Consider the numbers 4 n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero.
Step-by-Step Solution
Key Concept: A number ends with the digit zero iff it is divisible by 10. Since $10 = 2 \times 5$, a number must contain both the prime factors 2 and 5. The power $4^n$ contains only the prime factor 2.
Given: $4^n$ where $n \in \mathbb{N}$.
To Find: Whether there exists $n$ such that $4^n$ ends with 0 (i.e., $4^n$ is divisible by 10).
Step 1: Express $4^n$ in terms of its prime factors.
\[4^n = (2^2)^n = 2^{2n}.\]
Thus $4^n$ contains only the prime factor 2.
Step 2: For a number to be divisible by 10, it must contain the factor 5 as well (since $10 = 2 \times 5$).
Step 3: Observe that $2^{2n}$ has no factor 5 for any natural $n$ because the only prime factor present is 2.
Therefore $4^n$ can never be divisible by 5, and consequently it can never be divisible by 10.
Conclusion: No natural number $n$ makes $4^n$ end with the digit zero.
Correct Answer: There is no natural number $n$ for which $4^n$ ends with the digit zero.