Problem 1247

To determine the number of zeros at the right end of the decimal representation of $k$, we need to count the number of factors of 10 in $k$. Since $10 = 2 \times 5$, we need to count the number of pairs of factors of 2 and 5 in $k$.

Given that $k$ is the product of every third positive integer from 2 to 2006, i.e.,
$$k = 2 \cdot 5 \cdot 8 \cdot 11 \cdots 2006,$$
we observe that there are more factors of 2 than factors of 5. Therefore, the number of zeros at the end of $k$ is determined by the number of factors of 5 in $k$.

1. Identify the sequence:
The sequence of numbers is $2, 5, 8, 11, \ldots, 2006$. This is an arithmetic sequence with the first term $a = 2$ and common difference $d = 3$.

2. Find the number of terms in the sequence:
The $n$-th term of the sequence is given by:
$$a_n = 2 + (n-1) \cdot 3.$$
Setting $a_n = 2006$, we solve for $n$:
$$2006 = 2 + (n-1) \cdot 3$$
$$2004 = (n-1) \cdot 3$$
$$n-1 = 668$$
$$n = 669.$$
So, there are 669 terms in the sequence.

3. Count the factors of 5:
We need to count the multiples of 5, 25, 125, etc., in the sequence.

- Multiples of 5:
The multiples of 5 in the sequence are $5, 20, 35, \ldots, 2000$. This is another arithmetic sequence with the first term 5 and common difference 15.
The $m$-th term of this sequence is:
$$5 + (m-1) \cdot 15 = 2000$$
$$(m-1) \cdot 15 = 1995$$
$$m-1 = 133$$
$$m = 134.$$
So, there are 134 multiples of 5.

- Multiples of 25:
The multiples of 25 in the sequence are $50, 125, 200, \ldots, 2000$. This is another arithmetic sequence with the first term 50 and common difference 75.
The $m$-th term of this sequence is:
$$50 + (m-1) \cdot 75 = 2000$$
$$(m-1) \cdot 75 = 1950$$
$$m-1 = 26$$
$$m = 27.$$
So, there are 27 multiples of 25.

- Multiples of 125:
The multiples of 125 in the sequence are $125, 500, 875, \ldots, 2000$. This is another arithmetic sequence with the first term 125 and common difference 375.
The $m$-th term of this sequence is:
$$125 + (m-1) \cdot 375 = 2000$$
$$(m-1) \cdot 375 = 1875$$
$$m-1 = 5$$
$$m = 6.$$
So, there are 6 multiples of 125.

- Multiples of 625:
The multiples of 625 in the sequence are $1250$. There is only one such number.

4. Sum the factors:
The total number of factors of 5 is:
$$134 + 27 + 6 + 1 = 168.$$

The final answer is $\boxed{168}$.