`(4^(61) + 4^(62) + 4^(63))` is divisible by

Options

A. 3
B. 11
C. 13
D. 17

Answer

To solve the expression \(4^{61} + 4^{62} + 4^{63}\) and determine its divisibility, we can follow these steps:

### Step 1: Factor out the common term
We notice that all terms in the expression share a common factor of \(4^{61}\). We can factor this out:

\[ 4^{61} + 4^{62} + 4^{63} = 4^{61}(1 + 4 + 4^2) \]

### Step 2: Simplify the expression inside the parentheses
Next, we simplify the expression inside the parentheses:

\[ 1 + 4 + 4^2 = 1 + 4 + 16 = 21 \]

### Step 3: Rewrite the expression
Now we can rewrite the entire expression as:

\[ 4^{61} \times 21 \]

### Step 4: Determine the divisibility
Now we need to check the divisibility of \(4^{61} \times 21\).

- \(4^{61}\) is clearly divisible by \(4\).
- \(21\) can be factored into \(3 \times 7\), which means it is divisible by \(3\) and \(7\).

### Conclusion
Thus, the expression \(4^{61} + 4^{62} + 4^{63}\) is divisible by \(3\), \(7\), and \(4\).

Since the question asks for a specific divisor, we can conclude that the expression is divisible by \(3\).

### Final Answer
The expression \(4^{61} + 4^{62} + 4^{63}\) is divisible by \(3\).

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