)
  • 133 (ends with 3)

Textbook Page 5

Q. Which of the following numbers have the digit 6 in the units place?
(i) 382 (ii) 342 (iii) 462 (iv) 562 (v) 742 (vi) 822
Solution:
(i) 382
38 β†’ Units digit β†’ 8
8 Γ— 8 = 64 (ends in 4)
So, 382 does not end in 6.

(ii) 342
34 β†’ Units digit β†’ 4
4 Γ— 4 = 16 (ends in 6)
So, 342 ends in 6.

(iii) 462
46 β†’ Units digit β†’ 6
6 Γ— 6 = 36 (ends in 6)
So, 462 ends in 6.

(iv) 562
56 β†’ Units digit β†’ 6
6 Γ— 6 = 36 (ends in 6)
So, 562 ends in 6.

(v) 742
74 β†’ Units digit β†’ 4
4 Γ— 4 = 16 (ends in 6)
So, 742 ends in 6.

(vi) 822
82 β†’ Units digit β†’ 2
2 Γ— 2 = 4 (ends in 4)
So, 822 ends in 4.

Thus, if a number ends with 4 or 6, its square will have 6 in the units place.
Therefore, numbers like 342, 462, 562, and 742 will have 6 as the unit digit of their squares.

Q. Consider the following numbers and their squares.

If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Solution:

The number of zeros at the end of a square is always twice the number of zeros at the end of the original number. Hence, if a number ends with 3 zeros, its square will end with 6 zeros.

Q. What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
Solution:

The square of a number always has twice as many zeros at the end as the original number. This rule holds true in every case, meaning a square can only end with an even number of zeros.

Q. What can you say about the parity of a number and its square?
Solution:

The square of an even number is always even, while the square of an odd number is always odd.

Textbook Page 7

Q. Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Solution:

There are exactly 2n numbers between n2 and (n+1)2.
For example, between 32 and 42, there are 2Γ—3=6 numbers.

Q. How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?

Solution:

The largest square less than 1000 is given by  312 = 961.

Q. Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.

Solution:

Textbook Page 9

Q. Find whether 1156 and 2800 are perfect squares using prime factorisation.
Solution:
(i) 1156 = (2 Γ— 2) Γ— (17 Γ— 17)
= 22 Γ— 172  = (2 Γ— 17)2  = (34)2
∴ √1156 = 34. (ii) 2800 = (2 Γ— 2) Γ— (2 Γ— 2) Γ— (5 Γ— 5) Γ— 7
= 22 Γ— 22 Γ— 52 Γ— 7
2800 is not a perfect square because the factors cannot be paired.

Textbook Page 10 – 11

Figure it Out

1. Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Solution:

A perfect square always ends with 0, 1, 4, 5, 6, or 9 in the unit’s place.

(i) 2032 ends with 2, so it is not a perfect square.
(ii) 2048 ends with 8, so it is not a perfect square.
(iii) 1027 ends with 7, so it is not a perfect square.
(iv) 1089 ends with 9, so it is a perfect square.

2. Which one among 642, 1082, 2922, 362 has the last digit 4?
Solution:
(i) 642
The unit digit of 64 is 4.
4Γ—4=16 β†’ last digit is 6.

(ii) 1082
The unit digit of 108 is 8.
8Γ—8=64 β†’ last digit is 4.

(iii) 2922
The unit digit of 292 is 2.
2Γ—2=4β†’ last digit is 4.

(iv) 362
The unit digit of 36 is 6.
6Γ—6=36β†’ last digit is 6.

Therefore, 1082 and 2922 have 4 as their last digits.

3. Given 1252 = 15625, what is the value of 1262?
(i) 15625 + 126
(ii) 15625 + 262
(iii) 15625 + 253
(iv) 15625 + 251
(v) 15625 + 252
Solution:
1252 = 15625
This means 15625 is the sum of the first 125 odd numbers.
1262 = 15625 + 127th odd number
127th odd number = (2 Γ— 127) – 1 = 252 – 1 = 251.
∴ 1262 = 15625 + 251
Hence, the correct answer is (iv) 15625 + 251.

4. Find the length of the side of a square whose area is 441 m2.
Solution:

Area of the square = side Γ— side = 441 m2
(side)2=441
side=441=Β±21 m

Since the side of a square cannot be negative, the length of the side is 21 m.

