Jan. 1, and Dec. 31.
Divide the coins into three groups of three. Balance two groups and take the lighter group, or the third group if the first two weigh the same. Apply the same procedure for the three coins in the selected group.
Divide the coins into three groups of four. Balance two groups. If they weigh the same, the fake coin is in the remaining group of four. Balance two of them on one side, and one on the other plus one real coin. If they are equal then it's the last unused coin, just balance it against a real coin to tell if it's lighter or heavier. If the second weighing does not balance, the fake coin is one of the three suspect coins used. Take note which side was heavier and which lighter, then take the two suspect coins on one side and balance them against each other. If they're equal, it's the third coin and it's lighter or heavier as noted in the previous weighing. If they're not equal, it's the ligther or heavier coin depending on whether the two together were lighter or heavier as noted in the previous weighing.
If The first weighing does not balance, take two from the lighter group plus two from the heavier group. Balance them against one coin from the lighter group plus one from the heavier group plus two real coins. If they balance, then it's either the remaining lighter coin or the remaining heavier coin. Just balance either one against a real coin. If the second weighing does not balance, then if the group of two lighter plus two heavier is lighter, it's one of the two lighter suspects or the heavier suspect on the other side. Balance the lighter suspects against each other, and the fake coin is the lighter one or, if they balance, the heavier third coin. If the group of two lighter plus two heavier is heavier, it's one of the two heavier suspects or the lighter suspect on the other side. Balance the heavier suspects against each other, and the fake coin is the heavier one or, if they balance, the lighter third coin.
Note: There is another solution (provided by joey sonza). It uses what might be called a 3-grouping technique as opposed to what might be called a 2-grouping as described above. This solution uses 3 lighter + 1 heavier aginst 1 lighter + 3 real coins on the second weighing.
The wife saved 20 minutes travelling the shorter roundtrip from home to the point where she met her husband and back to home. Therefore the point where she met her husband is 10 minutes away from the station by car; she saved travelling this distance twice which accounts for the saved 20 minutes. This means the man was met by his wife 10 minutes before the usual time he gets picked up at the train station. Since he arrived 60 minutes before the usual pick-up time, this means he has been walking 50 minutes.
Average speed for the whole trip is total distance over total time. The time required to average 30 mph for the whole trip is therefore D/30, where D is the total distance. Now consider how much time the man spent driving half the trip -- it is also D/30. The man spent all the required time at just half the trip, so he can never attain his desired average no matter how fast he drives the rest of the way.
This is solved by induction. Everybody is told the day the visiting witch came (day one) that there is at least one unfaithful husband. If a witch's own husband was faithful, then she would see the actual number of unfaithful husbands, while if her husband was unfaithful she would see one less than the actual total. If there is only one unfaithful husband, then the witch who sees zero unfaithful husbands will correctly conclude that her husband must be the unfaithful one.
If there are two unfaithful husbands, then two witches see one unfaithful husband while the rest sees two. Each of the witches who see only one will be waiting to see what will happen to the one husband they see. If tomorrow the one husband they see becomes a pig, then they can relax knowing that the wife of the one unfaithful husband they see must have seen zero to conclude that her husband was unfaithful. But both witches will be waiting so that nothing will happen the first night. Tomorrow they will correctly conclude that the other witch must be seeing one more that's why she waited. On the second night both witches will turn their husbands into pigs.
If there are three unfaithful husbands, then the witches who see only two will be waiting for the wives of those two to turn their husbands to pigs on the second night, otherwise there must be a third one which would be their own husbands. The same logic can be extended to four, five, and so on. So the answer is: the last of the unfaithful husband (they'll all becomes pigs at the same night actually) will become a pig on the 50th day. Note that if the witches were not told that there is at least one unfaithful husband, they could not make the above induction and the husbands would remain unpunished, even though all the while half of the witches actually see 50 unfaithful husbands and half see 49.
(x-x) is zero, so the product is zero.
