Solution

Let's call the second, third, and fourth numbers in the list $a, b$, and $c$, respectively. Since the first number in the list is 6 and the last is 8, the list can be represented as:

$$
6, a, b, c, 8
$$

Since the product of the first three terms is 648, the product of the three middle numbers is 432, and the product of the last three numbers is 288, we have:

$$
6 \times a \times b=648, \quad a \times b \times c=432 \quad \text { and } \quad b \times c \times 8=288
$$

Dividing both sides of the first equation by 6, we get:

$$
a \times b=108
$$

Substituting (1) into the second equation, we have $432=(a \times b) \times c=108 \times c$, so:

$$
108 \times c=432
$$

Dividing both sides of this equation by 108, we get $c=4$. Using the value of $c=4$ in the equation $b \times c \times 8=288$, we get $b \times 4 \times 8=288$, so

$$
b \times 32=288
$$

Dividing both sides by 32, we get $b=9$. Finally, we can substitute $b=9$ into equation (1) to conclude that

$$
a \times 9=108
$$

Dividing both sides by 9, we get $a=12$. Therefore, the list we are looking for is:

$$
6,12,9,4,8
$$
Aureliano writes a list containing five numbers, the first of which is 6 and the last of which is 08. The product of the first three numbers is 648, the product of the three central numbers is 432, and the product of the three last numbers is 288. What is Aureliano's list?

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