There are 3 consecutive odd integers that sum to 45. what is the least of the 3 integers?

Step-by-step explanation:

Let n = a number and 2n = an even number, then ...

Let 2n + 1 = the first of three consecutive odd numbers.

Let 2n + 3 = the second of three consecutive numbers.

Let 2n + 5 = the third of three consecutive odd numbers.

Since it's given that the sum of our three consecutive odd numbers is 45, then we can write the following equation to be solved for the variable n:

(2n + 1) + (2n + 3) + (2n + 5) = 45

2n + 1 + 2n + 3 + 2n + 5 = 45

Collecting like-terms on the left, we get:

6n + 9 = 45

Now, subtracting 9 from both sides in order to begin isolating the unknown, n, on the left side:  

6n + 9 - 9 = 45 - 9

6n + 0 = 36

6n = 36

Now, divide both sides by 6 in order to solve the equation for n:

(6n)/6 = 36/6

(6/6)n = 36/6

(1)n = 6

n = 6

Therefore, ...

2n + 1 = 2(6) + 1

= 12 + 1

= 13

2n + 3 = 2(6) + 3

my answer is$ 12

step-by-step explanation:

because 15 -2=12

the correct answer is 5 6 and 7