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  • 11-07-2017
  • Mathematics
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four consecutive odd intergers whos sum is 56

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LammettHash
LammettHash LammettHash
  • 11-07-2017
Let [tex]N[/tex] be the first of the four odd integers belonging to the sum. The next three integers are [tex]N+2[/tex], [tex]N+4[/tex], and [tex]N+6[/tex].

We then have

[tex]N+(N+2)+(N+4)+(N+6)=4N+12=56\implies 4N=44\implies N=11[/tex]

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