+ BURNING ROPES

Light both ends of rope A and one end of rope B. After 30 minutes, rope A will be completely burned up and there will be 30 minutes of rope B left. Light the other end of rope B; it will burn up in 15 minutes. Total time elapsed since starting the ropes on fire: 45 minutes.

+ TEN BAGS OF GOLD

Take one coin out of the first bag, two out of the second, ... ten out of the tenth bag. Weigh all of them, your remainder will tell you which bag has the counterfeit coins (.1oz = bag1, .2oz = bag2, etc). No remainder means the tenth bag.

+ 3 LIGHT SWITCHES

Flip Switch 1 on and wait several minutes. Then flip Switch 1 off, flip Switch 2 on, and enter the room. If the light is on, it is controlled by Switch 2. If the light is off but the bulb is warm, it is controlled by switch 1. If the light is off and the bulb is not warm, it is controlled by switch 3.

+ ONE THIRD TOSS

Toss the coin twice. Let TAILS+HEADS, HEADS+TAILS, and TAILS+TAILS correspond to the three choices. If Heads comes up twice, redo the two tosses.

+ HOTEL BILL

The trick here is to realize that this is not a sum of the money that the three people paid originally, as that would need to include the money the clerk has ($25).

The sum that we really desire is this one: $30 = $1 (inside Guest pocket) + $1 (inside Guest pocket) + $1 (inside Guest pocket) + $2 (inside bellhop's pocket) + $25 (hotel cash register)

+ THREE CANNIBALS

First, two cannibals go across to the other side of the river, then the rower gets called back. Next, the rowing cannibal takes the second across and then gets called back, so now there are two cannibals on the far side. Two anthropologists go over, then one anthropologist accompanies one cannibal back, so now there is one anthropologist and one cannibal on the far side.

The last two anthropologists go over to the far side, so now all the anthropologists are across the other side, along with the boat and one cannibal.

In two trips, the cannibal on the far side takes the boat and ferries the other two cannibals across the river.

+ BOSS'S CUBES

Either: Cube One: 0, 1, 2, 3, 4, 5 Cube Two: 0, 1, 2, 6, 7, 8

or Cube One: 1, 2, 4, 0, 5, 6 Cube Two: 3, 1, 2, 7, 8, 0

+ FIVE GALLONS

Pour the four-gallon bucket filled with water into the empty seven-gallon bucket. Fill the four gallon bucket up again and poor as much as you can into the seven-gallon bucket until the seven-gallon bucket is fill. Now there is one gallon left in the four-gallon bucket. Empty the seven-gallon bucket and transfer the one gallon of water into the seven-gallon bucket. Fill the four-gallon bucket one more time, then pour the four gallons into the seven-gallon bucket making which already has one gallon in it, making a total of five gallons.

+ THE MOST INTELLIGENT PRINCE

The king would not select two white hats and one black hat. This would mean two princes would see one black hat and one white hat. You would be at a disadvantage if you were the only prince wearing a black hat.

If you were wearing the black hat, it would not take long for one of the other princes to deduce he was wearing a white hat.

If an intelligent prince saw a white hat and a black hat, he would eventually realize that the king would never select two black hats and one white hat. Any prince seeing two black hats would instantly know he was wearing a white hat. Therefore, if a prince can see one black hat, he can work out he is wearing white.

Therefore, the only fair test is for all three princes to be wearing white hats. After waiting some time just to be sure, you can safely assert you are wearing a white hat.

+ GLASS HALF FULL

Hold your finger at the water level, and then flip the glass upside down. If the water level is now higher than your finger, the glass was more than half full, and vice versa.

+ ROW, ROW, ROW YOUR TREES

Make an equilateral triangle with 3 trees at the corners, 3 trees at midpoints, and 1 tree in the center. 1 2 3 4 5 6 7

The six rows go through trees: 1-2-5, 1-3-7, 1-4-6, 2-4-7, 3-4-5, 5-6-7

+ TRUTH IN PACKAGING

Remove an item from box 3, that item tells you what label to put on box 3. Move the “nails & screws” label to the box that is still labeled, and put that label on the remaining box.

