Find four consecutive multiples of 3 such that the product of the first, third, and fourth is 1080

Therefore, we have 6, 9, 12, and 15, that are the first, second, third and fourth consecutive multiples of 3, such the product of 6, 12 and 15 is 1,080.

Step-by-step explanation:

Let's find the factors of 1,080, as follows:

1,080 Dividing by 2 (1,080/2)

540 Dividing by 2 (540/2)

270 Dividing by 2 (270/2)

135 Dividing by 3 (135/3)

45 Dividing by 3 (45/3)

15 Dividing by 3 (15/3)

5 Dividing by 5 (5/5)

1

In consequence, we have:

5 * 3 * 3 * 3 * 2 * 2 * 2

Let's find out what multiples of 3 there are:

5 * 3 = 15 is a multiple of 3,

And 3 * 3 * 2 * 2 * 2 is remaining.

3 * 2 = 6 is also a multiple of 3,

And 3 * 2 * 2 is remaining

3 * 2 * 2 = 12 is also a multiple of 3.

Therefore, we have 6, 9, 12, and 15, that are the first, second, third and fourth consecutive multiples of 3, such the product of 6, 12 and 15 is 1,080.