Which of the following patterns could represent the perfect square of a 3-digit odd number? (Digit(s) suppressed with x.) Choose the valid pattern based on known properties of odd squares.

Introduction / Context:
Recognizing the terminal digits and general patterns of perfect squares helps identify which masked number could be a square of a 3-digit odd integer (from 101 to 999, odd). This tests number properties and place-value reasoning.

Given Data / Assumptions:

• We consider squares of odd 3-digit numbers.
• Options are pattern forms such as 9xx1, where x denotes some digit.

Concept / Approach:

• Square of an odd integer ends in 1 or 9 (and in 5 only if the number ends in 5, giving last two digits 25).
• Known example: 99^2 = 9801 starts with 9 and ends with 01, matching pattern 9xx1.
• Squares ending with single 5 (not 25) cannot occur.

Step-by-Step Reasoning:
Check 9xx1: Example 99^2 = 9801 fits the pattern 9xx1 (starts with 9, ends with 1). 65xxx1: This is a 6-digit pattern starting with 65 and ending with 1. While 3-digit odd squares can be 5- or 6-digit, there is no simple property ensuring a prefix 65; nothing guarantees this form. It is not a general property-supported pattern. 10xxx4: Ends with 4, but odd squares do not end with 4. 9xxxxxx5: Ends with 5, but any square ending in 5 must end with 25. A single trailing 5 is impossible.

Verification / Alternative check:
Examine several odd squares: 31^2 = 961, 39^2 = 1521, 99^2 = 9801. The pattern 9xx1 appears (e.g., 9801). Others contradict terminal-digit rules for odd squares.

Why Other Options Are Wrong:
65xxx1: No property-based guarantee; generally not identified as a standard pattern. 10xxx4: Ends in 4; odd squares do not. 9xxxxxx5: Ends in 5 (not 25); contradiction.

Common Pitfalls:
Ignoring last-digit rules for squares leads to incorrect choices. Remember the final digit of odd squares and the special 25 ending for numbers ending in 5.

Final Answer:
9xx1