The expression `2^(6n) - 4^(2n)`, where n is a natural number is always divisible by

Options

A. 15
B. 18
C. 36
D. 48

Answer

To solve the expression \(2^{6n} - 4^{2n}\) and determine what it is always divisible by, we can follow these steps:

### Step 1: Rewrite the expression
First, we can rewrite \(4^{2n}\) in terms of base 2:
\[ 4^{2n} = (2^2)^{2n} = 2^{4n} \]
Thus, the expression becomes:
\[ 2^{6n} - 2^{4n} \]

### Step 2: Factor the expression
Next, we can factor out the common term \(2^{4n}\):
\[ 2^{6n} - 2^{4n} = 2^{4n}(2^{2n} - 1) \]

### Step 3: Analyze the factor \(2^{2n} - 1\)
Now, we need to analyze the term \(2^{2n} - 1\). This expression can be factored further using the difference of squares:
\[ 2^{2n} - 1 = (2^n - 1)(2^n + 1) \]

### Step 4: Combine the factors
Now we have:
\[ 2^{6n} - 4^{2n} = 2^{4n} \cdot (2^n - 1)(2^n + 1) \]

### Step 5: Determine divisibility
To find the divisibility of the entire expression, we need to consider the factors:
1. \(2^{4n}\) is always divisible by \(16\) (since \(2^{4n} = 16^n\)).
2. The product \((2^n - 1)(2^n + 1)\) consists of two consecutive integers, which means one of them is always even. Therefore, the product is divisible by \(2\).

### Step 6: Conclusion on divisibility
Since \(2^{4n}\) contributes a factor of \(16\) and \((2^n - 1)(2^n + 1)\) contributes at least a factor of \(2\), we can conclude that:
\[ 2^{4n} \cdot (2^n - 1)(2^n + 1) \text{ is divisible by } 16 \cdot 2 = 32. \]

However, we can also check specific values of \(n\) (e.g., \(n = 1, 2, 3\)) to find a higher common divisor.

For \(n = 1\):
\[ 2^{6 \cdot 1} - 4^{2 \cdot 1} = 64 - 16 = 48 \]

For \(n = 2\):
\[ 2^{6 \cdot 2} - 4^{2 \cdot 2} = 4096 - 256 = 3840 \]

For \(n = 3\):
\[ 2^{6 \cdot 3} - 4^{2 \cdot 3} = 262144 - 65536 = 196608 \]

The greatest common divisor of these results is \(48\).

### Final Answer
Thus, the expression \(2^{6n} - 4^{2n}\) is always divisible by \(48\).

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