Find the number of Zeros at the end of the product -
`2^3 xx 5^4 xx 4^2 xx 10^8 xx 6^10 xx 15^12 xx 8^14 X 20^16 xx 10^18 xx 25^20`

Options

A. 80
B. 98
C. 94
D. 100

Answer

To find the number of zeros at the end of the product \( 2^3 \times 5^4 \times 4^2 \times 10^8 \times 6^{10} \times 15^{12} \times 8^{14} \times 20^{16} \times 10^{18} \times 25^{20} \), we need to determine how many times 10 can be factored out of the product. Since \( 10 = 2 \times 5 \), we need to find the minimum of the total number of factors of 2 and the total number of factors of 5 in the product.

### Step 1: Factor each term into prime factors

1. \( 2^3 \) remains \( 2^3 \).
2. \( 5^4 \) remains \( 5^4 \).
3. \( 4^2 = (2^2)^2 = 2^4 \).
4. \( 10^8 = (2 \times 5)^8 = 2^8 \times 5^8 \).
5. \( 6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10} \).
6. \( 15^{12} = (3 \times 5)^{12} = 3^{12} \times 5^{12} \).
7. \( 8^{14} = (2^3)^{14} = 2^{42} \).
8. \( 20^{16} = (2^2 \times 5)^{16} = 2^{32} \times 5^{16} \).
9. \( 10^{18} = (2 \times 5)^{18} = 2^{18} \times 5^{18} \).
10. \( 25^{20} = (5^2)^{20} = 5^{40} \).

### Step 2: Combine all the factors

Now, we will sum the powers of 2 and the powers of 5:

#### Total power of 2:
- From \( 2^3 \): \( 3 \)
- From \( 4^2 \): \( 4 \)
- From \( 10^8 \): \( 8 \)
- From \( 6^{10} \): \( 10 \)
- From \( 8^{14} \): \( 42 \)
- From \( 20^{16} \): \( 32 \)
- From \( 10^{18} \): \( 18 \)

Total power of 2:
\[ 3 + 4 + 8 + 10 + 42 + 32 + 18 = 117 \]

#### Total power of 5:
- From \( 5^4 \): \( 4 \)
- From \( 10^8 \): \( 8 \)
- From \( 15^{12} \): \( 12 \)
- From \( 20^{16} \): \( 16 \)
- From \( 10^{18} \): \( 18 \)
- From \( 25^{20} \): \( 40 \)

Total power of 5:
\[ 4 + 8 + 12 + 16 + 18 + 40 = 98 \]

### Step 3: Find the minimum of the total powers

The number of zeros at the end of the product is determined by the minimum of the total powers of 2 and 5:
\[ \text{Number of zeros} = \min(117, 98) = 98 \]

### Final Answer:
The number of zeros at the end of the product is **98**.