How many zeros will be there at the end of the expression
`(2!)^(2!)+(4!)^(4!)+(8!)^(8!)+(9!)^(9!)+(10!)^(10!)+(11!)^(11!)?`

Options

A. `(8!)^(8!)+(9!)^(9!)+(10!)^(10!)+(11!)^(11!)`
B. `10^(10!)`
C. `4!+6!+8!+2(10!)`
D. 1

Answer

To determine how many zeros are at the end of the expression

\[ (2!)^{(2!)} + (4!)^{(4!)} + (8!)^{(8!)} + (9!)^{(9!)} + (10!)^{(10!)} + (11!)^{(11!)} \]

we need to find the number of factors of 10 in the entire expression. A factor of 10 is made up of a factor of 2 and a factor of 5. Therefore, we need to find the minimum of the number of 2s and 5s in the prime factorization of each term.

### Step 1: Calculate the factorials and their powers

1. **Calculate \(2!\)**:
\[ 2! = 2 \]
\[ (2!)^{(2!)} = 2^2 = 4 \]

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

3. **Calculate \(8!\)**:
\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \]
\[ (8!)^{(8!)} = 40320^{40320} \]

4. **Calculate \(9!\)**:
\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880 \]
\[ (9!)^{(9!)} = 362880^{362880} \]

5. **Calculate \(10!\)**:
\[ 10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3628800 \]
\[ (10!)^{(10!)} = 3628800^{3628800} \]

6. **Calculate \(11!\)**:
\[ 11! = 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 39916800 \]
\[ (11!)^{(11!)} = 39916800^{39916800} \]

### Step 2: Count factors of 5 in each term

To find the number of trailing zeros, we focus on counting the factors of 5 in each term since there are always more factors of 2 than factors of 5.

1. **For \(2!\)**:
- Factors of 5: 0

2. **For \(4!\)**:
- Factors of 5: 0

3. **For \(8!\)**:
- Factors of 5: \( \left\lfloor \frac{8}{5} \right\rfloor = 1 \)

4. **For \(9!\)**:
- Factors of 5: \( \left\lfloor \frac{9}{5} \right\rfloor = 1 \)

5. **For \(10!\)**:
- Factors of 5: \( \left\lfloor \frac{10}{5} \right\rfloor + \left\lfloor \frac{10}{25} \right\rfloor = 2 + 0 = 2 \)

6. **For \(11!\)**:
- Factors of 5: \( \left\lfloor \frac{11}{5} \right\rfloor + \left\lfloor \frac{11}{25} \right\rfloor = 2 + 0 = 2 \)

### Step 3: Combine the factors

Now we combine the factors of 5 from each term:

- From \(2!\): 0
- From \(4!\): 0
- From \(8!\): 1
- From \(9!\): 1
- From \(10!\): 2
- From \(11!\): 2

Total factors of 5 = \(0 + 0 + 1 + 1 + 2 + 2 = 6\)

### Step 4: Conclusion

Since the number of factors of 2 will always be greater than the number of factors of 5, the number of trailing zeros in the entire expression will be determined by the number of factors of 5.

Thus, the total number of trailing zeros in the expression is **6**.

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