`17^22 - 1` is a multiple of

Options

A. 16
B. 44
C. 46
D. 27

Answer

To determine what `17^22 - 1` is a multiple of, we can apply the binomial theorem and analyze the expression step by step.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to evaluate the expression `17^22 - 1`. We can rewrite `17` as `16 + 1`. Thus, we have:
\[ 17^22 - 1 = (16 + 1)^{22} - 1 \]

**Hint**: Rewrite the base in a form that can be expanded using the binomial theorem.

2. **Applying the Binomial Theorem**:
According to the binomial theorem, we can expand `(p + q)^n` as:
\[ (p + q)^n = \sum_{k=0}^{n} \binom{n}{k} p^{n-k} q^k \]
Here, let \( p = 16 \) and \( q = 1 \), and \( n = 22 \):
\[ (16 + 1)^{22} = \sum_{k=0}^{22} \binom{22}{k} 16^{22-k} \cdot 1^k \]

**Hint**: Use the binomial expansion to express the power in terms of its components.

3. **Subtracting 1**:
Now we subtract 1 from the expansion:
\[ (16 + 1)^{22} - 1 = \sum_{k=0}^{22} \binom{22}{k} 16^{22-k} \cdot 1^k - 1 \]
The term corresponding to \( k = 0 \) in the expansion is \( 16^{22} \) and the term corresponding to \( k = 22 \) is \( 1 \). Thus:
\[ \sum_{k=1}^{22} \binom{22}{k} 16^{22-k} \cdot 1^k \]

**Hint**: Recognize that subtracting 1 removes the constant term from the expansion.

4. **Factoring Out 16**:
Notice that all terms in the sum for \( k = 1 \) to \( 22 \) contain \( 16 \) as a factor. We can factor \( 16 \) out:
\[ = 16 \left( \binom{22}{1} 16^{21} + \binom{22}{2} 16^{20} + \ldots + \binom{22}{22} \cdot 1 \right) \]

**Hint**: Factor out the common term to simplify the expression.

5. **Conclusion**:
Since we have factored out \( 16 \), we can conclude that:
\[ 17^{22} - 1 \text{ is a multiple of } 16. \]

**Final Verification**: Check the options provided:
- Option A: 16 (Correct)
- Option B: 44 (Not a multiple)
- Option C: 46 (Not a multiple)
- Option D: 27 (Not a multiple)

Thus, the answer is that `17^22 - 1` is a multiple of **16**.