An intelligence agency forms 2-digit codes from distinct digits 0–9, with the first digit non-zero. A handwritten code can be misread upside down if both digits are from {0,1,6,8,9} and the 180° rotation maps it to another valid 2-digit code (e.g., 91 ↔ 16). How many codes avoid such confusion?

Introduction / Context:
Under 180° rotation, the only digits that remain valid digits are {0,1,6,8,9} with mappings 0→0, 1→1, 6↔9, 8→8. A code is confusable if both digits come from this set and its rotated reading is a different valid 2-digit code (non-zero first digit, digits distinct).

Given Data / Assumptions:

• Total 2-digit codes with distinct digits and non-zero first digit: 9 × 9 = 81.
• Rotation map f: {0,1,6,8,9} → {0,1,9,8,6} and order reverses.

Concept / Approach:
Count confusable codes and subtract from 81. Confusion requires both digits from {1,6,8,9} (not 0 for the second digit), so that the rotated code also has a non-zero leading digit and remains a valid 2-digit code with distinct digits.

Step-by-Step Solution:
Confusable codes = ordered pairs (x, y) with x ∈ {1,6,8,9}, y ∈ {1,6,8,9}, x ≠ y, because the rotated code is f(y)f(x) with f(y) ∈ {1,6,8,9} ≠ 0 and f(x) ≠ f(y).Count = 4 × 3 = 12.Therefore, non-confusable codes = 81 − 12 = 69.

Verification / Alternative check:
Cases with y = 0 yield rotated codes starting with 0 and hence are not considered confusing under the “valid code” criterion.

Why Other Options Are Wrong:
80/78 assume excluding very few cases; 71 is close but overcounts confusable cases; 81 ignores the confusion constraint.

Common Pitfalls:
Treating 0 as allowable as the rotated leading digit; or allowing equal digits (not permitted).

Final Answer:
69