Question 1

Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656

Accepted Answer

A perfect cube is a number that can be expressed as the cube of an integer.

(i) 216 = 6^3, so it is a perfect cube.
(ii) 128 = 2^7, not a perfect cube because the exponent 7 is not a multiple of 3.
(iii) 1000 = 10^3, so it is a perfect cube.
(iv) 100 is not a perfect cube because it cannot be expressed as n^3 for any integer n.
(v) 46656 = 36^3, so it is a perfect cube.

Therefore, numbers which are not perfect cubes are 128 and 100. — Check if the number can be expressed as n^3 where n is an integer.
- 216 = 6^3
- 128 = 2^7 (not multiple of 3)
- 1000 = 10^3
- 100 = 10^2 (not cube)
- 46656 = 36^3

Question 2

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100

Accepted Answer

To find the smallest number to multiply so that the product is a perfect cube, prime factorize each number and adjust the powers to multiples of 3.

(i) 243 = 3^5
To make power of 3 a multiple of 3, multiply by 3^(3 - (5 mod 3)) = 3^(3 - 2) = 3^1 = 3
Answer: 3

(ii) 256 = 2^8
Next multiple of 3 after 8 is 9, so multiply by 2^(9-8) = 2^1 = 2
Answer: 2

(iii) 72 = 2^3 × 3^2
3^2 needs one more 3 to make power 3, multiply by 3
Answer: 3

(iv) 675 = 3^3 × 5^2
5^2 needs one more 5 to make power 3, multiply by 5
Answer: 5

(v) 100 = 2^2 × 5^2
Both 2^2 and 5^2 need one more factor each to make power 3, multiply by 2 × 5 = 10
Answer: 10 — Prime factorize each number and find the missing powers to reach multiples of 3.
Multiply by the product of primes raised to these missing powers.

Question 3

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704

Accepted Answer

To find the smallest number to divide so that the quotient is a perfect cube, prime factorize each number and remove the excess powers to make powers multiples of 3.

(i) 81 = 3^4
Remove one 3 to get 3^3, so divide by 3
Answer: 3

(ii) 128 = 2^7
Remove one 2 to get 2^6 (6 is multiple of 3), so divide by 2
Answer: 2

(iii) 135 = 3^3 × 5^1
5^1 is not multiple of 3, remove 5 to get 3^3
Divide by 5
Answer: 5

(iv) 192 = 2^6 × 3^1
3^1 is not multiple of 3, remove 3
Divide by 3
Answer: 3

(v) 704 = 2^6 × 11^1
11^1 is not multiple of 3, remove 11
Divide by 11
Answer: 11 — Prime factorize and divide by the prime factors whose powers are not multiples of 3 to get a perfect cube.

Question 4

Parikshit makes a cuboid of plasticine of sides 5cm, 2cm, 5cm. How many such cuboids will he need to form a cube?

Accepted Answer

Volume of one cuboid = 5 × 2 × 5 = 50 cm³
Let the side of the cube be 'a' cm.
Since the cube is formed by joining these cuboids, the volume of the cube must be a perfect cube and a multiple of 50.
Find the smallest cube volume divisible by 50.
Prime factorize 50 = 2 × 5²
To make a perfect cube, powers of prime factors must be multiples of 3.
For 2: power is 1, needs 2 more → 2^3
For 5: power is 2, needs 1 more → 5^3
So cube volume = 2^3 × 5^3 = 8 × 125 = 1000 cm³
Number of cuboids = Total volume / volume of one cuboid = 1000 / 50 = 20
Answer: 20 cuboids — Calculate volume of cuboid.
Find smallest cube volume divisible by cuboid volume.
Divide cube volume by cuboid volume to get number of cuboids.

Question 5

State true or false: for any integer m, m^2 < m^3. Why?

Accepted Answer

False. Explanation: - For m > 1, m^3 > m^2, so m^2 < m^3 is true. - For m = 1, m^2 = m^3 = 1, so m^2 < m^3 is false. - For m = 0, m^2 = m^3 = 0, so m^2 < m^3 is false. - For m < 0, consider m = -1: m^2 = 1, m^3 = -1, so m^2 > m^3. Therefore, the statement is not true for all integers m. — Check the inequality for different integer values: - Positive integers greater than 1: true - 0 and 1: false - Negative integers: false

Question 6

Find the cube root of each of the following numbers by prime factorisation method.

(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125

Accepted Answer

Solution:
To find the cube root of a number by prime factorisation, express the number as a product of prime factors and group the factors in triples. The cube root is the product of one factor from each group.

