The numbers 1, 3, 5,7 .., 99 and 128 are multiplied together. The number of zeros at the end of the product must be :

Options

A. 19
B. 22
C. 7
D. Nil

Answer

To find the number of zeros at the end of the product of the numbers 1, 3, 5, 7, ..., 99, and 128, we need to determine how many times the product can be divided by 10. Since 10 is the product of 2 and 5, we need to count the factors of 2 and 5 in the product and the limiting factor will be the smaller of the two counts.

### Step 1: Identify the odd numbers in the range
The odd numbers from 1 to 99 are: 1, 3, 5, 7, ..., 99. This is an arithmetic sequence where:
- First term (a) = 1
- Common difference (d) = 2
- Last term (l) = 99

To find the number of terms (n) in this sequence, we can use the formula for the nth term of an arithmetic sequence:
\[ l = a + (n-1) \cdot d \]
Substituting the values:
\[ 99 = 1 + (n-1) \cdot 2 \]
\[ 98 = (n-1) \cdot 2 \]
\[ n-1 = 49 \]
\[ n = 50 \]

So, there are 50 odd numbers from 1 to 99.

### Step 2: Count the factors of 5
Next, we need to count how many of these numbers contribute factors of 5. The odd multiples of 5 from 1 to 99 are: 5, 15, 25, 35, 45, 55, 65, 75, 85, 95.

To count the number of odd multiples of 5:
- The first odd multiple of 5 is 5.
- The last odd multiple of 5 is 95.
- This is also an arithmetic sequence where:
- First term (a) = 5
- Common difference (d) = 10
- Last term (l) = 95

Using the same formula for the nth term:
\[ 95 = 5 + (n-1) \cdot 10 \]
\[ 90 = (n-1) \cdot 10 \]
\[ n-1 = 9 \]
\[ n = 10 \]

Now, we also need to count the multiples of \(25\) (since \(25 = 5^2\)) among the odd numbers:
- The odd multiples of 25 from 1 to 99 are: 25, 75.
- This is again an arithmetic sequence:
- First term (a) = 25
- Common difference (d) = 50
- Last term (l) = 75

Using the same formula:
\[ 75 = 25 + (n-1) \cdot 50 \]
\[ 50 = (n-1) \cdot 50 \]
\[ n-1 = 1 \]
\[ n = 2 \]

Thus, the total number of factors of 5 is:
\[ 10 + 2 = 12 \]

### Step 3: Count the factors of 2
Now we need to count the factors of 2 in the product. Since all the numbers from 1 to 99 are odd, they do not contribute any factors of 2. The only even number in our product is 128.

To find the number of factors of 2 in 128:
\[ 128 = 2^7 \]
This means there are 7 factors of 2 in 128.

### Step 4: Determine the limiting factor
Now we have:
- Factors of 5 = 12
- Factors of 2 = 7

The number of zeros at the end of the product is determined by the smaller count, which is the count of factors of 2.

### Final Answer
Thus, the number of zeros at the end of the product is **7**.

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