Question 5

A number ending with 0 must have 2 and 5 as its prime factors.

Step 1 — Prime Factors of 6^n

Let's find the prime factors of the base number. The base number is 6. We can write 6 as a product of its prime factors.

$6 = 2 \times 3$

Now, let's find the prime factors of $6^n$. We raise the prime factors to the power n.

$6^n = (2 \times 3)^n$

$= 2^n \times 3^n$

$\boxed{6^n = 2^n \times 3^n}$

Step 2 — Checking for Factor 5

We use the Fundamental Theorem of Arithmetic. This theorem states that every composite number has a unique prime factorization. The prime factors of $6^n$ are 2 and 3. We can see that the prime factor 5 is not present. For $6^n$ to end with the digit 0, it must be divisible by 10. This means it must be divisible by both 2 and 5. Since 5 is not a prime factor of $6^n$, $6^n$ is not divisible by 5. Therefore, $6^n$ cannot end with the digit 0.

Answer

$6^n$ cannot end with the digit 0 for any natural number n.