Now, we determine the prime factors of (14)^n. Using the property of exponents, (14)^n can be written as (2 × 7)^n, which simplifies to 2^n × 7^n. This means the only prime factors of (14)^n are 2 and 7, regardless of the value of n.

Now, we determine the prime factors of (14)^n. Using the property of exponents, (14)^n can be written as (2 × 7)^n, which simplifies to 2^n × 7^n. This means the only prime factors of (14)^n are 2 and 7, regardless of the value of n.

Check whether there is any natural number 'n' for which $(14)^n$ ends with the digit '0' or '5'.

Answer: No, there is no natural number 'n' for which $(14)^n$ ends with the digit '0' or '5'.

Step-by-step solution

Step 1: Understand the condition for ending in 0 or 5

For a natural number to end with the digit '0', its prime factorization must include both 2 and 5. For a natural number to end with the digit '5', its prime factorization must include 5.

Step 2: Find the prime factors of the base number

First, we find the prime factorization of the base number, which is 14. The prime factors of 14 are 2 and 7.

Step 3: Determine the prime factors of $(14)^n$

Now, we determine the prime factors of $(14)^n$ . Using the property of exponents, $(14)^n$ can be written as $(2 \times 7)^n$ , which simplifies to $2^n \times 7^n$ . This means the only prime factors of $(14)^n$ are 2 and 7, regardless of the value of $n$ .

Step 4: Check for the presence of prime factor 5

Since the prime factorization of $(14)^n$ only contains the prime factors 2 and 7, it does not contain the prime factor 5. Therefore, $(14)^n$ cannot end with the digit '0' or '5' for any natural number $n$ .

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Check whether there is any natural number 'n' for which (14)n(14)^n(14)n ends with the digit '0' or '5'.