Find an irreducible fraction such that the product of its numerator by its denominator is $2 \times 3 \times 4 \times 5 \times \ldots \times 10$. How many of these irreducible fractions exist?

## List 8
For a fraction to be irreducible, the numerator and the denominator must not have a common factor.

Initially, let's see what the prime factors of \(N=2 \times 3 \times 4 \times 5 \times \ldots \times 10\) 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:

\[
N=2^{8} \times 3^{4} \times 5^{2} \times 7(*)
\]

We can choose various fractions that satisfy the problem. For example:

(i) The numerator has only 1 factor of \((*)\):

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

In this case, we have 4 fractions plus the 4 inverse fractions, with denominators having only 1 factor of \((*)\).

We cannot forget the number 1, obtaining the 2 fractions:

\[
\frac{1}{2^{8} \times 3^{4} \times 5^{2} \times 7} ; \frac{2^{8} \times 3^{4} \times 5^{2} \times 7}{1} .
\]

(ii) The numerator has 2 factors of \((*)\):

\[
\frac{2^{8} \times 3^{4}}{5^{2} \times 7} ; \frac{2^{8} \times 5^{2}}{3^{4} \times 7} ; \frac{2^{8} \times 7}{3^{4} \times 5^{2}} ; \frac{3^{4} \times 5^{2}}{2^{8} \times 7} ; \frac{3^{4} \times 7}{2^{8} \times 5^{2}} ; \frac{5^{2} \times 7}{2^{8} \times 3^{4}}
\]

In this case, we have 6 fractions.

(iii) The numerator has 3 factors of \((*)\):

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

In total, we have 16 irreducible fractions.

## List 8