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 break the problem into manageable steps.

### Step 1: Find the remainder of \( 2^{2020} \) when divided by 9.

To do this, we can use **Fermat's Little Theorem**, which states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[ a^{p-1} \equiv 1 \mod p \]

In our case, \( a = 2 \) and \( p = 9 \). However, since 9 is not prime, we will look for a pattern in the powers of 2 modulo 9.

Calculating the first few powers of 2 modulo 9:

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

We see that \( 2^6 \equiv 1 \mod 9 \). This means the powers of 2 repeat every 6 terms.

### Step 2: Reduce the exponent \( 2020 \) modulo 6.

Now, we need to find \( 2020 \mod 6 \):

\[ 2020 \div 6 = 336 \quad \text{(with a remainder of 4)} \]

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

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

Since \( 2020 \equiv 4 \mod 6 \):

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

From our earlier calculations:

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

### Step 4: Find the remainder of \( 2021 \) when divided by 9.

Now we calculate \( 2021 \mod 9 \):

\[ 2021 \div 9 = 224 \quad \text{(with a remainder of 5)} \]

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

### Step 5: Combine the results.

Now we add the two remainders:

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

Calculating this gives:

\[ 7 + 5 = 12 \]

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

\[ 12 \equiv 3 \mod 9 \]

### Final Answer:

The remainder when \( 2^{2020} + 2021 \) is divided by 9 is \( \boxed{3} \).

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