Example: you remove a nail from box 3. Move the “Nails” label from box 1 to box 3. Move the “nails & screws” label to box 2, and the “screws” label to box 1.

+ GLOBE TRAVERSAL

One place is, of course, at the North Pole. You can walk 1 mile south, then 1 mile west, then 1 mile north, and be back where you started. But there are more.

There is a latitude near the South Pole, where the circumference around the Earth is exactly 1 mile. Let us call this latitude C(1). Now start your journey 1 mile north of this latitude. Walk 1 mile south, 1 mile west (which brings us all the way back around), and then 1 mile north to the starting point.

Of course, C(1) isn't just 1 point, but an infinite ring of points on that latitude. But there are more!

There must be a latitude where the circumference around the Earth is exactly 1/2 mile. We'll call that C(1/2), and it works just as well in place of C(1) - the only difference being we circle the Earth twice instead of once. C(1/3) works too, as a latitude where Earth's circumference is 1/3 of a mile.....

So the full answer is: “an infinite number of circles of latitude near the South Pole, each containing an infinite number of starting points, plus one extra point for the North Pole.” Whew!

+ ROPE LADDER

Sir Edmund cuts the rope into two pieces, five and ten yards long. He ties one end of the five-yard rope to the top hook. At the other end of the five-yard rope he makes a loop, and threads in the ten-yard rope, but only halfway, so that when he folds the ten-yard rope in half, it adds five yards to the rope's inital five yards.

Using this improvised ten-yard rope, Sir Edmund then proceeds down to the first ledge. When he reaches the ledge, he pulls on the ten-yard rope to retrieve its full length. He then ties the ten-yard rope to the hook on the ledge and decends to the ground.

+ POISONED WINE BARRELS

To begin, label each bottle with both its decimal number and binary equivalent.

Bottle 1 = 0000000001 Bottle 2 = 0000000010 Bottle 3 = 0000000011 ...etc.

Now choose 10 prisoners and assign each to one of the binary bits. (Pris A)(Pris B)(Pris C)(Pris D)(Pris E)(Pris F)(Pris G)(Pris H)(Pris I)(Pris)

The binary bottle numbers serve as a code describing which prisoners are to drink from it. In this system, a one means the prisoner drinks from it, a zero means the prisoner doesn’t.

For example, only Prisoner J should drink from bottle one since its binary is 0000000001. Whereas, Prisoners I and G should drink from bottle ten whose binary is 0000001010 because it has 1's in the columns that match up with prisoners I and G.

Continue this process until you have given out sips of wine from every bottle. After 24 hours, line up the prisoners in order and note which ones have been poisoned with ones and mark the rest with zeros. Convert this number back into decimal to reveal which bottle was poisoned.

+ BLIND MAN AND CARDS

If the original pile has c cards with f cards facing up, then the blind man divides them into piles of size f and c - f. Then he flips all cards in the pile with f cards. Let's see why it works for c = 52 and f = 10. The blind man would divide the cards into two piles with 10 and 42 cards each. If there are k face-up cards in the 10-card pile, then there must be 10 - k face-up cards in the 42-card pile (because the total number of face-up cards is 10). So by flipping all cards in the 10-card pile, the number of face-up cards in both piles would become equal to 10 - k.

+ TRIWIZARD RIDDLE

A spider.

+ CHERYL'S BIRTHDAY

It helps to put the list of 10 dates into table form: May 15 16 19 June 17 18 July 14 16 August 14 15 17

Now let’s examine what Albert and Bernard say. Albert goes first: I don’t know when your birthday is, but I know Bernard doesn’t know, either.

The first half of the sentence is obvious — Albert only knows the month, but not the day — but the second half is the first critical clue.

