The number of zeroes at the end of (127)! Is

Options

A. 31
B. 30
C. 0
D. 10

Answer

To find the number of zeros at the end of \(127!\), we need to determine how many times \(10\) is a factor in \(127!\). Since \(10 = 2 \times 5\), and there are always more factors of \(2\) than \(5\) in factorials, we only need to count the number of times \(5\) is a factor in \(127!\).

The number of times a prime \(p\) divides \(n!\) can be calculated using the formula:

\[ \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]

In our case, \(n = 127\) and \(p = 5\). We will compute:

1. \(\left\lfloor \frac{127}{5} \right\rfloor\)
2. \(\left\lfloor \frac{127}{5^2} \right\rfloor\)
3. \(\left\lfloor \frac{127}{5^3} \right\rfloor\)
4. \(\left\lfloor \frac{127}{5^4} \right\rfloor\)

Let's calculate each term step by step:

### Step 1: Calculate \(\left\lfloor \frac{127}{5} \right\rfloor\)

\[ \frac{127}{5} = 25.4 \quad \Rightarrow \quad \left\lfloor 25.4 \right\rfloor = 25 \]

### Step 2: Calculate \(\left\lfloor \frac{127}{5^2} \right\rfloor\)

\[ \frac{127}{25} = 5.08 \quad \Rightarrow \quad \left\lfloor 5.08 \right\rfloor = 5 \]

### Step 3: Calculate \(\left\lfloor \frac{127}{5^3} \right\rfloor\)

\[ \frac{127}{125} = 1.016 \quad \Rightarrow \quad \left\lfloor 1.016 \right\rfloor = 1 \]

### Step 4: Calculate \(\left\lfloor \frac{127}{5^4} \right\rfloor\)

\[ \frac{127}{625} = 0.2032 \quad \Rightarrow \quad \left\lfloor 0.2032 \right\rfloor = 0 \]

### Step 5: Sum the results

Now we sum all the values we calculated:

\[ 25 + 5 + 1 + 0 = 31 \]

Thus, the number of zeros at the end of \(127!\) is \(31\).

### Final Answer:
The number of zeros at the end of \(127!\) is \(31\).

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