Check whether $6^n$ can end with the digit $0$ for any natural number $n$.
Step-by-Step Solution
Key Concept: A number ends with digit 0 if its prime factorisation contains both 2 and 5.
If $6^n$ ends with digit $0$, it must be divisible by $5$, meaning its prime factorisation must contain prime factor $5$. [1.0 Mark]
Prime factorisation of $6^n = (2 \times 3)^n = 2^n \times 3^n$. By the uniqueness of the Fundamental Theorem of Arithmetic, there are no other prime factors in $6^n$ besides $2$ and $3$. Since $5$ is not a prime factor, $6^n$ can never end with the digit $0$ for any $n \in \mathbb{N}$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Stating condition for ending in 0 (presence of prime factor 5): 1.0 Mark
Writing prime factorisation of $6^n$ and concluding via Fundamental Theorem of Arithmetic: 1.0 Mark
Correct Answer: