When `2^33` is divided by 10, the remainder will be

Options

A. 2
B. 3
C. 4
D. 8

Answer

To find the remainder when \(2^{33}\) is divided by 10, we can use the concept of cyclicity in powers of 2 modulo 10.

### Step-by-Step Solution:

1. **Identify the powers of 2 modulo 10**:
- Calculate the first few powers of 2 and their remainders when divided by 10:
- \(2^1 = 2\) → remainder is 2
- \(2^2 = 4\) → remainder is 4
- \(2^3 = 8\) → remainder is 8
- \(2^4 = 16\) → remainder is 6 (since \(16 \mod 10 = 6\))
- \(2^5 = 32\) → remainder is 2 (since \(32 \mod 10 = 2\))

2. **Observe the pattern**:
- From the calculations, we see the remainders are:
- \(2, 4, 8, 6\)
- After \(2^4\), the pattern repeats every 4 terms:
- \(2^5 \equiv 2 \mod 10\)
- \(2^6 \equiv 4 \mod 10\)
- \(2^7 \equiv 8 \mod 10\)
- \(2^8 \equiv 6 \mod 10\)
- The cycle of remainders is: \(2, 4, 8, 6\).

3. **Determine the position of \(2^{33}\) in the cycle**:
- Since the cycle length is 4, we need to find \(33 \mod 4\):
- \(33 \div 4 = 8\) remainder \(1\) (since \(33 = 4 \times 8 + 1\)).
- Thus, \(33 \mod 4 = 1\).

4. **Find the corresponding remainder**:
- From the cycle \(2, 4, 8, 6\), the remainder corresponding to \(2^1\) (the first position in the cycle) is \(2\).

5. **Conclusion**:
- Therefore, the remainder when \(2^{33}\) is divided by 10 is **2**.

### Final Answer:
The remainder when \(2^{33}\) is divided by 10 is **2**.