(i) 64
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^6
Grouping in triples: (2 × 2 × 2)(2 × 2 × 2)
Cube root = 2 × 2 = 4

(ii) 512
512 = 2^9
Grouping in triples: (2 × 2 × 2)(2 × 2 × 2)(2 × 2 × 2)
Cube root = 2 × 2 × 2 = 8

(iii) 10648
Prime factorisation:
10648 ÷ 2 = 5324
5324 ÷ 2 = 2662
2662 ÷ 2 = 1331
1331 = 11 × 11 × 11
So, 10648 = 2 × 2 × 2 × 11 × 11 × 11 = (2^3)(11^3)
Cube root = 2 × 11 = 22

(iv) 27000
27000 = 27 × 1000 = (3^3) × (10^3) = (3 × 10)^3
Cube root = 3 × 10 = 30

(v) 15625
15625 = 5^6
Grouping in triples: (5 × 5 × 5)(5 × 5 × 5)
Cube root = 5 × 5 = 25

(vi) 13824
Prime factorisation:
13824 ÷ 2 = 6912
6912 ÷ 2 = 3456
3456 ÷ 2 = 1728
1728 ÷ 2 = 864
864 ÷ 2 = 432
432 ÷ 2 = 216
216 ÷ 2 = 108
108 ÷ 2 = 54
54 ÷ 2 = 27
27 = 3 × 3 × 3
Counting powers:
Number of 2's = 9 (since divided by 2 nine times)
Number of 3's = 3
So, 13824 = 2^9 × 3^3 = (2^3)^3 × (3^1)^3
Cube root = 2^3 × 3 = 8 × 3 = 24

(vii) 110592
Prime factorisation:
110592 ÷ 2 = 55296
55296 ÷ 2 = 27648
27648 ÷ 2 = 13824 (already factorised above)
From (vi), 13824 = 2^9 × 3^3
So, 110592 = 2 × 13824 = 2^10 × 3^3
Grouping in triples:
2^9 × 3^3 is perfect cube, extra 2 remains
So, 110592 = 2^10 × 3^3 = 2^9 × 2^1 × 3^3
Cube root = 2^3 × 3 × cube root of 2 = 8 × 3 × cube root of 2
Since cube root of 2 is irrational, 110592 is not a perfect cube.
But question asks to find cube root by prime factorisation method, so assuming perfect cube:
Actually, 110592 = (48)^3
Because 48^3 = 48 × 48 × 48 = 110592

(viii) 46656
46656 = 6^6
Because 6^3 = 216, 6^6 = (6^3)^2 = 216^2 = 46656
Cube root = 6^2 = 36

(ix) 175616
Prime factorisation:
175616 ÷ 2 = 87808
87808 ÷ 2 = 43904
43904 ÷ 2 = 21952
21952 ÷ 2 = 10976
10976 ÷ 2 = 5488
5488 ÷ 2 = 2744
2744 = 14^3 (since 14 × 14 × 14 = 2744)
So, 175616 = 2^6 × 14^3 = (2^3)^2 × 14^3
Cube root = 2^2 × 14 = 4 × 14 = 56

(x) 91125
Prime factorisation:
91125 ÷ 3 = 30375
30375 ÷ 3 = 10125
10125 ÷ 3 = 3375
3375 ÷ 3 = 1125
1125 ÷ 3 = 375
375 ÷ 3 = 125
125 = 5^3
So, 91125 = 3^6 × 5^3 = (3^2)^3 × 5^3
Cube root = 3^2 × 5 = 9 × 5 = 45 — Step-by-step prime factorisation of each number, grouping prime factors in triples, and multiplying one factor from each group to find the cube root.

Question 7

State true or false.

(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.

Accepted Answer

Solutions:
(i) Cube of any odd number is even.
False. Cube of an odd number is always odd.
Example: 3^3 = 27 (odd)

(ii) A perfect cube does not end with two zeros.
True. A perfect cube ending with two zeros means it is divisible by 100 = 2^2 × 5^2. For a cube, powers of prime factors must be multiples of 3. Since 2 and 5 appear squared, not cubed, perfect cube cannot end with exactly two zeros.

(iii) If square of a number ends with 5, then its cube ends with 25.
True. If square ends with 5, number ends with 5.
Cube of number ending with 5 ends with 125, so last two digits are 25.

(iv) There is no perfect cube which ends with 8.
False. For example, 2^3 = 8 ends with 8.

(v) The cube of a two digit number may be a three digit number.
False. Cube of 10 = 1000 (4 digits), so cube of two digit number is at least 4 digits.

(vi) The cube of a two digit number may have seven or more digits.
True. For example, 50^3 = 125000 (6 digits), 100^3 = 1000000 (7 digits), so some cubes of two digit numbers have 7 or more digits.

(vii) The cube of a single digit number may be a single digit number.
True. For example, 1^3 = 1, 2^3 = 8 (both single digit). — Each statement is evaluated based on properties of cubes and digits, with examples and reasoning provided.

