Show that `2^(16)` divides 32 !. Find the highest power of 2 that divides 32 !.

Answer

To solve the problem of showing that \(2^{16}\) divides \(32!\) and finding the highest power of \(2\) that divides \(32!\), we will use the formula for finding the highest power of a prime \(p\) that divides \(n!\):

\[ \text{Highest power of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]

### Step-by-step Solution:

1. **Identify the values**: Here, \(n = 32\) and \(p = 2\).

2. **Calculate each term**:
- First term:
\[ \left\lfloor \frac{32}{2} \right\rfloor = \left\lfloor 16 \right\rfloor = 16 \]
- Second term:
\[ \left\lfloor \frac{32}{2^2} \right\rfloor = \left\lfloor \frac{32}{4} \right\rfloor = \left\lfloor 8 \right\rfloor = 8 \]
- Third term:
\[ \left\lfloor \frac{32}{2^3} \right\rfloor = \left\lfloor \frac{32}{8} \right\rfloor = \left\lfloor 4 \right\rfloor = 4 \]
- Fourth term:
\[ \left\lfloor \frac{32}{2^4} \right\rfloor = \left\lfloor \frac{32}{16} \right\rfloor = \left\lfloor 2 \right\rfloor = 2 \]
- Fifth term:
\[ \left\lfloor \frac{32}{2^5} \right\rfloor = \left\lfloor \frac{32}{32} \right\rfloor = \left\lfloor 1 \right\rfloor = 1 \]
- Sixth term:
\[ \left\lfloor \frac{32}{2^6} \right\rfloor = \left\lfloor \frac{32}{64} \right\rfloor = \left\lfloor 0.5 \right\rfloor = 0 \]

3. **Sum the terms**:
\[ 16 + 8 + 4 + 2 + 1 = 31 \]

4. **Conclusion**: The highest power of \(2\) that divides \(32!\) is \(31\). Since \(31 > 16\), we can conclude that \(2^{16}\) divides \(32!\).

### Final Answer:
- The highest power of \(2\) that divides \(32!\) is \(31\).
- Therefore, \(2^{16}\) divides \(32!\).