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Four consecutive odd intergers whos sum is 56

User DaveWalley
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Let
N be the first of the four odd integers belonging to the sum. The next three integers are
N+2,
N+4, and
N+6.

We then have


N+(N+2)+(N+4)+(N+6)=4N+12=56\implies 4N=44\implies N=11

and so the four integers are 11, 13, 15, and 17.
User EcologyTom
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