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find three consecutive odd integers such that the sum of all three is 36 lessthan the product of the smaller two

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3 Answers

x + (x+2) + (x+4) = [x * (x+2)] - 36

x + x + 2 + x + 4 = [x2 + 2x] - 36

3x + 6 = x2 + 2x -36

0 = x2 + 2x - 3x - 36 - 6

0 = x2 - x - 42

0 = (x - 7)(x + 6)

x = -6 or 7

We know x = 7 because -6 is not odd

When x = 7, our consecutive integers are 7, 9 (or x+2) , and 11 (or x+4)

Let's take a variable x which represents any integer.  To ensure that the three numbers are odd we take the three numbers to be 2x+1, 2x+3 and 2x+5

Now the sum of the three numbers is 6x+9

We know this sum must equal 36 less than the product of multiplying the first two numbers 2x+1 and 2x+3

6x+9=(2x+1)(2x+3)-36

6x+9=4x^2+8x+3-36

4x^2+2x-42=0

2x^2+x-21=0

x=3 or -3.5.  Since -3.5 is not an integer we discard that solution

so the numbers are 2*3+1, 2*3+3 and 2*3+5 or 7, 9 and 11

7+9+11=27

7*9=63

63-27=36