Check whether `6^n` can end with the digit 0 for any natural number `n` .

Answer

To determine whether \( 6^n \) can end with the digit 0 for any natural number \( n \), we need to analyze the factors of \( 6^n \).

### Step 1: Understand the condition for a number to end with 0
A number ends with the digit 0 if it is divisible by 10. The number 10 can be expressed as the product of its prime factors:
\[ 10 = 2 \times 5 \]
This means that for a number to end with 0, it must have at least one factor of 2 and one factor of 5.

### Step 2: Factor \( 6^n \)
Next, we express \( 6^n \) in terms of its prime factors:
\[ 6 = 2 \times 3 \]
Thus, we can write:
\[ 6^n = (2 \times 3)^n = 2^n \times 3^n \]
From this expression, we see that \( 6^n \) contains the factor \( 2^n \) but does not contain any factor of 5.

### Step 3: Check for the presence of the factor 5
Since \( 6^n \) consists of the factors \( 2^n \) and \( 3^n \), we need to check if there is any factor of 5 in \( 6^n \). As we can see, there is no factor of 5 in the expression \( 6^n \).

### Step 4: Conclusion
Since \( 6^n \) contains the factor \( 2 \) but lacks the factor \( 5 \), it cannot be divisible by 10. Therefore, \( 6^n \) cannot end with the digit 0 for any natural number \( n \).

### Final Answer:
\[ 6^n \text{ cannot end with the digit } 0 \text{ for any natural number } n. \]

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