If the product of `1 xx 2 xx 3 xx 4 xx … n` contains 68 zeros in the end of the number. Then the maximum possible number of values of n is:

Options

A. 1
B. 3
C. 5
D. 6

Answer

To find the maximum possible number of values of \( n \) such that the product \( 1 \times 2 \times 3 \times \ldots \times n \) (which is \( n! \)) contains exactly 68 zeros at the end, we can follow these steps:

### Step 1: Understanding the number of trailing zeros
The number of trailing zeros in \( n! \) is determined by the number of times \( n! \) can be divided by 10. Since \( 10 = 2 \times 5 \) and there are always more factors of 2 than factors of 5 in \( n! \), we only need to count the number of times 5 is a factor in the numbers from 1 to \( n \).

### Step 2: Using the formula for counting trailing zeros
The formula for counting 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 \( Z(n) = 68 \).

### Step 3: Estimating \( n \)
To estimate \( n \), we can start with a rough approximation. Since \( Z(n) \) is approximately \( \frac{n}{5} \), we can set up the equation:
\[ \frac{n}{5} \approx 68 \implies n \approx 340 \]
This gives us a starting point to check values around 340.

### Step 4: Checking values of \( n \)
We will calculate \( Z(n) \) for values around 340 to find when it equals 68.

1. **Calculate \( Z(340) \)**:
\[ Z(340) = \left\lfloor \frac{340}{5} \right\rfloor + \left\lfloor \frac{340}{25} \right\rfloor + \left\lfloor \frac{340}{125} \right\rfloor \]
\[ = 68 + 13 + 2 = 83 \quad (\text{too high}) \]

2. **Calculate \( Z(300) \)**:
\[ Z(300) = \left\lfloor \frac{300}{5} \right\rfloor + \left\lfloor \frac{300}{25} \right\rfloor + \left\lfloor \frac{300}{125} \right\rfloor \]
\[ = 60 + 12 + 2 = 74 \quad (\text{still high}) \]

3. **Calculate \( Z(275) \)**:
\[ Z(275) = \left\lfloor \frac{275}{5} \right\rfloor + \left\lfloor \frac{275}{25} \right\rfloor + \left\lfloor \frac{275}{125} \right\rfloor \]
\[ = 55 + 11 + 2 = 68 \quad (\text{exactly right}) \]

4. **Check values above and below 275**:
- **For \( n = 276 \)**:
\[ Z(276) = 55 + 11 + 2 = 68 \]
- **For \( n = 277 \)**:
\[ Z(277) = 55 + 11 + 2 = 68 \]
- **For \( n = 278 \)**:
\[ Z(278) = 55 + 11 + 2 = 68 \]
- **For \( n = 279 \)**:
\[ Z(279) = 55 + 11 + 2 = 68 \]
- **For \( n = 280 \)**:
\[ Z(280) = 56 + 11 + 2 = 69 \quad (\text{too high}) \]

### Step 5: Conclusion
The values of \( n \) that give exactly 68 trailing zeros are \( 275, 276, 277, 278, 279 \). Thus, the maximum possible number of values of \( n \) is **5**.