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Four consecutive odd integers whose sum is 336? - Answers

Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.



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Four consecutive odd integers whose sum is 336? - Answers

https://math.answers.com/other-math/Four_consecutive_odd_integers_whose_sum_is_336

Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.



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https://math.answers.com/other-math/Four_consecutive_odd_integers_whose_sum_is_336

Four consecutive odd integers whose sum is 336? - Answers

Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.

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      Any odd integer can be represented as 2n+1, where n is any integer. The integers, starting with one, always proceed as odd, even, odd, even, etc. Therefore, a consecutive odd would be two more than the first. So 2n+1, 2n+3, 2n+5, and 2n+7 summed to 336. Simple algebra gives n=40, and the four numbers as 81, 83, 85, and 87.

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