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 have a cyclicity of 1.
Any positive integral power of a number ending in 0, 1, 5, or 6 will always end in the same digit respectively.
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 Question:
This question concerns the unit digit behavior of powers of natural numbers.
We are asked to find the digit at the units place for the expression $6^n$, where $n$ is any positive integer.

Step 2: Key Formula or Approach:
We can approach this by analyzing the prime factorization of 6 or by examining the unit digit cycle of the base number 6.
The units digit of the product of any two numbers depends solely on the product of their units digits.
If we multiply a number ending in 6 by another number ending in 6, the resulting units digit is always determined by $6 \times 6 = 36$, which also ends in 6.

Step 3: Detailed Explanation:

• Let us write out the first few terms of the sequence $6^n$ for natural numbers $n$:
For $n = 1$:
\[ 6^1 = 6 \]
For $n = 2$:
\[ 6^2 = 36 \]
For $n = 3$:
\[ 6^3 = 216 \]
For $n = 4$:
\[ 6^4 = 1296 \]

• From these calculations, we observe that the last digit is consistently 6.

• Let us prove this by mathematical induction:
Assume that $6^k$ ends with the digit 6 for some natural number $k$.
This means we can write $6^k = 10m + 6$ for some non-negative integer $m$.
Now consider the next term, $6^{k+1}$:
\[ 6^{k+1} = 6^k \times 6 \]
Substitute the inductive hypothesis:
\[ 6^{k+1} = (10m + 6) \times 6 \]
\[ 6^{k+1} = 60m + 36 \]
\[ 6^{k+1} = 60m + 30 + 6 \]
\[ 6^{k+1} = 10(6m + 3) + 6 \]
This clearly shows that $6^{k+1}$ is also of the form $10M + 6$, meaning it ends in the digit 6.

• Thus, by mathematical induction, $6^n$ ends with the digit 6 for all natural numbers $n$.


Step 4: Final Answer:
The expression $6^n$ always ends with the digit 6, which matches option (B).