Perfect-square formation from selected digits – Using the 2nd, 6th, and 9th digits of 187642539, form a 3-digit number that is a perfect square of a two-digit odd number. What is the unit digit of that two-digit odd number?

Introduction / Context:
From specific positions in a larger number, we must permute the selected digits to make a 3-digit perfect square whose square root is an odd two-digit number. Then, report the unit digit of that root.

Given Data / Assumptions:

• Number: 187642539. Positions (1-based): 1=1, 2=8, 3=7, 4=6, 5=4, 6=2, 7=5, 8=3, 9=9.
• Chosen digits: 2nd, 6th, 9th → 8, 2, 9.

Concept / Approach:
List permutations of {8,2,9} and check perfect squares (particularly those consistent with odd roots). Notable 3-digit odd-root squares include 11^2=121, 13^2=169, 15^2=225, 17^2=289, 19^2=361, etc.

Step-by-Step Solution:
Permute digits: 289, 298, 829, 892, 928, 982.Check squares: 289 = 17^2, which uses digits 2,8,9 exactly once.The two-digit odd root is 17, whose unit digit is 7.

Verification / Alternative check:
None of the other permutations are perfect squares. 289 is uniquely valid here.

Why Other Options Are Wrong:
1 or 9 are unit digits of some odd roots in other squares (e.g., 11, 19) but not for a square formed from {8,2,9} here. “No such number” is incorrect since 289 exists.

Common Pitfalls:
Mis-indexing the positions in the original 9-digit number or overlooking 289 = 17^2.

Final Answer:
7