In `n!` has 17 zeros, then what is the value of n?

Options

A. A)95
B. B)85
C. C)80
D. D)No such value

Answer

To find the value of \( n \) such that \( n! \) has 17 trailing zeros, we can use the formula to count the number of trailing zeros in \( n! \). The number of trailing zeros in \( n! \) is given by:

\[ Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]

We need to find \( n \) such that \( Z(n) = 17 \).

### Step 1: Estimate \( n \)

Since each factor of 5 contributes to a trailing zero, we can start by estimating \( n \). A rough estimate can be made by multiplying the number of zeros by 5:

\[ n \approx 17 \times 5 = 85 \]

### Step 2: Calculate \( Z(85) \)

Now we will calculate \( Z(85) \):

\[ Z(85) = \left\lfloor \frac{85}{5} \right\rfloor + \left\lfloor \frac{85}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{85}{5} \right\rfloor = \left\lfloor 17 \right\rfloor = 17 \)
- \( \left\lfloor \frac{85}{25} \right\rfloor = \left\lfloor 3.4 \right\rfloor = 3 \)

So,

\[ Z(85) = 17 + 3 = 20 \]

### Step 3: Calculate \( Z(80) \)

Since \( Z(85) = 20 \) is greater than 17, we will check \( Z(80) \):

\[ Z(80) = \left\lfloor \frac{80}{5} \right\rfloor + \left\lfloor \frac{80}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{80}{5} \right\rfloor = \left\lfloor 16 \right\rfloor = 16 \)
- \( \left\lfloor \frac{80}{25} \right\rfloor = \left\lfloor 3.2 \right\rfloor = 3 \)

So,

\[ Z(80) = 16 + 3 = 19 \]

### Step 4: Calculate \( Z(75) \)

Since \( Z(80) = 19 \) is still greater than 17, we will check \( Z(75) \):

\[ Z(75) = \left\lfloor \frac{75}{5} \right\rfloor + \left\lfloor \frac{75}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{75}{5} \right\rfloor = \left\lfloor 15 \right\rfloor = 15 \)
- \( \left\lfloor \frac{75}{25} \right\rfloor = \left\lfloor 3 \right\rfloor = 3 \)

So,

\[ Z(75) = 15 + 3 = 18 \]

### Step 5: Calculate \( Z(70) \)

Since \( Z(75) = 18 \) is still greater than 17, we will check \( Z(70) \):

\[ Z(70) = \left\lfloor \frac{70}{5} \right\rfloor + \left\lfloor \frac{70}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{70}{5} \right\rfloor = \left\lfloor 14 \right\rfloor = 14 \)
- \( \left\lfloor \frac{70}{25} \right\rfloor = \left\lfloor 2.8 \right\rfloor = 2 \)

So,

\[ Z(70) = 14 + 2 = 16 \]

### Step 6: Calculate \( Z(72) \)

Since \( Z(70) = 16 \) is less than 17, we will check \( Z(72) \):

\[ Z(72) = \left\lfloor \frac{72}{5} \right\rfloor + \left\lfloor \frac{72}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{72}{5} \right\rfloor = \left\lfloor 14.4 \right\rfloor = 14 \)
- \( \left\lfloor \frac{72}{25} \right\rfloor = \left\lfloor 2.88 \right\rfloor = 2 \)

So,

\[ Z(72) = 14 + 2 = 16 \]

### Step 7: Calculate \( Z(73) \)

Since \( Z(72) = 16 \) is still less than 17, we will check \( Z(73) \):

\[ Z(73) = \left\lfloor \frac{73}{5} \right\rfloor + \left\lfloor \frac{73}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{73}{5} \right\rfloor = \left\lfloor 14.6 \right\rfloor = 14 \)
- \( \left\lfloor \frac{73}{25} \right\rfloor = \left\lfloor 2.92 \right\rfloor = 2 \)

So,

\[ Z(73) = 14 + 2 = 16 \]

### Step 8: Calculate \( Z(74) \)

Since \( Z(73) = 16 \) is still less than 17, we will check \( Z(74) \):

\[ Z(74) = \left\lfloor \frac{74}{5} \right\rfloor + \left\lfloor \frac{74}{25} \right\rfloor \]

Calculating each term:

- \( \left\lfloor \frac{74}{5} \right\rfloor = \left\lfloor 14.8 \right\rfloor = 14 \)
- \( \left\lfloor \frac{74}{25} \right\rfloor = \left\lfloor 2.96 \right\rfloor = 2 \)

So,

\[ Z(74) = 14 + 2 = 16 \]

### Step 9: Calculate \( Z(75) \)

We have already calculated \( Z(75) \) and found it to be 18.

### Conclusion

Since \( Z(75) = 18 \) and \( Z(74) = 16 \), we can conclude that the value of \( n \) such that \( n! \) has exactly 17 trailing zeros is:

\[ \boxed{75} \]