1. For the numerator and denominator of the fraction

$$
\frac{29: 28: 27: 26: 25: 24: 23: 22: 21: 20: 19: 18: 17: 16}{15: 14: 13: 12: 11: 10: 9: 8: 7: 6: 5: 4: 3: 2}
$$

we can repeatedly insert parentheses, always at the same places under each other.

a) Determine the smallest possible integer value of the resulting expression.

b) Find all possible integer values of the resulting expression.
Solution. a) The resulting expression can always be written (without reducing) as the ratio $A: B$ of two products $A$ and $B$ of natural numbers, for which

$$
A \cdot B=2 \cdot 3 \cdot 4 \cdot \ldots \cdot 29=29 !=2^{25} \cdot 3^{13} \cdot 5^{6} \cdot 7^{4} \cdot 11^{2} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29
$$

(The exponents of the prime numbers can be calculated directly from the factors, or according to the known rule: a prime number $p$ has in the factorization of the number $n$! an exponent equal to the sum

$$
\left[\frac{n}{p}\right]+\left[\frac{n}{p^{2}}\right]+\left[\frac{n}{p^{3}}\right]+\cdots,
$$

where $[x]$ denotes the integer part of the number $x$.) The prime numbers that have an odd exponent in the factorization of the number 29! cannot "disappear" from the ratio $A: B$ even after its reduction. Therefore, no integer value of the result is less than the number

$$
H=2 \cdot 3 \cdot 17 \cdot 19 \cdot 23 \cdot 29=1292646 .
$$

On the other hand,

$$
\begin{aligned}
& \frac{29:}{15:}(28: 27: 26: 25: 24: 23: 22: 21: 20: 19: 18: 17: 16) \\
& \quad=\frac{29 \cdot 14 \cdot 27 \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 20 \cdot 19 \cdot 18 \cdot 17 \cdot 16}{15 \cdot 28 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}= \\
& \quad=\frac{29 \cdot 14^{2}}{28} \cdot \frac{27 !}{(15 !)^{2}}=29 \cdot 7 \cdot \frac{2^{23} \cdot 3^{13} \cdot 5^{6} \cdot 7^{3} \cdot 11^{2} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23}{\left(2^{11} \cdot 3^{6} \cdot 5^{3} \cdot 7^{2} \cdot 11 \cdot 13\right)^{2}}=H .
\end{aligned}
$$

(Again, it was useful to calculate the exponents according to (1).) The number $H$ is thus the sought smallest value.

b) Let us now look in more detail at how the products $A$ and $B$ from the first sentence of the solution look if the numerator and denominator of the fraction are bracketed in the same way. From each of the fourteen pairs of consecutive numbers

$$
\{29,15\},\{28,14\},\{27,13\}, \ldots,\{16,2\}
$$

one number is a factor in the product $A$, the other number is a factor in the product $B$ (so $A, B$ each have 14 factors). The resulting value $V$ can then also be written as the product

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

where $\varepsilon_{i}= \pm 1$, and clearly $\varepsilon_{1}=1$ and $\varepsilon_{2}=-1$ regardless of the bracketing. Since the fractions $\frac{27}{13}, \frac{26}{12}, \frac{25}{11}, \ldots, \frac{16}{2}$ are greater than 1, the resulting value $V$ (whether it is an integer or not) must satisfy the estimate

$$
V \leqq \frac{29}{15} \cdot \frac{14}{28} \cdot \frac{27}{13} \cdot \frac{26}{12} \cdot \ldots \cdot \frac{17}{3} \cdot \frac{16}{2}=H
$$

where $H$ is the natural number determined in part a) of the solution as the smallest possible integer value of $V$. It follows that $H$ is the only possible integer value of $V$. (Moreover, we have shown that with the same bracketing of the numerator and denominator of the given fraction, each value of the resulting expression is less than the number $H$.)