SOLUTION: the product of two consecutive odd integers is less than 5 times their sum

Question 807631: the product of two consecutive odd integers is less than 5 times their sum

Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
The trick you need to learn about integer problems is "How do we represent them."
Let n = the first odd integer
Then the NEXT consecutive ODD integer is (n + 2). Not n+1, because if n is odd then n+1 would be even!
So
Let n+2 = the next odd integer
The problem statement is given by
(1) n*(n+2) < 5*(n + n +2) or
(2) n^2 + 2n < 10n + 10 or
(3) n^2 + 2n - 10n < 10 or
(4) n^2 - 8n < 10 or
(5) n*(n-8) < 10
The form of (5) is really arbitrary and not super important.
What we need to do is find the values of n (remember it is an odd integer) that satisfy the inequality in (5). I use trial and error.
First I start a -13, then we get
(6) -13*(-21) = 273 which is not less than 10
So let's move to a higher value like n = -3, where we get
(7) -3*(-11) = 33, stiil not less than 10.
Let's try n = -1, and get
(8) -18(-9) = 9. Aha, the least integer is
(9) n > -1
Now go to high positive odd integers, say n = 11 and get
(10) 11*(3) = 33 which is not less than 10. So try n = 9 and get
(11) 9*(1) = 9. Aha we hit another limit.
To be sure there's a range of n's, let's try n = 7 and get
(12) 7*(-1) = -7 which is less than 10, so is OK. Let's check the low value at n = 1 and get
(13) 1*(-7) = -7, so we have
(14)
Answer: The consecutive odd integers are (-1,1), (1,3), (3,5), (5,7), (7,9), (9,11).
Note: we can solve for the roots of the equation
(15) n^2 - 8n - 10 = 0 and get
(16) n = 9.1 and -1.1
Since we want intger values for n we choose
(17) n = [-1,9], the same as my lenghty trial and error.
PS I actually found the roots first.

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