Web Puzzles

A Farmer's Good Fortune Puzzle Solution

Here's one combination:

 94 sheep =  $470
  1 pig   =   $30
  5 cows  =  $500
-----------------
100 heads = $1000

Are there anymore?

Switching Logic Puzzle Solution

Turn switch #1 ON. After about five minutes or so, turn switch #1 OFF and turn switch #2 ON.
Then go upstairs and check on the bulbs.

The one ON is obviously #2. The other ones are OFF, but one of them should be very hot by having been ON for five minutes. That's #1, and the remaining bulb is #3.

U2 Gig Puzzle Solution

The trick is to get the two slowest people to cross at the same time. One solution is...

• Bono and Edge cross the bridge for which they take 2 mins (Total time = 2)
• Then Bono comes back with the torch (Total Time = 2 + 1 = 3)
• Then Adam and Larry cross the bridge (Total time = 3 + 10 = 13)
• Then Edge comes back (Total = 13+2 = 15)
• Then both Bono and Edge cross the bridge (Time = 15+2=17)

The Diagram Puzzle Solution

Joe counted how many times each of the 7 LEDs lit up for all 10 digits...

 _
|_|
|_|

Loadsa Coins! Puzzle Solution

You reach out and pick any 20 coins, and turn each of them over. The 20 coins is one group, and the remaining coins the other.

"Bugger," says Greed. "Why didn't I think of that?"

"It's so simple, yet so difficult to think of," you say. "Now can I have my eyesight back? Hello? Hello......!?"

Socks And Shops Puzzle Solution

For each of the twenty pairs of socks, they unclipped it and gave one sock each.

TV Show Puzzle Solution

This puzzle isn't particularly new, but it became very well known in 1990, thanks to the person that allegedly had "the highest IQ ever recorded" -- 228, according to The Guinness Book of WorldRecords. Miss Marilyn vos Savant (yes, it was a female) wrote a solution to this puzzle on her weekly column, published in a popular magazine. This solution led to waves of mathematicians, statisticians, and professors to very heated discussions about the validity of it.

The contestant, according to vos Savant, had a greater chance of winning the car by switching his pick to the 2nd door.

She claimed that by sticking to the first choice, the chances of winning were 1 out of 3, while the chances doubled to 2 out of 3 by switching choices. To convince her readers, she invited her readers to imagine 1 million doors instead of just 3. "You pick door number 1," she wrote, "and the host, who knows what's behind every door, and doesn't want you to win, opens all of the other doors, bar number 777,777. You wouldn't think twice about switching doors, right?"

Most of her readership didn't find it as obvious as she thought it was... She started receiving a lot of mail, much of it from mathematicians, who didn't agree at all. They argued that the chances were absolutely equal, whether or not the contestant switched choice.

The week after, she attempted to convince her readers of her reasoning, by creating a table where all 6 possible outcomes were considered:

Door 1	Door 2	Door 3	Outcome (sticking to door 1)
Car	Nothing	Nothing	Victory
Nothing	Car	Nothing	Loss
Nothing	Nothing	Car	Loss
Door 1	Door 2	Door 3	Outcome (switching to other door)
Car	Nothing	Nothing	Loss
Nothing	Car	Nothing	Victory
Nothing	Nothing	Car	Victory

The table, she explained, shows that "by switching choices you win 2 out of 3 times; on the other hand, by sticking to the first choice, you win only once out of of 3 times".

However, this wasn't enough to silence her critics. Actually, it was getting worse.

"When reality seems to cause such a conflict with good sense," wrote vos Savant, "people are left shaken." This time, she tried a different route. Let's imagine that, after the host shows that there's nothing behind one of the doors, the set becomes the landing pad for a UFO. Out of it comes a little green lady. Without her knowing which door was picked first by the contestant, she is asked to pick one of the remaining closed doors. The chances for her to find the car are 50%. "That's because she doesn't have the advantage enjoyed by the contestant, ie the host's help. If the prize is behind door 2, he will open door 3; if it is behind door 3, he will open door 2. Therefore, if you switch choice, you will win if the prize is behind door 2 or 3. YOU WIN IN EITHER CASES! If you DON'T switch, you'll win only if the car is behind door 1".

Apparently, she was absolutely right, because the mathematicians reluctantly admitted their mistake.

