# SOLUTION: Eight times the first of three consecutive odd integers is ten more than twice the second.

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 Question 92593: Eight times the first of three consecutive odd integers is ten more than twice the second. Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!Eight times the first of three consecutive odd integers is ten more than twice the second. ANY INTEGER IS N EVEN NUMBER IS 2N ODD INTEGER IS 2N+1 NEXT ODD INTEGER IS 2N+1+2=2N+3 THE NEXT ODD INTEGER IS 2N+3+2=2N+5 8 TIMES THE I = 8(2N+1)=16N+8 TWICE THE SECOND = 2(2N+3)=4N+6 TEN MORE =4N+6+10=4N+16.............HENCE 16N+8=4N+16 16N-4N=16-8 12N=8 N=8/12=......THERE IS NO SOLUTION.......PLEASE CHECK THE DATA IN THE PROBLEM FOR ANY MISTAKES IN TYPING.