Find number of zeros in the end of `1^20 xx 2^20 xx 3^20xx 4^20 ................. Xx38^20.`

Options

A. 160
B. 180
C. 150
D. 120

Answer

To find the number of zeros at the end of the product \(1^{20} \times 2^{20} \times 3^{20} \times \ldots \times 38^{20}\), we need to determine how many times the number 10 can be factored from this product. Since \(10 = 2 \times 5\), we will find the powers of 2 and 5 in the product and take the minimum of the two.

### Step-by-Step Solution:

1. **Identify the Product**:
The product we are considering is:
\[ 1^{20} \times 2^{20} \times 3^{20} \times \ldots \times 38^{20} \]
This can be rewritten as:
\[ (1 \times 2 \times 3 \times \ldots \times 38)^{20} \]

2. **Count the Factors of 5**:
We will count the number of factors of 5 in \(1 \times 2 \times 3 \times \ldots \times 38\) using the formula:
\[ \text{Number of factors of } p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
Here, \(n = 38\) and \(p = 5\).

- Calculate:
\[ \left\lfloor \frac{38}{5} \right\rfloor = 7 \]
\[ \left\lfloor \frac{38}{25} \right\rfloor = 1 \]
\[ \left\lfloor \frac{38}{125} \right\rfloor = 0 \]

- Total factors of 5:
\[ 7 + 1 + 0 = 8 \]

3. **Count the Factors of 2**:
Now, we will count the number of factors of 2 in \(1 \times 2 \times 3 \times \ldots \times 38\) using the same formula with \(p = 2\).

- Calculate:
\[ \left\lfloor \frac{38}{2} \right\rfloor = 19 \]
\[ \left\lfloor \frac{38}{4} \right\rfloor = 9 \]
\[ \left\lfloor \frac{38}{8} \right\rfloor = 4 \]
\[ \left\lfloor \frac{38}{16} \right\rfloor = 2 \]
\[ \left\lfloor \frac{38}{32} \right\rfloor = 1 \]
\[ \left\lfloor \frac{38}{64} \right\rfloor = 0 \]

- Total factors of 2:
\[ 19 + 9 + 4 + 2 + 1 + 0 = 35 \]

4. **Calculate the Total Factors in the Product**:
Since we have \(1 \times 2 \times 3 \times \ldots \times 38\) raised to the power of 20, we multiply the counts of factors by 20:
- Factors of 5 in the product:
\[ 8 \times 20 = 160 \]
- Factors of 2 in the product:
\[ 35 \times 20 = 700 \]

5. **Determine the Number of Trailing Zeros**:
The number of trailing zeros is determined by the minimum of the counts of factors of 2 and 5:
\[ \text{Trailing Zeros} = \min(160, 700) = 160 \]

### Final Answer:
The number of zeros at the end of the product \(1^{20} \times 2^{20} \times 3^{20} \times \ldots \times 38^{20}\) is **160**.