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?

Daily savings form an arithmetic progression: 1, 3, 5, 7, … (Day n saves 2n−1). The cumulative savings after n days is 1+3+5+…+(2n−1) = n². So after day n the total is n², which is always a perfect square. We need n² to also be a perfect cube, i.e., n² must be a sixth power. This requires n itself to be a perfect cube. The smallest n that is a perfect cube greater than 1 is n = 8 (since 8 = 2³, and 8² = 64 = 4³, both a perfect square and a perfect cube). Hence at the end of 8th January, 2023, the total savings ₹64 is both a perfect square and a perfect cube.