Find three consecutive odd integers whose sum is 72 more than the smallest integer.

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Find three consecutive odd integers whose sum is 72 more than the smallest integer.

Mathematics
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let smallest integer be x then x + x + 2 + x + 4 = x + 72
^^ and add -2 and -4 to both sides. That results in 3x=66+x Subtract x from each side, and you get 2x=66 Therefore, x=33!
Same thing we did last time, but use the smallest number on the right hand side of the equality.

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Okay so the smallest consecutive odd integer is X. but I'm still confused on the 2nd consecutive odd integer and the largest integer. Would the 2nd be x+2?
Yes. and the third integer would be (x+4).
the next odd number is 2 after the first Remember thers is an even number in between
I would be careful, @welshfella . Your expressions for the three integers does not require them to be odd. The solution will be odd only by luck. Generically, three consecutive odd integers are best represented by 2x+1, 2x+3, and 2x+5, where x is an integer. This GUARANTEES that the integers are odd.
Thank you all so much! I understand it clearly now
yes @ospreytriple - I took the chance. I'm lazy that's my problem!
Me too! :)
lol.

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