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.
Mathematical Intuition and Creativity
This section highlights the importance of mathematical intuition and creativity, as exemplified by Ramanujan's insight into the number 1729. It discusses how mathematical discoveries often come from deep thinking, pattern recognition, and creative problem solving rather than just memorization or routine calculations. The section encourages students to develop their intuition by exploring numbers, asking questions, and seeking patterns. It also emphasizes that mathematics is not just about finding answers but about understanding concepts and relationships. Ramanujan's story inspires students to appreciate the creative aspect of mathematics and to approach problems with curiosity and imagination.
📊 Diagram: An illustration showing a brain with light bulbs symbolizing ideas and creative thinking in mathematics.
🧪 Activity: Students are encouraged to solve puzzles and problems that require creative thinking and pattern recognition.
🔗 Connection: This section connects the story of 1729 to the broader theme of mathematical thinking and prepares students for engaging with challenging problems.
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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