Question

Find three consecutive odd integers such that 5 times the sum of all three is 42 more than the product of the first and second integers

Answer 1

The three integers are x, x+2, x+4

5(3x+6)=42+x(x+2)
15x+30=x^2+2x+42
x^2 - 13x +12=0
(x-12)(x-1) = 0
x = 12 or 1
since the integers have to be odd we exclude 12

integers are 1,3,5