Question 8

Which of the following numbers are not perfect cubes?
(i) 216
(ii) 128
(iii) 1000
(iv) 100
(v) 46656

Accepted Answer

A perfect cube is a number that can be expressed as the cube of an integer.

(i) 216 = 6^3, so it is a perfect cube.
(ii) 128 = 2^7, not a perfect cube because 7 is not divisible by 3.
(iii) 1000 = 10^3, perfect cube.
(iv) 100 is not a perfect cube (since 4^3=64 and 5^3=125).
(v) 46656 = 36^3, perfect cube.

Therefore, the numbers which are not perfect cubes are 128 and 100. — Check if the number can be written as n^3 for some integer n.
216 = 6^3, 128 is not a cube, 1000 = 10^3, 100 is not a cube, 46656 = 36^3.

Question 9

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256
(iii) 72
(iv) 675
(v) 100

Accepted Answer

To find the smallest number to multiply to get a perfect cube, prime factorize each number and adjust exponents to multiples of 3.

(i) 243 = 3^5
Exponent of 3 is 5, needs 1 more 3 to make 3^6 (6 is multiple of 3).
Smallest number = 3

(ii) 256 = 2^8
Exponent 8, needs 1 more 2 to make 2^9.
Smallest number = 2

(iii) 72 = 2^3 × 3^2
3 exponent is 2, needs 1 more 3 to make 3^3.
Smallest number = 3

(iv) 675 = 3^3 × 5^2
5 exponent is 2, needs 1 more 5 to make 5^3.
Smallest number = 5

(v) 100 = 2^2 × 5^2
Both exponents 2, need 1 more 2 and 1 more 5.
Smallest number = 2 × 5 = 10 — Make exponents multiples of 3 by multiplying with the missing prime factors.

Question 10

Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
(ii) 128
(iii) 135
(iv) 192
(v) 704

Accepted Answer

To find the smallest number to divide by to get a perfect cube, prime factorize and remove extra prime factors to make exponents multiples of 3.

(i) 81 = 3^4
Remove one 3 to get 3^3 = 27
Smallest number = 3

(ii) 128 = 2^7
Remove one 2 to get 2^6 = 64
Smallest number = 2

(iii) 135 = 3^3 × 5^1
5 exponent 1, remove 5 to get 3^3
Smallest number = 5

(iv) 192 = 2^6 × 3^1
3 exponent 1, remove 3
Smallest number = 3

(v) 704 = 2^6 × 11^1
11 exponent 1, remove 11
Smallest number = 11 — Divide by prime factors whose exponents are not multiples of 3 to reduce exponents to multiples of 3.

Question 11

Parikshit makes a cuboid of plasticine of sides $5\mathrm{cm}$, $2\mathrm{cm}$, $5\mathrm{cm}$. How many such cuboids will he need to form a cube?

Accepted Answer

Volume of one cuboid = 5 × 2 × 5 = 50 cm³
Let the cube formed have side length 'a' cm.
Volume of cube = a³
Since the cube is formed by joining these cuboids, volume of cube = n × 50, where n is number of cuboids.
We want a cube such that a³ is divisible by 50.
Prime factorization of 50 = 2 × 5²
To form a cube, each prime factor's exponent in volume must be multiple of 3.
Currently, 50 = 2^1 × 5^2
To make exponents multiples of 3:
2^1 needs 2 more 2's → 2^3
5^2 needs 1 more 5 → 5^3
So multiply 50 by 2² × 5¹ = 4 × 5 = 20
Therefore, number of cuboids needed = 20

Hence, Parikshit needs 20 such cuboids to form a cube. — Find volume of cuboid, then find smallest cube volume divisible by it by adjusting prime factors to multiples of 3.

Question 12

State true or false: for any integer $m$, $m^2 < m^3$. Why?

Accepted Answer

This statement is not always true. If m > 1, then m^3 > m^2, so the statement is true. If m = 1, then m^2 = m^3 = 1, so m^2 < m^3 is false. If m = 0, then m^2 = m^3 = 0, false. If m is negative, for example m = -1: (-1)^2 = 1 (-1)^3 = -1 So 1 < -1 is false. Therefore, the statement is true only for integers greater than 1. Hence, the statement is false in general. — Compare m^2 and m^3 for different integer values of m to check validity.

Question 13

Find the cube root of each of the following numbers by prime factorisation method.

(i) 64
(ii) 512
(iii) 10648
(iv) 27000
(v) 15625
(vi) 13824
(vii) 110592
(viii) 46656
(ix) 175616
(x) 91125

Accepted Answer

Solution:

To find the cube root of a number by prime factorisation, we express the number as a product of prime factors and then group the factors in triples. The cube root is the product of one factor from each triple.

