Question

For any natural number n, \(9^n\) cannot end with which one of the following digits?

• A. 1
• B. 2
• C. 9
• D. None of these

Answer 1

The Correct Option is B

The last digit of \( 9^n \) follows a repeating cycle of 9 and 1:

• \( 9^1 = 9 \) (last digit: 9) 
• \( 9^2 = 81 \) (last digit: 1)
• \( 9^3 = 729 \) (last digit: 9)
• \( 9^4 = 6561 \) (last digit: 1)

From this, we see that the last digit of \( 9^n \) alternates between 9 and 1 for all \( n \).

Since the last digit never becomes 2, the answer is 2.

Answer 2

To determine which digit cannot be the last digit of \(9^n\) for any natural number \(n\), we need to analyze the pattern of the last digits of powers of 9:

• For \(n = 1\), \(9^1 = 9\), so the last digit is 9.
• For \(n = 2\), \(9^2 = 81\), so the last digit is 1.
• For \(n = 3\), \(9^3 = 729\), so the last digit is 9.
• For \(n = 4\), \(9^4 = 6561\), so the last digit is 1.

We see a repeating pattern every 2 cycles: the sequence of last digits is 9, 1, 9, 1, etc.

This pattern shows that the last digit of \(9^n\) can only be 9 or 1. 
Thus, \(9^n\) cannot end with the digit 2. 
Therefore, the correct answer is: 2