एक व्यक्ति ने 1 जनवरी, 2023 को ₹ 1 की बचत की। 2 जनवरी, 2023 को उसने पूर्ववर्ती दिन की तुलना में ₹ 2 की अधिक बचत की। 3 जनवरी, 2023 को उसने पूर्ववर्ती दिन की तुलना में ₹ 2 की अधिक बचत की और यह क्रम आगे भी जारी रहता है। किस तारीख के अंत में उसकी कुल बचत एक पूर्ण वर्ग होने के साथ-साथ एक पूर्ण घन भी थी?
Read this question in English
On January 1st, 2023, a person saved ₹1. On January 2nd, 2023, he saved ₹2 more than that on the previous day. On January 3rd, 2023, he saved ₹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?
(a) 7th January, 2023 · (b) 8th January, 2023 · (c) 9th January, 2023 · (d) Not possible
उत्तर और व्याख्या · explanation in English
Answer: (b) 8 जनवरी, 2023
He saves 1, 3, 5, 7, … rupees, so the total after n days is n². A number that is both a square and a cube is a sixth power, so n must be a perfect cube; n = 8 gives 64 = 8² = 4³.
- Daily savings are the odd numbers 1, 3, 5, …; the sum of the first n odd numbers is n².
- n² is also a perfect cube only when it is a sixth power, i.e. when n itself is a perfect cube.
- The first cube after 1 is 8: total after 8 days = 64 = 8² = 4³.
- So the date is 8th January, 2023.
- Check: 7 days give 49 and 9 days give 81, neither a cube. (The ₹1 at the end of 1st January is trivially 1² = 1³, but that date is not among the options.)
Remember · Sum of the first n odd numbers = n². A number that is both a square and a cube is a sixth power.
Question and answer: UPSC's official GS Paper II (2024, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·