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laila [671]
3 years ago
12

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

Mathematics
1 answer:
Dovator [93]3 years ago
7 0

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>

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