(i) 64
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2^6
Grouping in triples: (2 × 2 × 2)(2 × 2 × 2) = (2^3)(2^3)
Cube root = 2 × 2 = 4

(ii) 512
512 = 2^9
Grouping in triples: (2^3)(2^3)(2^3)
Cube root = 2 × 2 × 2 = 8

(iii) 10648
Prime factorisation:
10648 ÷ 2 = 5324
5324 ÷ 2 = 2662
2662 ÷ 2 = 1331
1331 = 11 × 11 × 11 = 11^3
So, 10648 = 2^3 × 11^3
Cube root = 2 × 11 = 22

(iv) 27000
27000 = 27 × 1000 = (3^3) × (10^3) = (3 × 10)^3
Cube root = 3 × 10 = 30

(v) 15625
15625 = 5^6 (since 5^6 = 15625)
Grouping in triples: (5^3)(5^3)
Cube root = 5 × 5 = 25

(vi) 13824
Prime factorisation:
13824 ÷ 2 = 6912
6912 ÷ 2 = 3456
3456 ÷ 2 = 1728
1728 ÷ 2 = 864
864 ÷ 2 = 432
432 ÷ 2 = 216
216 ÷ 2 = 108
108 ÷ 2 = 54
54 ÷ 2 = 27
27 = 3^3
Counting 2's: 9 times 2's and 3 times 3's
So, 13824 = 2^9 × 3^3
Cube root = 2^3 × 3 = 8 × 3 = 24

(vii) 110592
Prime factorisation:
110592 = 2^13 × 3^3 (since 110592 = 2^13 × 3^3)
Grouping in triples:
2^12 × 3^3 × 2^1
Cube root = 2^4 × 3 = 16 × 3 = 48

(viii) 46656
46656 = 6^6 = (6^3)^2
6^3 = 216
So 46656 = 216^2
Prime factorisation:
46656 = 2^6 × 3^6
Cube root = 2^2 × 3^2 = 4 × 9 = 36

(ix) 175616
Prime factorisation:
175616 = 2^6 × 7^3
Cube root = 2^2 × 7 = 4 × 7 = 28

(x) 91125
Prime factorisation:
91125 = 3^3 × 5^4
Grouping triples:
3^3 × 5^3 × 5^1
Cube root = 3 × 5 = 15

Hence, the cube roots are:
(i) 4
(ii) 8
(iii) 22
(iv) 30
(v) 25
(vi) 24
(vii) 48
(viii) 36
(ix) 28
(x) 15 — The cube root of a number can be found by prime factorisation. Express the number as a product of prime factors, group the factors in triples, and multiply one factor from each triple to get the cube root.

Question 14

State true or false.

(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.

Accepted Answer

Solutions:

(i) Cube of any odd number is even.
False. Cube of an odd number is always odd.

(ii) A perfect cube does not end with two zeros.
True. Because if a cube ends with two zeros, it must be divisible by 1000 = 10^3, so the number itself must be divisible by 10, so the cube ends with three zeros, not two.

(iii) If square of a number ends with 5, then its cube ends with 25.
True. If square ends with 5, the number ends with 5. Cube of a number ending with 5 ends with 125, so cube ends with 25 (last two digits).

(iv) There is no perfect cube which ends with 8.
False. For example, 2^3 = 8 ends with 8.

(v) The cube of a two digit number may be a three digit number.
False. Cube of 10 is 1000 (four digits), so cube of two digit number is at least four digits.

(vi) The cube of a two digit number may have seven or more digits.
True. For example, 50^3 = 125000, which has 6 digits, 100^3 = 1,000,000 (7 digits). So yes, it can have seven or more digits.

(vii) The cube of a single digit number may be a single digit number.
True. For example, 1^3 = 1, 2^3 = 8, both single digit numbers. — Each statement is analyzed based on properties of cubes and digits of numbers. Odd numbers cubed remain odd, cubes ending with zeros must end with three zeros, cubes of numbers ending with 5 end with 125, etc.

Question 15

The number 1729 is known as the Hardy–Ramanujan Number because it is the smallest number expressible as the sum of two cubes in two different ways. Which of the following correctly represents these two ways?

Accepted Answer

1³ + 12³ and 9³ + 10³ — 1729 can be expressed as 1729 = 1³ + 12³ = 9³ + 10³. This unique property makes it the smallest Hardy–Ramanujan Number.

was 1729. While talking to Ramanujan, Hardy described this number

was 1729. While talking to Ramanujan, Hardy described this number — Study Notes

The Number 1729

Sum of Two Cubes

Properties of Cubes

Practice Questions — was 1729. While talking to Ramanujan, Hardy described this number

All 13 Chapters in Mathematics