Correct Option
The daily savings pattern forms an arithmetic progression of consecutive odd numbers:
- January 1st: ₹1
- January 2nd: ₹1 + ₹2 = ₹3
- January 3rd: ₹3 + ₹2 = ₹5
The series of daily savings is 1, 3, 5, ...
The sum of the first 'n' consecutive odd natural numbers is given by the formula n2.
The problem requires the total savings to be both a perfect square and a perfect cube. A number that satisfies both conditions must be a perfect sixth power. The smallest positive integer that is both a perfect square and a perfect cube is 64, as 64=82 and 64=43.
Setting the total savings equal to 64, we have n2=64. Solving for 'n', we get n=8.
This means the total savings of ₹64 is accumulated at the end of the 8th day. Counting 8 days from January 1st, the date is January 8th, 2023.
Incorrect Options
Option A (7th January 2023): At the end of 7 days, the total savings would be the sum of the first 7 odd numbers, which is 72=49. While 49 is a perfect square, it is not a perfect cube.
Option C (9th January 2023): At the end of 9 days, the total savings would be the sum of the first 9 odd numbers, which is 92=81. While 81 is a perfect square, it is not a perfect cube.
Option D (Not possible): This option is incorrect as a date (January 8th, 2023) has been identified where the total savings meet both conditions of being a perfect square and a perfect cube.