If `N = 10 ! + 20! + 30! + ……….. + 100 !`. Then find the highest power of 10 in `N^(N)`

Options

A. N
B. 2N
C. 4N
D. 124

Answer

To solve the problem of finding the highest power of 10 in \( N^N \), where \( N = 10! + 20! + 30! + \ldots + 100! \), we will follow these steps:

### Step 1: Factor out \( 10! \)
We can factor \( 10! \) out of the expression for \( N \):
\[ N = 10! \left( 1 + \frac{20!}{10!} + \frac{30!}{10!} + \ldots + \frac{100!}{10!} \right) \]
This simplifies to:
\[ N = 10! \left( 1 + 20 \times 19 \times 18 \times \ldots \times 11 + 30 \times 29 \times \ldots \times 11 + \ldots + 100 \times 99 \times \ldots \times 11 \right) \]

### Step 2: Analyze the terms
Each term \( k! \) for \( k = 20, 30, \ldots, 100 \) has at least one factor of 10 (since \( 20, 30, \ldots, 100 \) are all multiples of 10). Therefore, the unit digit of each of these terms will contribute to the overall sum, but they will not contribute additional factors of 10 beyond what is contributed by \( 10! \).

### Step 3: Calculate the number of trailing zeros in \( 10! \)
To find the highest power of 10 in \( 10! \), we need to count the number of trailing zeros. The number of trailing zeros in \( n! \) can be calculated using the formula:
\[ \text{Number of trailing zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]
For \( 10! \):
\[ \left\lfloor \frac{10}{5} \right\rfloor = 2 \]
So, \( 10! \) has 2 trailing zeros.

### Step 4: Determine the power of 10 in \( N \)
Since \( N \) is of the form \( 10! \times \text{(some integer)} \), and the integer does not contribute any additional factors of 10, the highest power of 10 in \( N \) is:
\[ \text{Power of 10 in } N = 2 \]

### Step 5: Calculate the highest power of 10 in \( N^N \)
To find the highest power of 10 in \( N^N \):
\[ \text{Power of 10 in } N^N = N \times \text{(Power of 10 in } N) = N \times 2 \]
Thus:
\[ \text{Power of 10 in } N^N = 2N \]

### Conclusion
The highest power of 10 in \( N^N \) is \( 2N \).