SOLUTION: find three consecutive odd integers such that the sum of all three is 42 less than the product of the larger two

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Question 865115: find three consecutive odd integers such that the sum of all three is 42 less than the product of the larger two
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let's call those numbers n-2 , n, and n%2B2 .
(Those are 3 numbers, with each greater than the one before by 2.
The approach could be the same for a problem involving 3 consecutive even numbers.
For this problem, when we solve for n , n needs to be an odd number.
If we found a value for is the sum of all three consecutive odd integers.
n%28n%2B2%29 is the product of the larger two consecutive odd integers.
The problem states that
%28n-2%29%2Bn%2B%28n%2B2%29=n%28n%2B2%29-42 or 3n=n%28n%2B2%29-42 .

3n=n%28n%2B2%29-42 --> 3n=n%5E2%2B2n-42 --> n%5E2%2B2n-42-3n=0 --> n%5E2-n-42=0
That last equation is easy to solve by factoring:
n%5E2-n-42=0 --> %28n%2B6%29%28n-7%29=0 --> system%28either%2Cn%2B6=0%2Cor%2Cn-7=0%29 --> system%28either%2Cn=-6%2Cor%2Cn=7%29 .
Since we are expecting n to be an odd integer, highlight%28n=7%29 ,
and the three consecutive odd integers are
n-2=7-2=highlight%285%29 , n=highlight%287%29 , and n%2B2=7%2B2=highlight%289%29 .