Find the number of zeroes in 129!

Options

A. 31
B. 33
C. 35
D. 37

Answer

To find the number of trailing zeros in \( 129! \), we can use the formula that counts the number of times \( 10 \) is a factor in the factorial. 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 \( 129! \).

The formula to find the number of trailing zeros in \( n! \) is given by:

\[ \text{Number of trailing zeros} = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{5^2} \right\rfloor + \left\lfloor \frac{n}{5^3} \right\rfloor + \ldots \]

We will apply this formula step by step for \( n = 129 \).

### Step 1: Calculate \( \left\lfloor \frac{129}{5} \right\rfloor \)

\[ \left\lfloor \frac{129}{5} \right\rfloor = \left\lfloor 25.8 \right\rfloor = 25 \]

### Step 2: Calculate \( \left\lfloor \frac{129}{25} \right\rfloor \)

\[ \left\lfloor \frac{129}{25} \right\rfloor = \left\lfloor 5.16 \right\rfloor = 5 \]

### Step 3: Calculate \( \left\lfloor \frac{129}{125} \right\rfloor \)

\[ \left\lfloor \frac{129}{125} \right\rfloor = \left\lfloor 1.032 \right\rfloor = 1 \]

### Step 4: Calculate \( \left\lfloor \frac{129}{625} \right\rfloor \)

Since \( 625 > 129 \), we have:

\[ \left\lfloor \frac{129}{625} \right\rfloor = 0 \]

### Step 5: Sum all the values calculated

Now, we sum all the values obtained in the previous steps:

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

Thus, the number of trailing zeros in \( 129! \) is \( 31 \).

### Final Answer

The number of trailing zeros in \( 129! \) is \( 31 \).

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