Answer:
<u>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.</u>
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.
<u>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.</u>