Golf Puzzle Solution

The clues given in the puzzle are:

• The Stock Broker is not Levi.
• Jack is not the Lawyer; neither Jack nor the Lawyer owns the Corvette.
• The Musician owns the Porsche; therefore no one else owns the Porsche and the Musician owns no other car.
• Seth is not the owner of the Cadillac, the Corvette, nor is he the Doctor; also, the Doctor does not own either the Cadillac or the Corvette either.

By Reasoning:

• The Doctor, by elimination, must own the Ferrari; the Lawyer must own the Cadillac; the Stock Broker must own the Corvette.
• Tne Stock Broker owns the Corvette. Neither Seth nor Jack own the Corvette, so the Stock Broker must be Robert; since the Stock Broker drives the Corvette, then Robert drives the Corvette.
• The Lawyer owns the Cadillac. Seth does not own the Cadillac; therefore Seth is not the Lawyer; therefore Seth is the Musician; the Musician owns the Porsche; therefore Seth owns the Porsche.
• By elimination, Jack is the Doctor; the Doctor owns the Ferrari; therefore Jack owns the Ferrari.
• By elimination, Levi owns the Cadillac.

The Solution:

• Seth is the Musician and owns the Porsche.
• Levi is the Lawyer and owns the Cadillac.
• Jack is the Doctor and owns the Ferrari.
• Robert is the Stock Broker and owns the Corvette.

General Manoeuvres Puzzle Solution

The manoeuvre was conducted this way:

1. The two boys cross to the opposite bank (10 mins)
   
2. One of them stays there while the other comes back (5 mins)
   
3. The boy gets off the boat, a soldier jumps on board and crosses the river (8 mins)
   
4. The soldier gets off, and the boat returns with the other boy (5 mins)
   

This operation required 28 minutes. The sequence had to be repeated as many times as the number of men in the platoon, ie 39 more times. However, it was needed to subtract 5 minutes from from the total: when the last man of the platoon crossed, the time (5 mins) taken by the second boy to cross back must not be counted, as the last soldier had already reached the other bank of the river.

The total time was therefore [(28 * 40) - 5] = 1115 minutes, which amounts to 18 hours and 35 minutes.

Mike Horan points out that it can be done faster if you leave both boys stranded on one side with the boat on the other. The first 39 soldiers cross at 28 minutes each (1092 minutes). You then have the two boys plus the last solider on one side. The final soldier then rows across himself, hence 1092 + 8 = 1100 minutes.

Snake Puzzle Solution

Counters 1 and 8 need to be in the middle since they each have only one number that they cannot be neighbours with. Hence 7 and 2 must go in the sidewings, and the rest is trivial.

	3	5
7	1	8	2
	4	6

Visit At The White House Puzzle Solution

Person	Buttons
President	X	X	X	X	X	X	X
Vice President	X	X	X	X	X	X	-
President of Senate	X	X	X	-	-	-	X
Secretary of State	X	-	-	X	X	-	X
Chief of Armed Forces	-	X	-	X	-	X	X
Dean of Harvard	-	-	X	-	X	X	X

Drama Galore Puzzle Solution

They needed seventeen trips, frenquently using the island in the middle of the river as temporary destination. If we denote the names of the men with their initial in upper case, and the names of the women with their initial in lower case (by complete chance, the initials of the men are the same as the initials of their girlfriends), we would get the following table:

Trip #	Departure bank	Direction	Island	Direction	Arrival bank
1	ABCDcd		-	=>	ab
2	ABCDbcd	<=	-		a
3	ABCDd	=>	bc		a
4	ABCDcd	<=	b		a
5	CDcd		b	=>	ABa
6	BCDcd	<=	b		Aa
7	BCD	=>	bcd		Aa
8	BCDd	<=	bc		Aa
9	Dd		bc	=>	ABCa
10	Dd		abc	<=	ABC
11	Dd		b	=>	ABCac
12	BDd	<=	b		ACac
13	d		b	=>	ABCDac
14	d		bc	<=	ABCDa
15	d		-	=>	ABCDabc
16	cd	<=	-		ABCDab
17	-		-	=>	ABCDabcd

Galactic Expedition Puzzle Solution

The table below shows all possible combinations (32 of them) of three different integers, the product of which is 900. Next to each combination there is the result of the sum A+B, then the result of the sum A+C, and finally the the number B for that particular combination:

