Bob suggested that Dave hold his finger in the air while someone wound a piece of string tightly around his finger just above the metal ring. The string forced the swelling down. As they unwounded the string from the end nearest the ring, someone else slid the ring up. They continued winding and unwinding the string until the ring could be easily removed.
First think who will operate each bulb, obviously person #2 will do all the even numbers, and say person #10 will operate all the bulbs that end in a zero. So who would operate for example bulb 48: Persons numbered: 1 & 48, 2 & 24, 3 & 16, 4 & 12, 6 & 8 ........ That is all the factors (numbers by which 48 is divisible) will be in pairs. This means that for every person who switches a bulb on there will be someone to switch it off. This willl result in the bulb being back at it's original state. So why aren't all the bulbs off? Think of bulb 36:- The factors are: 1 & 36, 2 & 13, 6 & 6 Well in this case whilst all the factors are in pairs the number 6 is paired with it's self. Clearly the sixth person will only flick the bulb once and so the pairs don't cancel. This is true of all the square numbers. There are 10 square numbers between 1 and 100 (1, 4, 9, 16, 25, 36, 49, 64, 81 & 100) hence 10 bulbs remain on.
The shepherd who had three loaves should get one coin and the shepherd who had five loaves should get seven coins. If there were eight loaves and three men, each man ate two and two-thirds loaves. So the first shepherd gave the hunter one-third of a loaf and the second shepherd gave the hunter two and one-third loaves. The shepherd who gave one-third of a loaf should get one coin and the one who gave seven-thirds of a loaf should get seven coins.
The sum of the numbers on the watch face is 78. If the two lines cross, there must be 4 equal parts, but 78 is not divisible by 4. Then the lines do not cross, giving three parts with the sum of 26 each. Once you see the pairs on the face that add to 13(12+1,11+2, and so far) the answer will be easy to find.
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.