The number of porper divisors of 1800 which are also divisible by 10, is

Options

A. 18
B. 34
C. 17
D. none of these

Explanation

To find the number of proper divisors of 1800 that are also divisible by 10, we can follow these steps:

### Step 1: Prime Factorization of 1800
First, we need to find the prime factorization of 1800.

1800 can be factored as:
- 1800 = 18 × 100
- 18 = 2 × 9 = 2 × 3²
- 100 = 10 × 10 = (2 × 5) × (2 × 5) = 2² × 5²

Combining these, we get:
\[ 1800 = 2^3 × 3^2 × 5^2 \]

### Step 2: Identify Divisibility by 10
Next, we need to find the divisors of 1800 that are divisible by 10. Since 10 = 2 × 5, any divisor that is divisible by 10 must include at least one factor of 2 and one factor of 5.

### Step 3: Adjust the Prime Factorization
To find the divisors of 1800 that are divisible by 10, we can adjust the prime factorization:
- We need at least \(2^1\) and \(5^1\). Therefore, we can reduce the powers of 2 and 5 by 1:
\[ 1800 = 2^{3-1} × 3^2 × 5^{2-1} = 2^2 × 3^2 × 5^1 \]

### Step 4: Count the Divisors
Now, we can find the number of divisors of \(2^2 × 3^2 × 5^1\). The formula to find the number of divisors is to take each of the exponents in the prime factorization, add 1 to each, and then multiply the results.

So, we have:
- For \(2^2\): \(2 + 1 = 3\)
- For \(3^2\): \(2 + 1 = 3\)
- For \(5^1\): \(1 + 1 = 2\)

Thus, the total number of divisors is:
\[ 3 × 3 × 2 = 18 \]

### Step 5: Proper Divisors
Since we are looking for proper divisors, we need to exclude the number itself (1800). Therefore, the number of proper divisors that are also divisible by 10 is:
\[ 18 - 1 = 17 \]

### Conclusion
Thus, the number of proper divisors of 1800 that are also divisible by 10 is **17**.