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 iff it is divisible by 10 = 2 × 5. Use prime factorisation of 6^n.
Given: $n \in \mathbb{N}$ (natural number).\
To Find: Whether there exists an $n$ such that the decimal representation of $6^{n}$ ends with 0.\
Step 1: Write the prime factorisation of $6^{n}$.
$$6^{n} = (2 \times 3)^{n} = 2^{n} \cdot 3^{n}.$$\
Step 2: For a number to end with 0 it must be a multiple of 10.
$$10 = 2 \times 5.$$\
Hence a number ending with 0 must contain both the prime factors 2 and 5 in its factorisation.\
Step 3: Examine the prime factors of $6^{n}$.
- $6^{n}$ contains the factor $2^{n}$ (so the factor 2 is present).\
- $6^{n}$ contains only the prime factor 3 besides 2; there is no factor 5 in $6^{n}$ for any $n$ because the only primes appearing are 2 and 3.
Step 4: Since the factor 5 never appears in $6^{n}$, $6^{n}$ can never be divisible by 10.
Therefore $6^{n}$ can never end with the digit 0 for any natural number $n$.\
Conclusion: No natural number $n$ makes $6^{n}$ end with 0.
Correct Answer: No. $6^{n}$ never ends with the digit 0 for any natural number $n$ because $6^{n}=2^{n}3^{n}$ lacks the factor 5 required for divisibility by 10.