Find the number of Zeros at the end of the given product
`1^5xx2^5xx3^5......32^5`

Options

A. a) 35
B. b) 40
C. c) 36
D. d) 32

Answer

To find the number of zeros at the end of the product \(1^5 \times 2^5 \times 3^5 \times \ldots \times 32^5\), we need to determine how many times the factor 10 appears in this product. Since \(10 = 2 \times 5\), the number of zeros at the end of the product will depend on the minimum of the number of factors of 2 and the number of factors of 5 in the entire product.

### Step 1: Calculate the total product
The product can be expressed as:
\[ (1 \times 2 \times 3 \times \ldots \times 32)^5 \]
This means we need to find the number of factors of 2 and 5 in \(1 \times 2 \times 3 \times \ldots \times 32\) (which is \(32!\)) and then raise those counts to the power of 5.

### Step 2: Count the number of factors of 5 in \(32!\)
To find the number of factors of 5 in \(32!\), we can use the formula:
\[ \text{Number of factors of } p \text{ in } n! = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
For \(n = 32\) and \(p = 5\):
\[ \left\lfloor \frac{32}{5} \right\rfloor + \left\lfloor \frac{32}{25} \right\rfloor = 6 + 1 = 7 \]
So, there are 7 factors of 5 in \(32!\).

### Step 3: Count the number of factors of 2 in \(32!\)
Using the same formula for \(p = 2\):
\[ \left\lfloor \frac{32}{2} \right\rfloor + \left\lfloor \frac{32}{4} \right\rfloor + \left\lfloor \frac{32}{8} \right\rfloor + \left\lfloor \frac{32}{16} \right\rfloor + \left\lfloor \frac{32}{32} \right\rfloor \]
Calculating each term:
\[ 16 + 8 + 4 + 2 + 1 = 31 \]
So, there are 31 factors of 2 in \(32!\).

### Step 4: Determine the limiting factor
Now we have:
- Factors of 5 in \(32!\) = 7
- Factors of 2 in \(32!\) = 31

The number of pairs of \(2\) and \(5\) that can form \(10\) is determined by the smaller count, which is 7.

### Step 5: Multiply by the power
Since we have \( (32!)^5 \), we multiply the counts of factors by 5:
- Factors of 5 in \( (32!)^5 \) = \( 7 \times 5 = 35 \)
- Factors of 2 in \( (32!)^5 \) = \( 31 \times 5 = 155 \)

### Step 6: Conclusion
The number of zeros at the end of the product \(1^5 \times 2^5 \times 3^5 \times \ldots \times 32^5\) is determined by the limiting factor, which is the number of factors of 5:
\[ \text{Number of zeros} = 35 \]

Thus, the final answer is:
\[ \boxed{35} \]