`4^(61)+4^(62)+4^(63)+4^(64)` is divisible by

Options

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

Answer

To solve the expression \(4^{61} + 4^{62} + 4^{63} + 4^{64}\) and determine what it is divisible by, we can factor out the smallest power of 4 from the expression.

### Step 1: Factor out the smallest power of 4

The smallest power in the expression is \(4^{61}\). We can factor this out:

\[ 4^{61} + 4^{62} + 4^{63} + 4^{64} = 4^{61}(1 + 4 + 4^2 + 4^3) \]

### 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:

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

So, we have:

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

### Step 3: Rewrite the expression

Now we can rewrite the entire expression:

\[ 4^{61}(85) \]

### Step 4: Determine divisibility

Now, we need to find what \(4^{61} \times 85\) is divisible by.

1. \(4^{61}\) is clearly divisible by \(4\) (since \(4^{61} = 2^{122}\), it is also divisible by \(2\)).
2. \(85\) can be factored as \(5 \times 17\).

Thus, \(4^{61} \times 85\) is divisible by:

- \(4\) (from \(4^{61}\))
- \(5\) (from \(85\))
- \(17\) (from \(85\))

### Final Answer

Therefore, \(4^{61} + 4^{62} + 4^{63} + 4^{64}\) is divisible by \(4\), \(5\), and \(17\).