`4^(11)+4^(12)+4^(13)+4^(14)` is divisible by

Options

A. 7
B. 14
C. 17
D. 9

Answer

To solve the expression \(4^{11} + 4^{12} + 4^{13} + 4^{14}\) and determine its divisibility, we can follow these steps:

### Step 1: Factor out the common term
We notice that each term in the expression has a common factor of \(4^{11}\). Therefore, we can factor it out:

\[ 4^{11} + 4^{12} + 4^{13} + 4^{14} = 4^{11}(1 + 4 + 4^2 + 4^3) \]

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

\[ 1 + 4 + 4^2 + 4^3 = 1 + 4 + 16 + 64 \]

Calculating this step-by-step:

- \(1 + 4 = 5\)
- \(5 + 16 = 21\)
- \(21 + 64 = 85\)

Thus, we have:

\[ 1 + 4 + 4^2 + 4^3 = 85 \]

### Step 3: Substitute back into the factored expression
Now we substitute back into our factored expression:

\[ 4^{11} \times 85 \]

### Step 4: Determine the divisibility of the expression
Next, we need to check the divisibility of \(4^{11} \times 85\).

1. **Divisibility by 5**: Since \(85 = 5 \times 17\), it is divisible by 5.
2. **Divisibility by 17**: As mentioned, \(85\) can also be expressed as \(17 \times 5\), so it is divisible by 17 as well.

### Conclusion
Thus, the expression \(4^{11} + 4^{12} + 4^{13} + 4^{14}\) is divisible by both 5 and 17. Among the options given (7, 14, 17, 9), the correct answer is:

\[ \text{Divisible by } 17 \]