If `S=1!+4!+7!+10!+ . . .+400!,` then Q. The last two digits in `(1!+4!+7!)!` is

Options

A. 1
B. 10
C. 0
D. None of these

Answer

To solve the problem, we need to find the last two digits of \( (1! + 4! + 7!)! \).

### Step 1: Calculate \( 1! \), \( 4! \), and \( 7! \)

1. **Calculate \( 1! \)**:
\[ 1! = 1 \]

2. **Calculate \( 4! \)**:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]

3. **Calculate \( 7! \)**:
\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \]

### Step 2: Sum the factorials

Now we sum these values:
\[ S = 1! + 4! + 7! = 1 + 24 + 5040 = 5065 \]

### Step 3: Calculate \( S! \)

Next, we need to find the last two digits of \( 5065! \).

### Step 4: Analyze \( 5065! \) for last two digits

To find the last two digits of \( 5065! \), we need to consider the factors of 10 in \( 5065! \). Each factor of 10 contributes a zero to the end of the number. A factor of 10 is made up of a factor of 2 and a factor of 5.

1. **Count the number of factors of 5 in \( 5065! \)**:
\[ \left\lfloor \frac{5065}{5} \right\rfloor + \left\lfloor \frac{5065}{25} \right\rfloor + \left\lfloor \frac{5065}{125} \right\rfloor + \left\lfloor \frac{5065}{625} \right\rfloor \]
- \( \left\lfloor \frac{5065}{5} \right\rfloor = 1013 \)
- \( \left\lfloor \frac{5065}{25} \right\rfloor = 202 \)
- \( \left\lfloor \frac{5065}{125} \right\rfloor = 40 \)
- \( \left\lfloor \frac{5065}{625} \right\rfloor = 8 \)

Total factors of 5:
\[ 1013 + 202 + 40 + 8 = 1263 \]

2. **Count the number of factors of 2 in \( 5065! \)** (not necessary for the last two digits, but for completeness):
\[ \left\lfloor \frac{5065}{2} \right\rfloor + \left\lfloor \frac{5065}{4} \right\rfloor + \left\lfloor \frac{5065}{8} \right\rfloor + \ldots \]
This will yield a much larger number than the count of factors of 5.

### Conclusion

Since \( 5065! \) has at least 1263 factors of 10, it means that \( 5065! \) ends with at least 1263 zeros. Therefore, the last two digits of \( 5065! \) are:
\[ \boxed{00} \]