SOLUTION: Find three consecutive odd intergers such that the sum of the first and second is equal to 23 less than the third

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Question 756101: Find three consecutive odd intergers such that the sum of the first and second is equal to 23 less than the third
Found 2 solutions by tommyt3rd, solver91311:
Answer by tommyt3rd(5050)   (Show Source): You can put this solution on YOUR website!
Find three consecutive odd integers: n+(n+2)+(n+4)
such that the sum of the first and second is equal to 23 less than the third:
n+(n+2)=(n+4)-23

2n+2=n-19
n=-21



numbers:

-21, -19, -17

:)
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


So what's an interger? I know what an integer is, and can solve the problem presuming that's what you meant.

Let represent the first of three consecutive integers, then the second must be and the third, .

We are told that:



Solve for , then calculate and

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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