5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Solution:

The LCM of 4, 9, and 10 is 180.
Thus, the smallest number divisible by 4, 9, and 10 is 180.

Prime factorization:
180=2Γ—2Γ—3Γ—3Γ—5
180=22Γ—32Γ—5

Here, 5 is unpaired, so 180 is not a perfect square.
To make it a perfect square, we multiply by 5.

Hence, the required smallest perfect square is:
180Γ—5=900.

6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Solution:

9408=2Γ—2Γ—2Γ—2Γ—2Γ—2Γ—3Γ—7Γ—7
9408=22Γ—22Γ—22Γ—72Γ—3

Here, 3 is unpaired.
So, 9408 is not a perfect square. To make it a perfect square, we multiply by 3.

9408Γ—3=28224 which is a perfect square.

28224=22Γ—22Γ—22Γ—32Γ—72 so, square root of 28224=2Γ—2Γ—2Γ—3Γ—7=8Γ—21=168

Thus, 28224=168.

7. How many numbers lie between the squares of the following numbers?
(i) 16 and 17 (ii) 99 and 100
Solution:
There are 2n numbers between n2 and (n + 1)2 .
(i) 162 and 172
Here, n = 16 and (n + 1) = 17
Therefore, the numbers between 162 and 172 = 2n = 2 Γ— 16 = 32.

(ii) 992 and 1002
Here, n = 99 and (n + 1) = 100
Therefore, the numbers between the squares 992 and 1002 = 2n = 2 Γ— 99 = 198.

8. In the following pattern, fill in the missing numbers:
12 Γ— 22 Γ— 22 = 32
22 Γ— 32 Γ— 62 = 72
32 Γ— 42 Γ— 122 = 132
42 Γ— 52 Γ— 202 = (___)2
92 Γ— 102 Γ— (___)2 = (___)2
Solution:
12 Γ— 22 Γ— 22 = 32
22 Γ— 32 Γ— 62 = 72
32 Γ— 42 Γ— 122 = 132
42 Γ— 52 Γ— 202 = (21)2
92 Γ— 102 Γ— (90)2 = (91)2

9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.

Solution:

Number of squares in each row = 9
Number of squares in each column = 9
Total squares in the figure = 9Γ—9=81

Each square contains 5Γ—5=25 tiny squares.
So, the total number of tiny squares in the figure = 25Γ—81=2025.

Prime factorization of 2025:
2025 = (3 Γ— 3) Γ— (3 Γ— 3) Γ— (5 Γ— 5)= 32 Γ— 32 Γ— 52
2= 34 Γ— 52.

Textbook Page 12

Q. Complete the table below.

Solution:

Q. What patterns do you notice in the table above?
Solution:

(i) The cube of a number ending in 1 also ends in 1.
(ii) A number ending in 2 has a cube ending in 8, and a number ending in 8 has a cube ending in 2.
(iii) A number ending in 3 has a cube ending in 7, and a number ending in 7 has a cube ending in 3.
(iv) Numbers ending in 4, 5, 6, 9, or 0 retain the same unit digit when cubed.

Q. We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?
Solution:
The last digits of cubes can be any digit from 0 to 9.

Textbook Page 13

Q. Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Solution:

1-digit cubes: 13=1, 23=8
Count: 2 cubes β†’ (1, 8)

2-digit cubes: 33=27, 43=64
Count: 2 cubes β†’ (27, 64)

3-digit cubes: 53=125, 63=216, 73=343, 83=512, 93=729
Count: 5 cubes β†’ (125, 216, 343, 512, 729)

Thus, the number of perfect cubes increases with larger numbers.
Unlike squares, cubes grow faster, so fewer of them are found in smaller ranges.

Q. Can a cube end with exactly two zeroes (00)? Explain.
Solution:

No, a cube cannot end with exactly two zeros, because zeros in a cube always appear in multiples of three. If a number ends with one zero, its cube will end with three zeros.

Q. The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes.
Solution:
4104:
23 + 163 = 8 + 4096 = 4104
93 + 153 = 729 + 3375 = 4104 13832:
23 + 243 = 8 + 13824 = 13832
183 + 203 = 5832 + 8000 = 13832

Textbook Page 14

Q. Can you tell what this sum is without doing the calculation?
91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109.
Solution:

This series consists of 10 consecutive odd numbers, and their sum is 103=1000.

