The number of zeros at end of the product of
`222^(111) xx 35^(53) + (7!)^(6!) xx (10!)^(5!) + 42^(42) xx 25^(25)` is :

Options

A. 42
B. 53
C. 1055
D. none of these

Answer

To find the number of zeros at the end of the expression \( 222^{111} \times 35^{53} + (7!)^{6!} \times (10!)^{5!} + 42^{42} \times 25^{25} \), we need to determine how many pairs of factors of 2 and 5 are present in the product. Each pair contributes to one trailing zero.

### Step 1: Analyze the first term \( 222^{111} \times 35^{53} \)

1. **Factorization**:
- \( 222 = 2 \times 111 = 2 \times 3 \times 37 \) (only the factor 2 is relevant for trailing zeros)
- \( 35 = 5 \times 7 \)

2. **Count factors of 2 and 5**:
- In \( 222^{111} \):
- Number of 2s = \( 111 \) (from \( 2^{111} \))
- In \( 35^{53} \):
- Number of 5s = \( 53 \) (from \( 5^{53} \))

3. **Determine the number of trailing zeros**:
- Minimum of the number of 2s and 5s = \( \min(111, 53) = 53 \)

### Step 2: Analyze the second term \( (7!)^{6!} \times (10!)^{5!} \)

1. **Factorials**:
- Calculate \( 7! \) and \( 10! \):
- \( 7! = 5040 \) (factors: \( 2^4, 3^2, 5^1, 7^1 \))
- \( 10! = 3628800 \) (factors: \( 2^8, 3^4, 5^2, 7^1 \))

2. **Count factors of 2 and 5**:
- In \( (7!)^{6!} \):
- Number of 2s = \( 4 \times 6! \)
- Number of 5s = \( 1 \times 6! \)
- In \( (10!)^{5!} \):
- Number of 2s = \( 8 \times 5! \)
- Number of 5s = \( 2 \times 5! \)

3. **Total counts**:
- Total number of 2s = \( 4 \times 720 \) + \( 8 \times 120 \) = \( 2880 + 960 = 3840 \)
- Total number of 5s = \( 1 \times 720 \) + \( 2 \times 120 \) = \( 720 + 240 = 960 \)

4. **Determine the number of trailing zeros**:
- Minimum of the number of 2s and 5s = \( \min(3840, 960) = 960 \)

### Step 3: Analyze the third term \( 42^{42} \times 25^{25} \)

1. **Factorization**:
- \( 42 = 2 \times 3 \times 7 \) (only the factor 2 is relevant)
- \( 25 = 5^2 \)

2. **Count factors of 2 and 5**:
- In \( 42^{42} \):
- Number of 2s = \( 42 \) (from \( 2^{42} \))
- In \( 25^{25} \):
- Number of 5s = \( 2 \times 25 = 50 \)

3. **Determine the number of trailing zeros**:
- Minimum of the number of 2s and 5s = \( \min(42, 50) = 42 \)

### Step 4: Combine results

Now we have the number of trailing zeros from each term:
- From the first term: 53
- From the second term: 960
- From the third term: 42

The total number of trailing zeros in the entire expression is determined by the term with the minimum number of trailing zeros:
- \( \min(53, 960, 42) = 42 \)

### Final Answer
The number of zeros at the end of the product is **42**.

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