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 \(n!\).

### Step-by-Step Solution:

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

2. **Use the formula**:
The highest power of a prime \(p\) that divides \(n!\) is given by:
\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]
For our case, it becomes:
\[ \sum_{k=1}^{\infty} \left\lfloor \frac{32}{2^k} \right\rfloor \]

3. **Calculate each term**:
- For \(k = 1\):
\[ \left\lfloor \frac{32}{2^1} \right\rfloor = \left\lfloor \frac{32}{2} \right\rfloor = \left\lfloor 16 \right\rfloor = 16 \]
- For \(k = 2\):
\[ \left\lfloor \frac{32}{2^2} \right\rfloor = \left\lfloor \frac{32}{4} \right\rfloor = \left\lfloor 8 \right\rfloor = 8 \]
- For \(k = 3\):
\[ \left\lfloor \frac{32}{2^3} \right\rfloor = \left\lfloor \frac{32}{8} \right\rfloor = \left\lfloor 4 \right\rfloor = 4 \]
- For \(k = 4\):
\[ \left\lfloor \frac{32}{2^4} \right\rfloor = \left\lfloor \frac{32}{16} \right\rfloor = \left\lfloor 2 \right\rfloor = 2 \]
- For \(k = 5\):
\[ \left\lfloor \frac{32}{2^5} \right\rfloor = \left\lfloor \frac{32}{32} \right\rfloor = \left\lfloor 1 \right\rfloor = 1 \]
- For \(k = 6\):
\[ \left\lfloor \frac{32}{2^6} \right\rfloor = \left\lfloor \frac{32}{64} \right\rfloor = \left\lfloor 0.5 \right\rfloor = 0 \]
Since all further terms will also be \(0\), we can stop here.

4. **Sum the results**:
Now, we sum all the values we calculated:
\[ 16 + 8 + 4 + 2 + 1 = 31 \]

5. **Conclusion**:
Therefore, the highest power of \(2\) that divides \(32!\) is \(31\). Since \(31 > 16\), it follows that \(2^{16}\) also divides \(32!\).

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