Square triangular savings day: On March 1, 2016, Sherry saves ₹1. From March 2 onward, each day he saves ₹1 more than the previous day. Find the first date after March 1, 2016 on which his total savings becomes a perfect square.

Introduction / Context:
The daily savings form an arithmetic sequence starting at 1 with common difference 1. The cumulative savings after n days equals the n-th triangular number n(n + 1)/2. The question asks for the earliest n > 1 such that this total is a perfect square.

Given Data / Assumptions:

• Day 1 (March 1): save 1.
• Day d: save d rupees.
• Total after n days: T_n = n(n + 1)/2.

Concept / Approach:
We need T_n to be a square (a square triangular number). The smallest solutions are well known: n = 1 gives T_1 = 1 (but this is not “after March 1”), and the next is n = 8 giving T_8 = 36, a perfect square.

Step-by-Step Solution:

Verification / Alternative check:
Known square triangular sequence starts 1, 36, 1225, … corresponding to n = 1, 8, 49, … So the earliest date after March 1 is March 8.

Why Other Options Are Wrong:

• 17th, 18th, 26th March: These correspond to n = 17, 18, 26; their triangular totals are not perfect squares. Hence “None of these” is correct.

Common Pitfalls:
Confusing “after March 1” (n > 1) or mistaking daily deposit with cumulative total. Always test triangular numbers explicitly.

Final Answer: