SOLUTION: find three consecutive odd positive integers such that 5 times the sum of all three is 42 more than the product of the first and second interger

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Question 634601: find three consecutive odd positive integers such that 5 times the sum of all three is 42 more than the product of the first and second interger
Answer by sachi(548) About Me  (Show Source):
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find three consecutive odd positive integers such that 5 times the sum of all three is 42 more than the product of the first and second interger

ans:=
let the three consecutive odd positive integers are x,x+2,x+4
as per question
5(x+x+2+x+4)=x(x+2)+42
or 5(3x+6)=x^2+2x+42
or x^2-13x+12=0
or x^2-x-12x+12=0
or (x-1)(x-12)=0
or x=1or12
so the integers 1,3,5 as we have to select odd integers ans