Consider the number `N=2016`. Find the Number of cyphers at the end of `.^(N)C_(N//2)` is

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem of finding the number of zeros at the end of \( \binom{2016}{1008} \), we can follow these steps:

### Step 1: Understand the Problem
We need to find the number of trailing zeros in the binomial coefficient \( \binom{2016}{1008} \). The number of trailing zeros in a number is determined by the number of times 10 is a factor in that number, which is the product of the factors 2 and 5. Since there are generally more factors of 2 than factors of 5, we only need to count the number of factors of 5.

### Step 2: Use the Formula for Binomial Coefficient
The binomial coefficient \( \binom{n}{r} \) is given by the formula:
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
For our case:
\[ \binom{2016}{1008} = \frac{2016!}{1008! \times 1008!} \]

### Step 3: Count the Factors of 5 in Factorials
To find the number of trailing zeros in \( \binom{2016}{1008} \), we need to calculate the number of factors of 5 in \( 2016! \) and in \( 1008! \).

#### Step 3.1: Calculate \( E_5(2016!) \)
Using the formula for counting the number of factors of a prime \( p \) in \( n! \):
\[ E_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]
For \( n = 2016 \) and \( p = 5 \):
\[ E_5(2016!) = \left\lfloor \frac{2016}{5} \right\rfloor + \left\lfloor \frac{2016}{25} \right\rfloor + \left\lfloor \frac{2016}{125} \right\rfloor + \left\lfloor \frac{2016}{625} \right\rfloor \]
Calculating each term:
- \( \left\lfloor \frac{2016}{5} \right\rfloor = 403 \)
- \( \left\lfloor \frac{2016}{25} \right\rfloor = 80 \)
- \( \left\lfloor \frac{2016}{125} \right\rfloor = 16 \)
- \( \left\lfloor \frac{2016}{625} \right\rfloor = 3 \)

Adding these together:
\[ E_5(2016!) = 403 + 80 + 16 + 3 = 502 \]

#### Step 3.2: Calculate \( E_5(1008!) \)
Now we do the same for \( n = 1008 \):
\[ E_5(1008!) = \left\lfloor \frac{1008}{5} \right\rfloor + \left\lfloor \frac{1008}{25} \right\rfloor + \left\lfloor \frac{1008}{125} \right\rfloor + \left\lfloor \frac{1008}{625} \right\rfloor \]
Calculating each term:
- \( \left\lfloor \frac{1008}{5} \right\rfloor = 201 \)
- \( \left\lfloor \frac{1008}{25} \right\rfloor = 40 \)
- \( \left\lfloor \frac{1008}{125} \right\rfloor = 8 \)
- \( \left\lfloor \frac{1008}{625} \right\rfloor = 1 \)

Adding these together:
\[ E_5(1008!) = 201 + 40 + 8 + 1 = 250 \]

### Step 4: Calculate the Number of Trailing Zeros
Now we can find the number of trailing zeros in \( \binom{2016}{1008} \):
\[ \text{Trailing Zeros} = E_5(2016!) - 2 \times E_5(1008!) \]
Substituting the values we found:
\[ \text{Trailing Zeros} = 502 - 2 \times 250 = 502 - 500 = 2 \]

### Final Answer
Thus, the number of trailing zeros (ciphers) at the end of \( \binom{2016}{1008} \) is \( \boxed{2} \).