Question

For any natural number n, \(6^n\) ends with the digit :

• A. 0
• B. 6
• C. 3
• D. 2

Hint:

The digits 0, 1, 5, and 6 always maintain their own value at the units place for any power \( n>0 \). (e.g., \( 5^n \) always ends in 5).

Answer 1

The Correct Option is B

To determine what digit \(6^n\) ends with for any natural number \(n\), we need to examine the pattern of the last digit of successive powers of 6.

1. First, let's calculate the first few powers of 6 and observe the last digit:
   • \(6^1 = 6\), the last digit is 6.
   • \(6^2 = 36\), the last digit is 6.
   • \(6^3 = 216\), the last digit is 6.
   • \(6^4 = 1296\), the last digit is 6.
2. From the calculations above, we can see that the last digit of \(6^n\) is always 6.
3. Let's understand why this pattern occurs:
   • When multiplying by 6, the factor of 6 in the ones place of any number implies that the way multiplication carries forward does not change the last digit from 6.
   • This is similar to how multiplication by 0 in the tens place returns a cycle that always ends in a predictable manner, here with a 6.
4. Thus, regardless of the value of \(n\), \(6^n\) will always end with the digit 6.

Therefore, the correct answer is that \(6^n\) ends with the digit 6.