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 result in the same unit digit (0, 1, 5, and 6 respectively) when raised to any positive integer power.
For example, \(5^{n}\) always ends in 5, and \(6^{n}\) always ends in 6.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The unit digit of a number raised to any power depends on the cycle of the unit digit of the base.
For any natural number \(n\), we need to find the last digit of \(6^{n}\).
Step 2: Key Formula or Approach:
We observe the pattern of powers of 6:
\[ 6^{1} = 6 \]
\[ 6^{2} = 36 \]
\[ 6^{3} = 216 \]
\[ 6^{4} = 1296 \]
Step 3: Detailed Explanation:
From the pattern above, it is clear that for any power of 6, the unit digit is always 6.
Mathematically, if the unit digit of a number is \(k\), then the unit digit of \(k \times k\) will determine the next power's unit digit.
Since \(6 \times 6 = 36\), which again ends in 6, the product will always end in 6 regardless of how many times it is multiplied by itself.
Step 4: Final Answer:
Therefore, for any natural number \(n\), \(6^{n}\) always ends with the digit 6.