If `2^(2020)+2021` is divided by 9, then the remainder obtained is

Options

A. 0
B. 1
C. 3
D. 7

Answer

To find the remainder when \( 2^{2020} + 2021 \) is divided by 9, we can follow these steps:

### Step 1: Find \( 2^{2020} \mod 9 \)

To find \( 2^{2020} \mod 9 \), we can use **Fermat's Little Theorem**, which states that if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \mod p \).

Here, \( p = 9 \) is not prime, but we can find the powers of 2 modulo 9 directly:

- \( 2^1 \equiv 2 \mod 9 \)
- \( 2^2 \equiv 4 \mod 9 \)
- \( 2^3 \equiv 8 \mod 9 \)
- \( 2^4 \equiv 7 \mod 9 \)
- \( 2^5 \equiv 5 \mod 9 \)
- \( 2^6 \equiv 1 \mod 9 \)

Notice that \( 2^6 \equiv 1 \mod 9 \), which means the powers of 2 repeat every 6 terms.

### Step 2: Reduce the exponent modulo 6

Now, we need to reduce \( 2020 \) modulo \( 6 \):

\[ 2020 \div 6 = 336 \quad \text{remainder } 4 \]

So, \( 2020 \equiv 4 \mod 6 \).

### Step 3: Calculate \( 2^{2020} \mod 9 \)

Using the result from Step 2:

\[ 2^{2020} \equiv 2^4 \mod 9 \]

From our earlier calculations:

\[ 2^4 \equiv 7 \mod 9 \]

### Step 4: Add 2021 and find the result modulo 9

Now we calculate \( 2^{2020} + 2021 \mod 9 \):

First, find \( 2021 \mod 9 \):

\[ 2021 \div 9 = 224 \quad \text{remainder } 5 \]

So, \( 2021 \equiv 5 \mod 9 \).

Now, combine the results:

\[ 2^{2020} + 2021 \equiv 7 + 5 \mod 9 \]

\[ 7 + 5 = 12 \]

Now, find \( 12 \mod 9 \):

\[ 12 \div 9 = 1 \quad \text{remainder } 3 \]

### Final Answer

Thus, the remainder when \( 2^{2020} + 2021 \) is divided by 9 is:

\[ \boxed{3} \]