There exist several positive integers x such that 1/(x^2 + 2x) is a terminating decimal. What is the second smallest such integer?
+483 We can first factor the denominator as \(\frac{1}{(x)(x+2)}\). In order for this to be a terminating decimal, both \(x\) and \(x+2\) have to only have factors of 2 and/or 5. Testing small values of x, we find that \(x = \{2,8, etc.\}\). Looking at this, we can see that the second smallest integer value of x is \(\boxed{8}\)
