Question 5

Let's use proof by contradiction. We will assume the opposite of what we want to prove.

Step 1 — Assume the opposite

Let's assume that for some natural number $n$, $6^n$ ends with the digit zero. If a number ends with zero, it must be a multiple of 10. So, we can write $6^n$ as 10 multiplied by some integer $k$.

$6^n = 10k$

$\boxed{6^n \text{ is a multiple of } 10}$

Step 2 — Prime factorization

Let's look at the prime factors of $6^n$. The number 6 can be written as 2 multiplied by 3.

$6 = 2 \times 3$

So, $6^n$ is $(2 \times 3)^n$.

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

Now, let's look at the prime factors of 10. The number 10 can be written as 2 multiplied by 5.

$10 = 2 \times 5$

So, $10k$ is $(2 \times 5)k$.

$10k = 2 \times 5 \times k$

$\boxed{6^n = 2^n \times 3^n \text{ and } 10k = 2 \times 5 \times k}$

Step 3 — Find the contradiction

From Step 1, we assumed $6^n = 10k$. So, we can write the prime factorizations as equal.

$2^n \times 3^n = 2 \times 5 \times k$

The Fundamental Theorem of Arithmetic says every number has a unique set of prime factors. On the left side, the prime factors are only 2 and 3. The number 5 is not a prime factor of $6^n$. On the right side, the prime factors include 2, 5, and the prime factors of $k$. For the two sides to be equal, they must have the exact same prime factors. The left side ($2^n \times 3^n$) does not have 5 as a prime factor. The right side ($2 \times 5 \times k$) must have 5 as a prime factor. This is a contradiction. The number 5 cannot be a prime factor of $6^n$. Therefore, our initial assumption must be false.

$\boxed{\text{The assumption leads to a contradiction}}$

Answer

There is no value of $n$ for which $6^n$ ends with the digit zero.