2024 CSAT — Q10

Daily savings form the AP $1, 3, 5, 7, \dots$ (first term $1$, common difference $2$) — the odd numbers. The cumulative total after $n$ days is the sum of the first $n$ odd numbers, a classic identity:

So at the end of day $n$ the savings are exactly $n^2$ — already a perfect square for every $n$. We need it to be a perfect cube too, i.e. $n^2$ must be a perfect sixth power, which forces $n$ itself to be a perfect cube.

The first non-trivial day is $n = 8$ ($=2^3$): total $= 8^2 = 64 = 4^3$, a perfect square and a perfect cube. That is 8th January, 2023. (Day $7$ gives $49=7^2$, not a cube; day $9$ gives $81=9^2$, not a cube.)

Answer: (b) 8th January, 2023.

The killer simplification is the identity sum of the first $n$ odds $= n^2$ — it turns “add up the AP” into “is $n^2$ a cube?”. Since the total is always a square, the binding constraint is the cube: $n^2$ is a perfect cube $\iff n$ is a perfect cube. The smallest cube past the trivial $n=1$ is $8$, so day $8$ ($64=2^6$) is the first date where savings are simultaneously a square and a cube. Spotting the odd-sum identity is worth far more than grinding the running total day by day.