If highest power of 8 in N! is 19 find highest power of 8 in (N+1)!.

Options

A. 19 or 20
B. 21 or 22
C. 19 or 21
D. None of these

Answer

To find the highest power of 8 in (N+1)!, given that the highest power of 8 in N! is 19, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Relationship Between Powers of 8 and 2**:
Since \(8 = 2^3\), the highest power of 8 in \(N!\) can be found by determining the highest power of 2 in \(N!\) and then dividing that by 3. Therefore, if the highest power of 8 in \(N!\) is 19, we can express this as:
\[ \frac{\text{highest power of 2 in } N!}{3} = 19 \]
This implies:
\[ \text{highest power of 2 in } N! = 19 \times 3 = 57 \]

2. **Finding the Highest Power of 2 in (N+1)!**:
The highest power of 2 in \((N+1)!\) can be calculated using the formula:
\[ \text{highest power of 2 in } (N+1)! = \text{highest power of 2 in } N! + \text{number of factors of 2 in } (N+1) \]
The number of factors of 2 in \((N+1)\) can be found using:
\[ \left\lfloor \frac{N+1}{2} \right\rfloor + \left\lfloor \frac{N+1}{4} \right\rfloor + \left\lfloor \frac{N+1}{8} \right\rfloor + \ldots \]

3. **Calculating for N = 62 or N = 63**:
From the video transcript, we find that \(N\) could be either 62 or 63. We will calculate for both cases.

- **Case 1: N = 62**:
\[ \text{highest power of 2 in } (62 + 1)! = 57 + \left\lfloor \frac{63}{2} \right\rfloor + \left\lfloor \frac{63}{4} \right\rfloor + \left\lfloor \frac{63}{8} \right\rfloor + \left\lfloor \frac{63}{16} \right\rfloor + \left\lfloor \frac{63}{32} \right\rfloor \]
\[ = 57 + 31 + 15 + 7 + 3 + 1 = 114 \]

- **Case 2: N = 63**:
\[ \text{highest power of 2 in } (63 + 1)! = 57 + \left\lfloor \frac{64}{2} \right\rfloor + \left\lfloor \frac{64}{4} \right\rfloor + \left\lfloor \frac{64}{8} \right\rfloor + \left\lfloor \frac{64}{16} \right\rfloor + \left\lfloor \frac{64}{32} \right\rfloor + \left\lfloor \frac{64}{64} \right\rfloor \]
\[ = 57 + 32 + 16 + 8 + 4 + 2 + 1 = 120 \]

4. **Finding the Highest Power of 8 in (N+1)!**:
Now, we convert the highest power of 2 back to the highest power of 8:
- For \(N = 62\):
\[ \text{highest power of 8 in } (N+1)! = \frac{114}{3} = 38 \]
- For \(N = 63\):
\[ \text{highest power of 8 in } (N+1)! = \frac{120}{3} = 40 \]

### Conclusion:
The highest power of 8 in \((N+1)!\) can be either 38 or 40 depending on whether \(N\) is 62 or 63.