Question

For any natural number n, \(5^n\) ends with the digit :

• A. 0
• B. 5
• C. 3
• D. 2

Hint:

The numbers ending in 0, 1, 5, and 6 always retain their unit digits when raised to any positive integer power.
Specifically:
- \(0^n\) ends in 0
- \(1^n\) ends in 1
- \(5^n\) ends in 5
- \(6^n\) ends in 6
Recognizing this cyclical property saves time during exponentiation questions!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to determine the unit digit of \(5^n\) where \(n\) is any natural number (\(n \in \{1, 2, 3, \dots\}\)).

Step 2: Key Formula or Approach:
1. The unit digit of a number can be analyzed by looking at the cycle of units digits under exponentiation.
2. Alternatively, we can use mathematical induction or modular arithmetic modulo 10:
\[ 5^n \equiv x \pmod{10} \]
where \(x\) represents the ending digit.

Step 3: Detailed Explanation:
1. Let us compute the values of \(5^n\) for the first few natural numbers:
- For \(n = 1\):
\[ 5^1 = 5 \quad (\text{ends with } 5) \]
- For \(n = 2\):
\[ 5^2 = 25 \quad (\text{ends with } 5) \]
- For \(n = 3\):
\[ 5^3 = 125 \quad (\text{ends with } 5) \]
- For \(n = 4\):
\[ 5^4 = 625 \quad (\text{ends with } 5) \]
2. Observe that in every calculation, the unit digit is consistently 5.
3. Let us prove this generally using induction:
Assume that \(5^k\) ends with the digit 5 for some natural number \(k\). This means:
\[ 5^k = 10m + 5 \]
for some non-negative integer \(m\).
Now, find the value for \(n = k+1\):
\[ 5^{k+1} = 5^k \times 5 = (10m + 5) \times 5 = 50m + 25 = 10(5m + 2) + 5 \]
Since \(10(5m + 2) + 5\) is of the form \(10k' + 5\), \(5^{k+1}\) must also end with the digit 5.
Therefore, by induction, \(5^n\) always ends with the digit 5 for any natural number \(n\).

Step 4: Final Answer:
For any natural number \(n\), \(5^n\) ends with the digit 5, which corresponds to option (B).