. In the fraction:

$$
\frac{29 \div 28 \div 27 \div \cdots \div 16}{15 \div 14 \div 13 \div \cdots \div 2}
$$

we place the same parentheses in the numerator and the denominator and observe that the result of the operation obtained is an integer. What is this number?
. After placing the parentheses, the value of the fraction is:

$$
A=\left(\frac{29}{15}\right)^{\varepsilon_{1}}\left(\frac{28}{14}\right)^{\varepsilon_{2}}\left(\frac{27}{13}\right)^{\varepsilon_{3}}\left(\frac{26}{12}\right)^{\varepsilon_{4}} \cdots\left(\frac{18}{4}\right)^{\varepsilon_{12}}\left(\frac{17}{3}\right)^{\varepsilon_{13}}\left(\frac{16}{2}\right)^{\varepsilon_{14}}
$$

where the $\varepsilon_{i}$ are either 1 or -1 and, necessarily, $\varepsilon_{1}=1$ and $\varepsilon_{2}=-1$. Moreover, since the result is an integer by assumption, the prime numbers that appear only once, namely 29, 23, 19, and 17, must be in the numerator. This gives $\varepsilon_{7}=\varepsilon_{11}=\varepsilon_{13}=1$. After simplifying the fractions, we get:

$$
A=\frac{29 \cdot 23 \cdot 19 \cdot 17}{2 \cdot 3^{4} \cdot 5^{2}} \cdot\left(\frac{3^{3}}{13}\right)^{\varepsilon_{3}}\left(\frac{13}{2 \cdot 3}\right)^{\varepsilon_{4}}\left(\frac{5^{2}}{11}\right)^{\varepsilon_{5}}\left(\frac{2^{2} \cdot 3}{5}\right)^{\varepsilon_{6}}\left(\frac{11}{2^{2}}\right)^{\varepsilon_{8}} 3^{\varepsilon_{9}}\left(\frac{2 \cdot 5}{3}\right)^{\varepsilon_{10}}\left(\frac{3^{2}}{2}\right)^{\varepsilon_{12}}\left(2^{3}\right)^{\varepsilon_{14}}
$$

It is impossible for $\varepsilon_{3}$ to be -1, because otherwise, there would not be enough 3s to compensate for such a denominator. Thus, $\varepsilon_{3}=1$, and to eliminate the 13 that then appears in the denominator, we must have $\varepsilon_{4}=1$. By counting the powers of 5 now, we get $\varepsilon_{5}=1$, and then by looking at the factor 11, it follows that $\varepsilon_{8}=1$. We are now at:

$$
A=\frac{29 \cdot 23 \cdot 19 \cdot 17}{2^{4} \cdot 3^{2}} \cdot\left(\frac{2^{2} \cdot 3}{5}\right)^{\varepsilon_{6}} 3^{\varepsilon_{9}}\left(\frac{2 \cdot 5}{3}\right)^{\varepsilon_{10}}\left(\frac{3^{2}}{2}\right)^{\varepsilon_{12}}\left(2^{3}\right)^{\varepsilon_{14}}
$$

By looking at the exponent of 3 again, it follows that $\varepsilon_{12}=1$, and then by looking at the exponent of 2, we get $\varepsilon_{6}=\varepsilon_{14}=1$. The prime number 5 then gives $\varepsilon_{10}$, and finally, by looking at the exponent of 3 one last time, we get $\varepsilon_{9}=\varepsilon_{12}=1$. In the end:

$$
A=2 \times 3 \times 17 \times 19 \times 23 \times 29=1292646
$$