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Two consecutive odd integers must differ by 2. Call them (x) and (x+2).

(I know those aren't necessarily odd, but it won't matter, because we'll quickly discover that
there's no solution anyway.)


You said that 8(x) = 2(x+2)

4 (x) = (x+2)

4x = x+2

3x = 2

x = 2/3

That's not an integer. So no two integers that differ by 2 can satisfy the given conditions.

The question has no solution.

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Q: A. Eight times the first of three consecutive odd integers is twice the second?
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