Find an irreducible fraction such that the product of its numerator by its denominator is $2 \times 3 \times 4 \times 5 \times \cdots \times 10$. How many of these irreducible fractions exist?
To ensure that a fraction is irreducible, the numerator and the denominator must not have a common factor. We start by calculating the prime factors of \(N=2 \times 3 \times 4 \times 5 \times \cdots \times 10\), which are

\[
2 \times 3 \times \underbrace{4}_{2^{2}} \times 5 \times \underbrace{6}_{2 \times 3} \times 7 \times \underbrace{8}_{2^{3}} \times \underbrace{9}_{3^{2}} \times \underbrace{10}_{2 \times 5}.
\]

Thus, the prime factorization of \(N\) is given by

\[
N=2^{8} \times 3^{4} \times 5^{2} \times 7.
\]

We can choose various fractions that satisfy the problem, as follows.

(i) If the numerator is 1, we have the fraction \(\frac{1}{2^{8} \times 3^{4} \times 5^{2} \times 7}\).

(ii) If the numerator has only one factor of \(N\), we have the four fractions

\[
\frac{2^{8}}{3^{4} \times 5^{2} \times 7} ; \quad \frac{3^{4}}{2^{8} \times 5^{2} \times 7} ; \quad \frac{5^{2}}{2^{8} \times 3^{4} \times 7} \quad \text { and } \frac{7}{2^{8} \times 3^{4} \times 5^{2}}.
\]

(iii) If the numerator has two factors of \(N\), we have the six fractions

\[
\frac{2^{8} \times 3^{4}}{5^{2} \times 7} ; \frac{2^{8} \times 5^{2}}{3^{4} \times 7} ; \frac{2^{8} \times 7}{3^{4} \times 5^{2}} ; \frac{3^{4} \times 5^{2}}{2^{8} \times 7} ; \frac{3^{4} \times 7}{2^{8} \times 5^{2}} \quad \text { and } \quad \frac{5^{2} \times 7}{2^{8} \times 3^{4}}.
\]

(iv) If the numerator has three factors of \(N\), we have the four fractions

\[
\frac{2^{8} \times 3^{4} \times 5^{2}}{7} ; \quad \frac{2^{8} \times 3^{4} \times 7}{5^{2}} ; \quad \frac{2^{8} \times 5^{2} \times 7}{3^{4}} \quad \text { and } \frac{3^{4} \times 5^{2} \times 7}{2^{8}}.
\]

(v) If the numerator is \(N\), we have the fraction \(\frac{2^{8} \times 3^{4} \times 5^{2} \times 7}{1}\).

Thus, in total, we have sixteen of these irreducible fractions.