[Algebra]
Problem: The sequence $2,5,10,50,500, \ldots$ is formed so that each term after the second is the product of the two previous terms. The 15 th term ends with exactly $k$ zeroes. What is the value of $k$ ?

Solution: We calculate the first 15 terms, writing each as an integer times a power of 10:

$$
\begin{gathered}
2,5,10,5 \times 10,5 \times 10^{2}, 5^{2} \times 10^{3}, 5^{3} \times 10^{5}, 5^{5} \times 10^{8}, 5^{8} \times 10^{13}, 5^{13} \times 10^{21}, 5^{21} \times 10^{34} \\
5^{34} \times 10^{55}, 5^{55} \times 10^{89}, 5^{89} \times 10^{144}, 5^{144} \times 10^{233}
\end{gathered}
$$

Since the 15 th term equals an odd integer times $10^{233}$, then the 15 th term ends with 233 zeroes.
To obtain the 6 th term, we calculate $50 \times 500=25 \times 1000$.

Each of the 4th and 5th terms equals an odd integer followed by a number of zeroes, so the 6th term also equals an odd integer followed by a number of zeroes, where the number of zeroes is the sum of the numbers of zeroes at the ends of the 4th and 5th terms.

This pattern will continue. Thus, starting with the 6th term, the number of zeroes at the end of the term will be the sum of the number of zeroes at the ends of the two previous terms.

This tells us that, starting with the 4th term, the number of zeroes at the ends of the terms is

$$
1,2,3,5,8,13,21,34,55,89,144,233
$$

Therefore, the 15 th term ends with 233 zeroes.

Final answer: 233