Odd one out among squares: 25, 36, 49, 81, 121, 169, 225

Introduction / Context:All listed numbers are perfect squares, but there is a deeper structural contrast: one is the square of an even integer while the rest are squares of odd integers. Identifying that unique parity distinguishes the odd term.

Given Data / Assumptions:

• Squares listed: 25 (5^2), 36 (6^2), 49 (7^2), 81 (9^2), 121 (11^2), 169 (13^2), 225 (15^2).
• Only one is an even square.

Concept / Approach:Classify each term by the parity of its square root. If exactly one is even and all others are odd, that even square is the odd man out.

Step-by-Step Solution:25 = 5^2 (odd)36 = 6^2 (even) ← unique49 = 7^2 (odd)81 = 9^2 (odd)121 = 11^2 (odd)169 = 13^2 (odd)225 = 15^2 (odd)

Verification / Alternative check:Other properties (e.g., prime vs composite roots) do not segregate uniquely. Parity does: only 36 arises from an even root, 6.

Why Other Options Are Wrong:

• 49 / 121 / 169: all are odd squares like the majority of the list.

Common Pitfalls:

• Searching for complex relations when a simple structural difference (even vs odd square) suffices.

Final Answer:36