## Task B-1.3.

Can the number 24024 be written in the form of a fraction $\frac{m!}{n!}$ where $m!$ denotes the product of the first $m$ natural numbers, and $n!$ denotes the product of the first $n$ natural numbers? Explain.
## Solution.

If the numbers $m$ and $n$ exist, then

$$
\frac{1 \cdot 2 \cdot 3 \cdot \ldots \cdot m}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}=24024
$$

For the fraction to be simplified, it must be $m>n$. Therefore,

$$
\frac{1 \cdot 2 \cdot 3 \cdot \ldots n \cdot(n+1)(n+2) \ldots \cdot m}{1 \cdot 2 \cdot 3 \cdot \ldots \cdot n}=24024
$$

or after simplification,

$(n+1)(n+2) \ldots m=24024$.

We conclude that the desired representation is possible if the number 24024 can be written as the product of consecutive natural numbers.

The prime factorization of 24024 is $24024=2^{3} \cdot 3 \cdot 7 \cdot 11 \cdot 13$.

We can arrange the obtained factors as follows:

$2^{2} \cdot 3 \cdot 2 \cdot 7 \cdot 11 \cdot 13=11 \cdot 12 \cdot 13 \cdot 14$

We conclude that the number 24024 can be written as the product of consecutive numbers 11, 12, 13, and 14, or $24024=\frac{14!}{10!}$.