`(2^19+1)` is divisible by :

Options

A. 3
B. 4
C. 6
D. 3 & 6 both

Answer

To determine whether \(2^{19} + 1\) is divisible by 3, 4, 6, or both 3 and 6, we can evaluate \(2^{19} + 1\) modulo these numbers.

### Step 1: Check divisibility by 3
To check if \(2^{19} + 1\) is divisible by 3, we can evaluate \(2^{19} \mod 3\).

1. Notice that \(2 \equiv -1 \mod 3\).
2. Therefore, \(2^{19} \equiv (-1)^{19} \mod 3\).
3. Since 19 is odd, \((-1)^{19} \equiv -1 \mod 3\).
4. Thus, \(2^{19} \equiv 2 \mod 3\) (because \(-1 \mod 3\) is equivalent to 2).
5. Now, \(2^{19} + 1 \equiv 2 + 1 \equiv 3 \equiv 0 \mod 3\).

**Conclusion**: \(2^{19} + 1\) is divisible by 3.

### Step 2: Check divisibility by 4
Next, we check if \(2^{19} + 1\) is divisible by 4.

1. Notice that \(2^{19} \equiv 0 \mod 4\) (since any power of 2 greater than or equal to 2 is divisible by 4).
2. Thus, \(2^{19} + 1 \equiv 0 + 1 \equiv 1 \mod 4\).

**Conclusion**: \(2^{19} + 1\) is not divisible by 4.

### Step 3: Check divisibility by 6
Now, we check if \(2^{19} + 1\) is divisible by 6. Since 6 is the product of 2 and 3, we need to check divisibility by both.

1. We already found that \(2^{19} + 1\) is divisible by 3.
2. However, we found that \(2^{19} + 1\) is not divisible by 2 (since \(2^{19} + 1 \equiv 1 \mod 2\)).

**Conclusion**: \(2^{19} + 1\) is not divisible by 6.

### Final Conclusion
From our checks:
- \(2^{19} + 1\) is divisible by 3.
- \(2^{19} + 1\) is not divisible by 4.
- \(2^{19} + 1\) is not divisible by 6.

Thus, the answer is that \(2^{19} + 1\) is divisible by **3 only**.

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