Problema Solution
Find three consecutive odd integers such that the sum of the three numbers is 110 less than the product of the larger two integers.
Answer provided by our tutors
Every odd number can be written in a form 2k +1, where k is integer.
Thus the three consecutive odd integers can be written as
2k - 1, 2k + 1 and 2k + 3
the sum of the three numbers is 110 less than the product of the larger two integers
2k - 1 + 2k + 1 + 2k + 3 = (2k + 1)(2k + 3) - 110
by simplifying and solving we find and consider only the integer roots since k is integer thus
k = 5
click here to see the step by step solution of the equations
2k - 1 = 2*5 - 1 = 9
2k + 1 = 2*5 + 1 = 11
2k + 1 = 2*5 + 3 = 13
the three consecutive odd numbers are 9, 11 and 13.