3. On the board, the numbers 1, 2, .., 33 are written. In one step, we choose two numbers on the board, one of which is a divisor of the other, erase both, and write their (integer) quotient on the board. We continue this process until only numbers remain on the board, none of which is a divisor of another. (In one step, we can also erase two identical numbers and replace them with the number 1.) How few numbers can remain on the board?
SOLUTION. The numbers on the board will likely remain only from the set $M=\{1,2, \ldots, 33\}$. The prime numbers 17, 19, 23, 29, and 31 will always be written, each exactly once, because they have no divisors other than 1, and the set $M$ does not contain any of their multiples (so they can never disappear from the board, nor appear in another instance).

We will now explain why, in addition to the five mentioned primes, there will always be at least two more numbers on the board. The product $S$ of all numbers written on the board is initially equal to

$$
S=33 !=2^{31} \cdot 3^{15} \cdot 5^{7} \cdot 7^{4} \cdot 11^{3} \cdot 13^{2} \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31
$$

In each step, we choose a pair of numbers $(x, y)$ with the property $x \mid y$, i.e., numbers of the form $x=a$ and $y=k a$, and replace them with a single number $y / x=k$. The product of all numbers on the board will then change from its previous value $S$ to a new value $S / a^{2}$, because two factors $x, y$ with a product $x y=k a^{2}$ will transition to one new factor $k$ (and the other factors will remain unchanged). It is clear that when changing $S \rightarrow S / a^{2}$, the exponent of any prime factor $p$ in the factorization of $S$ will either remain the same (if $p \nmid a$) or decrease by an even number (equal to the exponent of $p$ in the factorization of $a^{2}$). In no case will the parity (even-odd) of the exponent of any prime factor change. Therefore, each of the primes that initially had an odd exponent in the factorization (1) will have an odd exponent in the changing $S$ after any number of steps. These include (in addition to 17, 19, 23, 29, and 31) the primes $2,3,5$, and 11. This means that there will always be numbers on the board (not necessarily four different ones) that are divisible by these four individual primes. However, it cannot be just one number (since $2 \cdot 3 \cdot 5 \cdot 11>33$), so it must be at least two numbers, for example, 10 and 33 (or 11 and 30, or 15 and 22; other possibilities do not exist with a total of seven numbers on the board). Thus, we have proven that there will indeed always be at least 7 numbers on the board.

It remains to describe a sequence of steps that will leave 7 numbers on the board. There are many possibilities; for example, we can "set aside" the primes $17,19,23,29,31$ and the numbers 10 and 33, and with the remaining numbers perform the following steps:

$$
\begin{gathered}
32,16 \rightarrow 2, \quad 30,15 \rightarrow 2, \quad 28,14 \rightarrow 2, \quad 26,13 \rightarrow 2, \quad 24,12 \rightarrow 2, \quad 22,11 \rightarrow 2, \\
27,9 \rightarrow 3, \quad 21,7 \rightarrow 3, \quad 18,6 \rightarrow 3, \quad 25,5 \rightarrow 5, \quad 20,4 \rightarrow 5, \quad 8,2 \rightarrow 4, \\
5,5 \rightarrow 1, \quad 4,2 \rightarrow 2, \quad 3,3 \rightarrow 1, \quad 3,3 \rightarrow 1, \quad 2,2 \rightarrow 1, \quad 2,2 \rightarrow 1, \quad 2,2 \rightarrow 1 .
\end{gathered}
$$

After these steps, there are already 7 ones on the board (in addition to the seven numbers set aside), which we all remove in six steps $1,1 \rightarrow 1$ and the last step, for example, $10,1 \rightarrow 10$.

## GUIDING AND ADDITIONAL PROBLEMS:

N1. Determine how many zeros the decimal representation of the number 33! ends with. [Seven zeros. It is sufficient to determine the exponents of the prime numbers 2 and 5 in the factorization of the given factorial into prime factors. This can be done by a detailed analysis of the product $1 \cdot 2 \cdot 3 \cdot \ldots \cdot 33$, or by using a known rule: the number of occurrences of a prime number $p$ in the factorization of the number $n$! into prime factors is equal to the sum

$$
\left\lfloor\frac{n}{p^{1}}\right\rfloor+\left\lfloor\frac{n}{p^{2}}\right\rfloor+\left\lfloor\frac{n}{p^{3}}\right\rfloor+\ldots
$$

where $\lfloor x\rfloor$ denotes the integer part of the number $x$ and the summation continues as long as the power $p^{k}$ in the denominator of the fraction does not exceed the numerator $n$. The complete factorization of the number 33! is given in the solution to the competition problem.]

N2. Prove that the number $N=46 ! \cdot 47 ! \cdot 48 ! \cdot 49$! is not a perfect square, and then find its largest divisor that is a perfect square. [The number $N$ is not a perfect square because the prime number $47$ appears with an odd exponent 3 in its factorization into prime factors. From the expression $N=(46 !)^{4} \cdot 47 \cdot(47 \cdot 48) \cdot(47 \cdot 48 \cdot 49)=(46 !)^{4} \cdot 47^{3} \cdot\left(48^{2}\right) \cdot 7^{2}$, it follows that the largest perfect square that is a divisor of the number $N$ is the number $(46 !)^{4} \cdot 47^{2} \cdot 48^{2} \cdot 7^{2}=N / 47$.]

N3. Find the smallest positive integer $