When ` 2^(31)` is divided by 5 the remainder is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To find the remainder when \( 2^{31} \) is divided by 5, we can use the concept of cyclicity in modular arithmetic. Here’s a step-by-step solution:

### Step 1: Identify the pattern of \( 2^n \mod 5 \)

We will calculate the first few powers of 2 modulo 5 to find a repeating pattern:

- \( 2^1 = 2 \) → \( 2 \mod 5 = 2 \)
- \( 2^2 = 4 \) → \( 4 \mod 5 = 4 \)
- \( 2^3 = 8 \) → \( 8 \mod 5 = 3 \)
- \( 2^4 = 16 \) → \( 16 \mod 5 = 1 \)
- \( 2^5 = 32 \) → \( 32 \mod 5 = 2 \)

From the calculations, we can see that the remainders repeat every 4 terms: \( 2, 4, 3, 1 \).

### Step 2: Determine the cyclicity

The cyclicity of \( 2 \mod 5 \) is 4. This means that every 4th power of 2 will have the same remainder when divided by 5.

### Step 3: Reduce the exponent modulo the cyclicity

Now we need to find \( 31 \mod 4 \) to determine the equivalent exponent within the cycle:

\[ 31 \div 4 = 7 \quad \text{(quotient)} \]
\[ 31 - (4 \times 7) = 3 \quad \text{(remainder)} \]

So, \( 31 \mod 4 = 3 \).

### Step 4: Find \( 2^3 \mod 5 \)

Now we can use the remainder we found to determine \( 2^{31} \mod 5 \):

\[ 2^{31} \equiv 2^3 \mod 5 \]

From our earlier calculations, we know:

\[ 2^3 = 8 \quad \Rightarrow \quad 8 \mod 5 = 3 \]

### Step 5: Conclusion

Thus, the remainder when \( 2^{31} \) is divided by 5 is:

\[ \boxed{3} \]