Find the number of zeros in the following multiplication:
`5xx10xx15xx20xx25xx30xx35xx40xx45xx50`.

Options

A. 6
B. 7
C. 8
D. 9

Answer

To find the number of zeros in the multiplication \(5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50\), we need to determine how many times 10 can be formed from the factors of this multiplication. Since \(10 = 2 \times 5\), we will count the number of factors of 2 and 5 in the product.

### Step-by-Step Solution:

1. **Factor each number into its prime factors:**
- \(5 = 5^1\)
- \(10 = 5^1 \times 2^1\)
- \(15 = 5^1 \times 3^1\)
- \(20 = 5^1 \times 2^2\)
- \(25 = 5^2\)
- \(30 = 5^1 \times 3^1 \times 2^1\)
- \(35 = 5^1 \times 7^1\)
- \(40 = 5^1 \times 2^3\)
- \(45 = 5^1 \times 3^2\)
- \(50 = 5^2 \times 2^1\)

2. **Count the total number of factors of 2:**
- From \(10\): \(1\)
- From \(20\): \(2\)
- From \(30\): \(1\)
- From \(40\): \(3\)
- From \(50\): \(1\)
- Total for \(2\): \(1 + 2 + 1 + 3 + 1 = 8\)

3. **Count the total number of factors of 5:**
- From \(5\): \(1\)
- From \(10\): \(1\)
- From \(15\): \(1\)
- From \(20\): \(1\)
- From \(25\): \(2\)
- From \(30\): \(1\)
- From \(35\): \(1\)
- From \(40\): \(1\)
- From \(45\): \(1\)
- From \(50\): \(2\)
- Total for \(5\): \(1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 2 = 12\)

4. **Determine the number of pairs of \(2\) and \(5\):**
- The number of zeros at the end of the product is determined by the minimum of the counts of \(2\) and \(5\).
- We have \(8\) factors of \(2\) and \(12\) factors of \(5\).
- Thus, the number of pairs (and hence the number of zeros) is \(\min(8, 12) = 8\).

### Final Answer:
The number of zeros in the multiplication \(5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50\) is **8**.