Divide the numbers 2, 3, 5, 7, 11, 13, and 17 into two groups in such a way that, by multiplying all the numbers in one group and all in the other, we find consecutive numbers.
Solution 1: Since the products are consecutive numbers, we can denote them by $p$ and $p+1$. Then,

$$
p^{2}+p=p(p+1)=2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17=510510
$$

Solving the equation $p^{2}+p-510510=0$, we find a single positive root, $p=714$. Thus, $p+1=715$ and, factoring, we obtain

$$
714=2 \times 3 \times 7 \times 17 \quad \text { and } \quad 715=5 \times 11 \times 13
$$

Solution 2: The given numbers are 2, 3, 5, 7, 11, 13, and 17. If 2 and 5 are in the same group, then one of the products ends in 0 and the other, being consecutive, ends in 1 or 9. The possible products ending in 1 are $3 \times 7=21,3 \times 17=51$, $7 \times 13=91,13 \times 17=221,3 \times 7 \times 11=231,3 \times 11 \times 17=561$ and $7 \times 11 \times 13=1001$. It is verified that these groups do not constitute a solution and, similarly, those ending in 9. We conclude that 2 and 5 are in different groups. Therefore, one product ends in 5 and the other in 4 or 6. Since it is not possible to form a product ending in 6 with the given numbers, necessarily one of the products ends in 4 and the other in 5. By trial, we obtain the solution

$$
714=2 \times 3 \times 7 \times 17 \quad \text { and } \quad 715=5 \times 11 \times 13
$$