was 1729. While talking to Ramanujan, Hardy described this number | Class 8 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 min read
was 1729. While talking to Ramanujan, Hardy described this number – this guide gives you a concise, exam-ready overview of was 1729. While talking to Ramanujan, Hardy described this number from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
Sum of Two Cubes
This section explains the mathematical concept of expressing numbers as the sum of two cubes. A cube of a number n is defined as n³ = n × n × n. The sum of two cubes is then the addition of two such cubes, for example, a³ + b³, where a and b are integers. The section elaborates on how certain numbers can be expressed in this form and how some numbers can be expressed as the sum of two cubes in more than one way. The number 1729 is the smallest such number. Understanding sums of cubes is important in number theory and helps in exploring properties of numbers, factorization, and Diophantine equations. The section also briefly discusses how to verify if a number can be expressed as a sum of two cubes by checking all possible cube pairs less than the number.
📊 Diagram: A table showing cubes of numbers 1 to 12 and their sums to illustrate possible sums of two cubes.
🧪 Activity: Students are asked to find other numbers less than 2000 that can be expressed as the sum of two cubes in two different ways.
🔗 Connection: This section prepares students to understand the uniqueness of 1729 and introduces the idea of exploring numbers through their cube sums, leading to the next section on properties of cubes.
Frequently asked questions
1. Which of the following numbers are not perfect cubes? (i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656
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.
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
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, mul
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
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^
4. Parikshit makes a cuboid of plasticine of sides 5cm, 2cm, 5cm. How many such cuboids will he need to form a cube?
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
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