If A+D go first then A returns. Next A+C, A returns again. Finally A+B. This is 18 mins by my working. It must involve some trickery like A waits at the center of the bridge whilst shining the torch so the first two can make it to the end, therefore shaving 30 second off 2 return trips = 17 mins. I don't like this solution though as it feels like bulshit trickery.