The answer is 3616. There are different ways to arrive at the answer. First, it's possible to use brute force (good luck). Then it's possible to have a computer do the brute force by writing a computer program. Then it's also possible to look for a pattern to the answer when the number of prisoners is 1, 2, and so on. If you do this third method, you'll soon see that the answer apparently can be computed by subtracting the difference of the smallest power of two greater than or equal to the number of prisoners and the number of prisoners from the number of prisoners. Huh? In other words, the safe post is given by x - (2^n - x), where x is the number of prisoners and 2^n is the smallest power of two greater than or equal to x. Expressed another way, the answer is
2x - 2^n, where n is an integer such that 2^(n-1) < x <= 2^nor, as stated by nj devil,
2(x - 2^int(log x/log 2)).Seeing this pattern is of course only half the solution. The other half is proving that the formula is actually true for all x (greater than 0). This can be done by mathematical induction, by first proving that the formula is correct for x=1, then proving that if the formula is correct for x, it is also correct for x+1. This can probably be another puzzle by itself. The intriguing question is why does the formula work. Is the formula actually derivable from a solution method? MJordan described a method for solving the problem that looks like a formula can be derived from it. Here is his answer verbatim (if anyone is able to derive the formula from this, I'd be interested in seeing your solution, please send me a note):
"I assume I am included in the 10,000 mutinous pirates. After the first round, half will be spent, the remaining 5000 posts area all even numbered. Try to renumber them by dividing each by 2, then we'll have #1, 2, 3,... again.The second round will skip #10000 which is now #5000 and start at #2 which is now #1. Then #3, and so on. See a pattern here?
So you would say, the number to choose is the last one if you continue dividing it in 2 in this manner. Well, not until you have 625 remaining. The #625 (originally #10000) will be dead after the 5th round.
After the 5th round, there will be 312 remaining. Renumber again, but then start at #2 ( since the last man down was #625 in the previous round) and execute each even numbered post up to #312.
After 6th round, there will be 156 remaining. Continue the same process. After 7th round, there will be 78 remaining. 8th round, there will be 39 remaining.
After 9th round, there will be 20 remaining. Reverse the process again by killing odd-numbered posts up to #19.
After 10th round, 10 remain. After 11th round, 5 remain. After 12th round, 2 remain. Reverse the process again, killing even-numbered posts. #2 will be dead and #1 remains.
So, after 13th round, only 1 remains
Whew! I'm no longer sure if this is the right way to solve this.
Now, back-track the process to get the original number of the remaining post. There are 13 rounds of shooting. Make sense, 2 raise to 13th is 8192 (closest to 10000).
After Round # Original # of #1 13 1 12 1 of the 2 11 2 of the 5 10 4 of the 10 9 8 of the 20 8 15 of the 39 7 29 of the 78 6 57 of the 156 5 113 of the 312 4 226 of the 625 3 452 of the 1250 2 904 of the 2500 1 1808 of the 5000 0 3616 of the original 10,000
They can do it in 2.75 hours (Sorry, I originally wrote 2.5). The key to the solution is that they should all arrive at the same time so that no time is wasted waiting. The girl (or boy it doesn't matter) takes the bike first going a certain distance (point A), then leaves the bike to walk the rest of the way. The dog trotting will arrive at the bike, then rides it back closer to the boy (at point B). The dog trots the rest of the way. The boy walks up to the bike then rides the rest of the way. Calculate where along the 10-mile route should the points A and B be so that they all arrive at the same time. This is the shortest time it will take them to complete the trip.
These are the three equations.
girl: T = A/12 + (10-A)/2
boy: T = B/2 + (10-B)/12
dog: T = A/4 + (A-B)/16 + (10-B)/4
Solving for the three unknowns should give you A = 5.4, B = 4.6, and T = 2.75.
4 weights: 1, 3, 9, 27.
"The person in front is wearing a red hat. The third person must be seeing one blue hat/one red hat or two red hats since he can't determine what color of hat he's wearing. The second person knows that since the third person cant determine the color of his hat, the third person must be seeing one blue/one red or two red hats. Since the second person cannot determine the color of his hat, then he must be seeing a red hat. Using the same logic, the first person will be able to conclude that he is wearing a red hat." - chary
$1.20, $1.25, $1.50, $3.16