Which of the following options is correct?

Options

A. If n is any natural number, then `6^n – 5^n` always ends with 1.
B. For any integer t, every even integer is of the form 2t + 1.
C. Both A and B
D. Neither A nor B

Answer

To solve the question, we need to determine which of the given options is correct based on the information provided. Let's analyze the first option step by step.

### Step 1: Understand the first option
The first option states that for any natural number \( n \), \( 6^n - 5^n \) always ends with the digit 1.

### Step 2: Test with \( n = 1 \)
Let's calculate \( 6^1 - 5^1 \):
\[ 6^1 - 5^1 = 6 - 5 = 1 \]
**Hint:** Check the result of \( 6^n - 5^n \) for small values of \( n \).

### Step 3: Test with \( n = 2 \)
Now, let's calculate \( 6^2 - 5^2 \):
\[ 6^2 - 5^2 = 36 - 25 = 11 \]
The last digit is 1.

### Step 4: Test with \( n = 3 \)
Next, we calculate \( 6^3 - 5^3 \):
\[ 6^3 - 5^3 = 216 - 125 = 91 \]
The last digit is 1.

### Step 5: Generalize the pattern
Now, we can observe a pattern. Let's analyze the last digits of \( 6^n \) and \( 5^n \):
- The last digit of \( 6^n \) is always 6 for any \( n \).
- The last digit of \( 5^n \) is always 5 for any \( n \).

Thus, the last digit of \( 6^n - 5^n \) will always be:
\[ 6 - 5 = 1 \]

### Conclusion for the first option
Since we have shown that \( 6^n - 5^n \) ends with the digit 1 for any natural number \( n \), the first option is correct.

### Step 6: Analyze the second option
The second option states that every even integer can be expressed in the form \( 2t + 1 \) for any integer \( t \).

### Step 7: Test with \( t = 0 \)
If we take \( t = 0 \):
\[ 2t + 1 = 2(0) + 1 = 1 \]
1 is not an even integer.

### Conclusion for the second option
Since we found a case where \( 2t + 1 \) does not yield an even integer, the second option is incorrect.

### Final Answer
The only correct option is the first one.