On January 1^st, 2023, a person saved ₹1. On January 2^nd, 2003, he save ₹2 more than that on the previous day. On January 3^rd, 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?

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 $n^2$.

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 $64=8^2$ and $64=4^3$.

Setting the total savings equal to 64, we have $n^2=64$. 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 $7^2=49$. 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 $9^2=81$. 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.