10. On January 1st, 2023, a person saved Rs 1. On January 2nd, 2023, he saved Rs. 2 more than that on the previous day. On January 3rd, 2023, he saved Rs. 2 more than that on the previous day and so on. At the end of which date was his total savings a perfect square as well a perfect cube?

1. A. 7th January, 2023
2. B. 8th January, 2023
3. C. 9th January, 2023
4. D. Not possible

Answer: B

Explanation

The savings pattern is 1, 3, 5, 7, … which is a sequence of odd numbers. The sum of the first ‘n’ odd numbers is given by n^2. We are looking for a total savings amount that is both a perfect square and a perfect cube, meaning it must be a perfect sixth power (e.g., k^6). Let’s calculate the cumulative savings:
– End of Jan 1st: 1 (1^2, 1^3)
– End of Jan 2nd: 1 + 3 = 4 (2^2, not a cube)
– End of Jan 3rd: 1 + 3 + 5 = 9 (3^2, not a cube)
– End of Jan 4th: 1 + 3 + 5 + 7 = 16 (4^2, not a cube)
– End of Jan 5th: 1 + 3 + 5 + 7 + 9 = 25 (5^2, not a cube)
– End of Jan 6th: 1 + 3 + 5 + 7 + 9 + 11 = 36 (6^2, not a cube)
– End of Jan 7th: 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 (7^2, not a cube)
– End of Jan 8th: 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64. Here, 64 is 8^2 (a perfect square) and 4^3 (a perfect cube). Thus, the condition is met at the end of January 8th, 2023. This question combines arithmetic progression with properties of numbers.