Textbook Page 15

Q. Find the cube roots of these numbers:
(i) βˆ›64 = (ii) βˆ›512 = (iii) βˆ›729 =
Solution:
(i) βˆ›64
64 = (2 Γ— 2 Γ— 2) Γ— (2 Γ— 2 Γ— 2) = 23 Γ— 23 = (2 Γ— 2)3 = 43
∴ βˆ›64 = 4. (ii) βˆ›512
512 = (2 Γ— 2 Γ— 2) Γ— (2 Γ— 2 Γ— 2) Γ— (2 Γ— 2 Γ— 2)  = 23 Γ— 23 Γ— 23 = (2 Γ— 2 Γ— 2)3 = 83
∴ βˆ›512 = 8. (iii) βˆ›729
729 = (3 Γ— 3 Γ— 3) Γ— (3 Γ— 3 Γ— 3) = 33 Γ— 33 = (3 Γ— 3)3 = 93
∴ βˆ›729 = 9.

Textbook Page 16 – 17

Figure it Out

1. Find the cube roots of 27000 and 10648.
Solution:
(i) βˆ›27000
Prime factorization of 27000 = 2 Γ— 2 Γ— 2 Γ— 3 Γ— 3 Γ— 3 Γ— 5 Γ— 5 Γ— 5
27000 = 23 Γ— 33 Γ— 53  = (2 Γ— 3 Γ— 5)3 = 303
∴ βˆ›27000 = 30.

(ii) βˆ›10648
Prime factorisation of 10648 = 2 Γ— 2 Γ— 2 Γ— 11 Γ— 11 Γ— 11
10648 = 23 Γ— 113 = (2 Γ— 11)3 = 223
∴ βˆ›10648 = 22.

2. What number will you multiply by 1323 to make it a cube number?
Solution:

Prime factorization of 1323=3Γ—3Γ—3Γ—7Γ—7.
Here, 7 does not form a complete triplet.

So, 1323 is not a perfect cube. To make it a cube, we multiply by 7:

1323Γ—7=3Γ—3Γ—3Γ—7Γ—7Γ—7=9261

which is a perfect cube.

Hence, the number that must be multiplied by 1323 to make it a cube is 7.

3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
Solution:

StatementTrue/FalseReason
(i)FalseCube of odd number is odd.
(ii)FalseCubes of numbers ending in 2 or 8 can end with 8.
(iii)FalseCube of 2-digit number β‰₯ 1000 β†’ 4 digits.
(iv)FalseCube of 2-digit number ≀ 970,299 β†’ max 6 digits.
(v)FalseOnly perfect squares have odd number of factors.

4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
Solution:

Yes! We can guess cube roots without full factorisation by looking at the last digit and the approximate size of the number.

  1. 1331
    • Last digit is 1 β†’ cube root must end in 1.
    • 103=1000, 113=1331.
      β†’ Cube root = 11.
  2. 4913
    • Last digit is 3 β†’ cube root must end in 7 (since 73=343 ends in 3).
    • 173=4913.
      β†’ Cube root = 17.
  3. 12167
    • Last digit is 7 β†’ cube root must end in 3 (since 33=27 ends in 7).
    • 233=12167.
      β†’ Cube root = 23.
  4. 32768
    • Last digit is 8 β†’ cube root must end in 2 (since 23=8 ends in 8).
    • 323=32768.
      β†’ Cube root = 32.

So the cube roots are: 1331 β†’ 11, 4913 β†’ 17, 12167 β†’ 23, 32768 β†’ 32.

5. Which of the following is the greatest? Explain your reasoning.
(a) 673 β€“ 663
(b) 433 β€“ 423
(c) 672 β€“ 662
(d) 432 β€“ 422
Solution:

(a) 673βˆ’663 is the greatest.

Reason: For consecutive integers n and nβˆ’1,

  • n3βˆ’(nβˆ’1)3=3n2βˆ’3n+1
  • n2βˆ’(nβˆ’1)2=2nβˆ’1

Evaluating:

  • (a) 673βˆ’663=3β‹…672βˆ’3β‹…67+1=13,267
  • (b) 433βˆ’423=3β‹…432βˆ’3β‹…43+1=5,419
  • (c) 672βˆ’662=133
  • (d) 432βˆ’422=85

So the order is (a) > (b) > (c) > (d).

error: Content is protected !!