Riddle:

Answer:

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.

Riddle:

Answer:

The difference between the real time and the time of the mirror image is two hours and ten minutes (two and a half hours, minus the twenty minutes of cycling). Therefore, the original time on the clock at home that morning could only have been five minutes past seven: The difference between these clocks is exactly 2 hours and ten minutes (note that also five minutes past one can be mirrored in a similar way, but this is not in the morning!). Conclusion: The boy reaches school at five minutes past seven plus twenty minutes of cycling, which is twenty-five minutes past seven!...

Riddle:

Answer:

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.

Riddle:

Answer:

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.

thanks

Our adult riddles may be a little over the heads of the kids. Some have knowledge that you have to have been around a bit to understand, others may have content that is target for the adult mind. If you are a kid, head over to the kids riddles section or even the difficult riddles or math riddles to challenge your young brain.