The remainder left out when 8 2n −(62) 2n+1 is divided by 9 is

Options

A. a. 1
B. b. `a - 1`
C. c. `n`
D. d. none of these

Answer

To find the remainder left when \( 8^{2n} - 62^{2n+1} \) is divided by 9, we can follow these steps:

### Step 1: Rewrite the bases in terms of modulo 9
First, we simplify \( 8 \) and \( 62 \) modulo \( 9 \):
- \( 8 \equiv -1 \mod 9 \)
- \( 62 \equiv 8 \mod 9 \) (since \( 62 - 9 \times 6 = 62 - 54 = 8 \))

### Step 2: Substitute the simplified values
Now we can rewrite the expression:
\[ 8^{2n} - 62^{2n+1} \equiv (-1)^{2n} - 8^{2n+1} \mod 9 \]

### Step 3: Simplify the powers
Since \( (-1)^{2n} = 1 \) for any integer \( n \):
\[ 1 - 8^{2n+1} \mod 9 \]

Next, we simplify \( 8^{2n+1} \) modulo \( 9 \):
\[ 8^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \mod 9 \]

### Step 4: Substitute back into the expression
Now substituting this back into our expression:
\[ 1 - (-1) \mod 9 \]

### Step 5: Simplify the final expression
This simplifies to:
\[ 1 + 1 = 2 \mod 9 \]

### Conclusion
Thus, the remainder when \( 8^{2n} - 62^{2n+1} \) is divided by \( 9 \) is \( 2 \).

### Final Answer
The remainder is \( 2 \).

---