The initial reaction is, how could Bernard know? Cheryl only whispered the day, so how could he have more information than Albert? But if Cheryl had whispered “19,” then Bernard would indeed know the exact date — May 19 — because there is only one date with 19 in it. Similarly, if Cheryl had told Bernard, “18,” then Bernard would know Cheryl’s birthday was June 18.

Thus, for this statement by Albert to be true means that Cheryl did not say to Albert, “May” or “June.” (Again, for logic puzzles, the possibility that Albert is lying or confused is off the table.)

Then Bernard replies: I didn’t know originally, but now I do. So from Albert’s statement, Bernard now also knows that Cheryl’s birthday is not in May or June, eliminating half of the possibilities, leaving July 14, July 16, Aug. 14, Aug. 15 and Aug. 17. But Bernard now knows. If Cheryl had told him “14,” he would not know, because there would still be two possibilities: July 14 and Aug. 14. Thus we know the day is not the 14th.

Now there are only three possibilities left: July 16, Aug. 15 and Aug. 17.

Albert again: Well, now I know too! The same logical process again: For Albert to know, the month has to be July, because if Cheryl had told him, “August,” then he would still have two possibilities: Aug. 15 and Aug. 17.

The answer is July 16.

+ UNUSUAL SUSPECTS

Jim robbed the bank.

+ CHAMELEONS

Let (p, y, m) denote a population of p purple, y yellow and m maroon chameleons. Can population (13, 15, 17) can be transformed into (45, 0, 0) or (0, 45, 0) or (0, 0, 45) through a series of pairwise meetings? Define function X(p, y, m) = (0p + 1y + 2m) mod 3. An interesting property of X is that its value does not change after any pairwise meeting because X(p, y, m) = X(p-1, y-1, m+2) = X(p-1, y+2, m-1) = X(p+2, y-1, m-1). Now X(13, 15, 17) equals 1. However, X(45, 0, 0) = X(0, 45, 0) = X(0, 0, 45) = 0. This means that there is no sequence of pairwise meetings after which all chameleons will have identical color.

+ ZEBRA PUZZLE

House: 1 2 3 4 5 Color: Yellow Blue Red Ivory Green Nationality: Norwegian Ukrainian Englishman Spaniard Japanese Drink: Water Tea Milk O.J. Coffee Smoke: Kools Chestr. Old Gold Lucky Sk. Parliament Pet: Fox Horse Snails Dog Zebra

+ WATER AND WINE

The proportions are the same. Imagine that after you have mixed the water and wine, the two liquids separate, like water and oil. Visualize the layers of water and wine in the two glasses at the end.

+ ALICE AND BOB

Alice is 36, Bob is 27.

+ CORK, BOTTLE, COIN

Push the cork into the bottle and then shake the coin out.

+ ST. IVES RIDDLE

One.

+ BLUE EYES

If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he's the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: (THEOREM 1) If there is one blue-eyed person, he leaves the first night.

If there are two blue-eyed people, they will each look at the other. They will each realize that "if I don't have blue eyes (HYPOTHESIS 1), then that guy is the only blue-eyed person. And if he's the only person, by THEOREM 1 he will leave tonight." They each wait and see, and when neither of them leave the first night, each realizes "My HYPOTHESIS 1 was incorrect. I must have blue eyes." And each leaves the second night.

So: (THEOREM 2): If there are two blue-eyed people on the island, they will each leave the 2nd night.

If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities -- "I have blue eyes" or "I don't have blue eyes." He will know that if he doesn't have blue eyes, there are only two blue-eyed people on the island -- the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes -- THEOREM 2 says that if he didn't, the other guys would have left. When he sees that they didn't, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.

This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they'll each wait 99 days, see that the rest of the group hasn't gone anywhere, and on the 100th night, they all leave.

The answer is that on the 100th day, all 100 blue-eyed people will leave.