Find the sum of 3 consecutive positive odd integers such that the product of the first and third is 4 less than 7 times the second.

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Check out the set:
2016 Problems
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The answer is 21.

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Let:

$x=1st\quad number\\ x+2=2nd\quad number\\ x+4=3rd\quad number\\$

$\left( x \right) \left( x+4 \right) =7\left( x+2 \right) -4\\ { x }^{ 2 }+4x=7x+14-4\\ { x }^{ 2 }+4x=7x+10\\ { x }^{ 2 }-3x=10\\ { x }^{ 2 }-3x-10=0\\ \left( x-5 \right) \left( x+2 \right) =0\\ x=5\quad |\quad x=-2$

Since we are only looking for positive integers:

$\\ x=5\\ x+2=5+2=7\\ x+4=5+4=9\\ \\ 5+7+9=21\\ \\$