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 ✓ UPSC's answer
(c) 9th January, 2023
(d) Not possible

Why the answer is (b)

• The daily savings form an arithmetic progression: ₹1, ₹3, ₹5, ..., where the amount saved on day $n$ is $2n - 1$.

• The total savings after $n$ days is the sum of the first $n$ odd numbers, which equals $n^2$.

• For the total savings to be both a perfect square and a perfect cube, it must be a perfect sixth power ($k^6$).

• Since the total is $n^2$, we need $n^2 = k^6$, which implies $n = k^3$.

• The smallest positive integer $k$ is 1, giving $n = 1^3 = 1$ (Total = 1, which is $1^2$ and $1^3$). However, the question implies a non-trivial accumulation or checks specific dates. Let's re-evaluate the condition 'perfect square as well as a perfect cube'.

• Wait, $n^2$ is always a perfect square. For it to be a perfect cube, $n^2$ must be divisible by $3^3, 5^3$, etc., or simply $n$ must be a perfect cube. If $n=1$, Total=1. If $n=8$, Total=$8^2=64$. 64 is $8^2$ (square) and $4^3$ (cube). Thus, on the 8th day, the total is 64, which satisfies both conditions.

Why the other options are wrong

(a) 7th January, 2023: On the 7th day, the total savings is $7^2 = 49$, which is a perfect square but not a perfect cube.
(c) 9th January, 2023: On the 9th day, the total savings is $9^2 = 81$, which is a perfect square but not a perfect cube.
(d) Not possible: It is possible, as demonstrated by the 8th day where the total savings of 64 is both a square and a cube.

Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2024, held on 16 June 2024. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.