Find the no. of zeros in expression :
`10 xx 20 xx 30 xx ……xx 1000`.

Options

A. 124
B. 130
C. 249
D. 150

Answer

To find the number of zeros in the expression \(10 \times 20 \times 30 \times \ldots \times 1000\), we can follow these steps:

### Step 1: Rewrite the expression
The expression can be rewritten as:
\[ 10 \times 20 \times 30 \times \ldots \times 1000 = 10^n \times (1 \times 2 \times 3 \times \ldots \times 100) \]
where \(n\) is the number of terms in the product. The terms are \(10, 20, 30, \ldots, 1000\), which can be expressed as \(10 \times 1, 10 \times 2, 10 \times 3, \ldots, 10 \times 100\).

### Step 2: Count the number of terms
The terms from \(10\) to \(1000\) in steps of \(10\) can be counted:
\[ n = \frac{1000 - 10}{10} + 1 = 100 \]
So, there are \(100\) terms.

### Step 3: Factor out the powers of 10
From the expression, we can factor out \(10^{100}\):
\[ 10^{100} \times (1 \times 2 \times 3 \times \ldots \times 100) = 10^{100} \times 100! \]

### Step 4: Count the number of trailing zeros in \(100!\)
To find the total number of trailing zeros in \(100!\), we need to count the number of pairs of \(2\) and \(5\) in its prime factorization. Since there are generally more \(2\)s than \(5\)s, we only need to count the number of \(5\)s.

The formula to count the number of \(5\)s in \(n!\) is:
\[ \text{Number of } 5s = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \ldots \]
For \(n = 100\):
\[ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor = 20 + 4 = 24 \]

### Step 5: Combine the results
Now, we combine the number of zeros from \(10^{100}\) and \(100!\):
\[ \text{Total zeros} = 100 + 24 = 124 \]

### Final Answer
Thus, the total number of trailing zeros in the expression \(10 \times 20 \times 30 \times \ldots \times 1000\) is:
\[ \boxed{124} \]