Let k = the smallest odd integer
So, k + 2 = the next odd integer (since each consecutive odd integer is always 2 greater than the odd integer before it)
And k + 4 = the next odd integer
And k + 6 = the next odd integer
And k + 8 = the next odd integer
And k + 10 = the last (6th) odd integer

Since the average of the six integers is \(x+2\), we can write: \(\frac{k + (k+2)+(k+4)+(k+6)+(k+8)+(k+10)}{6}=x+2\)

Simplify: \(\frac{6k + 30}{6}=x+2\)

Multiply both sides of the equation by 6 to get: \(6k + 30=6x+12\)

What is the smallest of six consecutive odd integers?
Since k is the smallest odd integer, we need to solve this equation for k.

Take: \(6k + 30=6x+12\)
Subtract 30 from both sides: \(6k=6x-18\)
Divide both sides by 6 to get: \(k = x-3\)

Answer: B

Cheers,
Brent