Finding the date when an arithmetic progression sum hits a perfect square and cube

Identify the daily savings pattern: Day 1: 1 Day 2: 1 + 2 = 3 Day 3: 3 + 2 = 5 This is the sequence of consecutive odd numbers. The sum of the first ⟨MATH⟩n⟨/MATH⟩ consecutive odd numbers is a well-known algebraic identity: ⟨MATH⟩S_n = n²⟨/MATH⟩.

We need the total savings (n²) to be both a perfect square and a perfect cube. A number is a perfect square and a perfect cube simultaneously if and only if its prime exponents are multiples of LCM(2,3) = 6, meaning the total sum must look like k^6.

Since S_n = n², for n² to equal k^6, we take the square root of both sides: n = k³. Thus, the number of days n must be a perfect cube. Looking at our options, the day counts corresponding to the dates are n = 7, 8, 9. The only perfect cube among these values is n = 8 (2³ = 8). On January 8th, total savings = 8² = 64, which is 8² (square) and 4³ (cube).

Answer: (b).