Question 544329
find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18.


Let the first of the four consecutive odd integers = F


Then others are: F + 2, F + 4, and F + 6


We therefore have: F + F + 2 + F + 4 = F + 6 + 18


3F + 6 = F + 24


3F - F = 24 - 6


2F = 18


F, or first of the three odd integers = {{{18/2}}}, or {{{highlight_green(9)}}}


Other three are: {{{highlight_green(11_13_and_15)}}}


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