Question

For any natural number $n$, $6^n$ ends with the digit :

• A. 0
• B. 6
• C. 3
• D. 2

Hint:

The numbers ending with digits 0, 1, 5, and 6 have a cyclicity of 1.
This means any positive integral power of these numbers will always end with the same unit digit.
For example, $5^n$ always ends in 5, and $6^n$ always ends in 6.

Answer 1

The Correct Option is B

Step 1: What we are checking.
We just need the units digit of $6^n$ for any natural number $n$, so let's watch what happens to the last digit each time we multiply by 6.
Step 2: Multiply out the first few powers.
$6^1 = 6$. Multiplying by 6 again, $6 \times 6 = 36$, so $6^2$ also ends in 6. Multiply once more and the last digit is still governed by $6 \times 6 = 36$, so it stays 6.
Step 3: See why this never changes.
Once a number ends in 6, multiplying it by 6 always gives a units digit of $6 \times 6 = 36$, which again ends in 6. So the last digit is stuck at 6 no matter how large $n$ gets.
\[ \boxed{6} \]