Question

Find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the largest number.

Answer 1

Answer:

17, 19, 21

Step-by-step explanation:

If we denote the smallest number as x, then the middle number will be x + 2 and the largest number will be x + 2 + 2.

The equation is:

x + (x + 2) = 3 * (x + 2 + 2) - 27

2x + 2 = 3x + 12 - 27

2x - 3x = 12 - 27 - 2

-x = -17

x = 17

The numbers are: 17, 19, 21

Let's check to be sure:

17 + 19 = 36

3 * 21 - 27 = 63 - 27 = 36

Answer 2

We want to find 3 consecutive odd integers such that the sum between the first two is 27 less than 3 times the largest one.

We will find the 3 odd numbers:

• (2*8 + 1) = 17
• (2*8 + 3) = 19
• (2*8 + 5) = 21

First, we can write an odd number as: (2*n + 1)

Where n is an integger.

The next odd number is (2*n + 3)

The next odd number is (2*n + 5)

We want that the sum of the first two to be 27 less than 3 times the last one, so we can write:

(2*n + 1) + (2*n + 3) = 3*(2*n + 5) - 27

So we can solve this for n.

4*n + 4 = 6*n + 15 - 27

4 - 15 + 27 = 6*n - 4*n = 2*n

16 = 2*n

16/2 = n = 8

Then the odd numbers are:

• (2*8 + 1) = 17
• (2*8 + 3) = 19
• (2*8 + 5) = 21

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