Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
(b) 6
Since $6 = 2 \times 3$, we have $6^n = 2^n \times 3^n$. The prime factorisation contains no factor of 5, so $6^n$ never ends in 0; it always ends with the digit 6 (e.g., $6^1=6,\ 6^2=36,\ 6^3=216$).
Source: Chapter 1, Section 1.2 (Exercise 1.1, Q5)
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Explanation
- The key reasoning mirrors Exercise 1.1 Q5: for $6^n$ to end in 0, it must be divisible by both 2 and 5. Since $6^n = 2^n \times 3^n$ has no factor of 5, it can never end in 0.
- A quick pattern check (6, 36, 216, 1296…) confirms the units digit is always 6.
- Examiners expect you to cite the prime factorisation argument, not just pattern observation.