As the problem says the apprentice mixed up the hands so that the minute hand was short and the hour hand was long.
The first time the apprentice returned to the client was about 2 hours and 10 minutes after he had set the clock at six.The long had moved olny from twelve to a little past two. The little made two whole circles and an additional 10 minutes. Thus the clock showed the correct time.
The next day around 7:o5 a.m.he came a second time,13 hours and 15 minutes after he had set the clock for six. The long had, acting as the hour hand,covered 13 hours to reach 1. The short hand made 13 full circles and 5 minutes, reaching 7, So the clock showed the correct time again.
The numbers can be grouped by pairs:
999,999,999 and 0;
999,999,998 and 1'
999,999,997 and 2;
and so on....
There are half a billion pairs, and the sum of the digits in each pair is 81. The digits in the unpaired number, 1,000,000,000, add to 1. Then:
(500,000,000 X 81) + 1= 40,500,000,001.
Inless some one can tell me a way that 2 prisoners, at some point, don't out number the guards whether they are just dropping off and still in the boat or actually on land (because even if they are just dropping off and remain in the boat they are still on the other side of the river) I conclude this to be impossible. Please let me know an alternative if you figure one out because i'm stumped.
First he split them in to piles of 3, 3, and 2 bricks. Then he weighs both groups of 3 with each other. If they balance he knows the brick is one of the 2 unweighed bricks and he can weigh them to find the heaver one. If the the stacks of 3 bricks do not balance, he will weigh 2 of the 3 bricks. If they balance he will know the brick left unweighed is heavier, or if they do not balance, he will find the heavier one.