Classification – Odd one out (unique multiple of 13) Among the following numbers that all end with the digit 5, exactly one is a multiple of 13. Identify that number and mark it as the odd one out.

Introduction / Context:
When several options share a surface feature (here, all end with 5 → all are multiples of 5), the true classification hinge is often a secondary factor. In this item, that hidden factor is divisibility by 13. Only one number in the set carries 13 as a factor, which makes it the odd one out even though all options are multiples of 5.

Given Data / Assumptions:

• Candidates: 115, 145, 195, 75
• Shared surface property: each ends with 5 (therefore divisible by 5)
• Hidden discriminant: divisibility by 13

Concept / Approach:
First, acknowledge the commonality (×5). Next, test each value for 13. Efficient checks: 13 * 15 = 195; 13 * 11 = 143 (near 145 but not equal); 13 * 9 = 117 (near 115 but not equal); 13 does not divide 75. Thus 195 uniquely contains 13 as a factor.

Step-by-Step Solution:
115 = 5 * 23 → not divisible by 13.145 = 5 * 29 → not divisible by 13 (13 * 11 = 143, 13 * 12 = 156).75 = 3 * 5 * 5 → not divisible by 13.195 = 13 * 15 → divisible by 13 → odd one out.

Verification / Alternative check:
Modular test: for k ending in 5, k ≡ 0 (mod 5). Compute k (mod 13) to isolate the one ≡ 0 (mod 13): 195 (mod 13) = 0; the others produce nonzero residues.

Why Other Options Are Wrong:

• 115: Not divisible by 13.
• 145: Not divisible by 13.
• 75: Not divisible by 13.
• None of these: A single 13-multiple exists (195).

Common Pitfalls:
Stopping after noticing that all end with 5, which would make the set appear homogeneous. Always look for a deeper factor that can create a unique outlier.

Final Answer:
195