Combination #	Factors	A+B	A+C	B
1	450 * 2 * 1	452	451	2
2	300 * 3 * 1	303	301	3
3	225 * 4 * 1	229	226	4
4	180 * 5 * 1	185	181	5
5	150 * 6 * 1	156	151	6
6	100 * 9 * 1	109	101	9
7	90 * 10 * 1	100	91	10
8	75 * 12 * 1	87	76	12
9	60 * 15 * 1	75	61	15
10	50 * 18 * 1	68	51	18
11	45 * 20 * 1	65	46	20
12	36 * 25 * 1	61	37	25
13	150 * 3 * 2	153	152	3
14	90 * 5 * 2	95	92	5
15	75 * 6 * 2	81	77	6
16	50 * 9 * 2	59	52	9
17	45 * 10 * 2	55	47	10
18	30 * 15 * 2	45	32	15
19	25 * 18 * 2	43	27	18
20	75 * 4 * 3	79	78	4
21	60 * 5 * 3	65	63	5
22	50 * 6 * 3	56	53	6
23	30 * 10 * 3	40	33	10
24	25 * 12 * 3	37	28	12
25	20 * 15 * 3	35	23	15
26	45 * 5 * 4	50	49	5
27	25 * 9 * 4	34	29	9
28	30 * 6 * 5	36	35	6
29	20 * 9 * 5	29	25	9
30	18 * 10 * 5	28	23	10
31	15 * 12 * 5	27	20	12
32	15 * 10 * 6	25	21	10

If the first question was asked to the person (Wadzru) that was given A+B or A+C and he answers "don't know", that means that the number he received appears in the A+B and A+C columns more than once. On the other hand, if that number appears in those columns only once, then the solution would be found immediately, as that number would relate to only one of the 32 combinations.

Therefore the first candidate was given one of the following numbers: 65, 61, 37, 65, 29, 28, 27, 25, 23. These numbers appear more than once in columns A+B and A+C, and because of this, they do not yet allow to find the correct combination. But we can now discard, from the 32 combinations, the ones where the 9 numbers listed above are not included in columns A+B and A+C. The combinations left were then:

Combination #	Factors	A+B	A+C	B
9	60 * 15 * 1	75	61	15
11	45 * 20 * 1	65	46	20
12	36 * 25 * 1	61	37	25
19	25 * 18 * 2	43	27	18
21	60 * 5 * 3	65	63	5
24	25 * 12 * 3	37	28	12
25	20 * 15 * 3	35	23	15
27	25 * 9 * 4	34	29	9
28	30 * 6 * 5	36	35	6
29	20 * 9 * 5	29	25	9
30	18 * 10 * 5	28	23	10
31	15 * 12 * 5	27	20	12
32	15 * 10 * 6	25	21	10

The second candidate (Zaxre) followed the same reasoning, so she then knew that these were the only possible combinations left, after Wadzru gave his answer. The situation for Zaxre was the same though: if the number she was given appeared only once in column B, then the right combination would be the one containing that number; if the number appeared more than once in column B, then the solution would not yet be within reach. And this is what happened, since she answered "don't know". But the elimination of certain combinations could go on nevertheless, because after she said "don't know", combinations # 11, 12, 19, 21, 28 were automatically discarded. Then it was Wadzru's second turn, and the current situation was:

Combination #	Factors	A+B	A+C	B
9	60 * 15 * 1	75	61	15
24	25 * 12 * 3	37	28	12
25	20 * 15 * 3	35	23	15
27	25 * 9 * 4	34	29	9
29	20 * 9 * 5	29	25	9
30	18 * 10 * 5	28	23	10
31	15 * 12 * 5	27	20	12
32	15 * 10 * 6	25	21	10

At this point, the first contestant answered again "don't know", and automatically discarded combinations 9 and 31. Then the second candidate, after following, again, the logical reasoning of Wadzru, answered "don't know", therefore discarding combinations 24 and 25.The third turn started with the first contestant facing the following combinations:

Combination #	Factors	A+B	A+C	B
27	25 * 9 * 4	34	29	9
29	20 * 9 * 5	29	25	9
30	18 * 10 * 5	28	23	10
32	15 * 10 * 6	25	21	10

Wadzru, by answering "don't know" on his third turn, discarded the only combination with no alternatives, ie 30, so Zaxre was presented only with combinations 27, 29, 32; her answer "don't know" left only two combinations: 27 and 29. Then it was the first contestant again, and he was faced with the following situation:

Combination #	Factors	A+B	A+C	B
27	25 * 9 * 4	34	29	9
29	20 * 9 * 5	29	25	9

If the first contestant now answered "don't know", then the number he was given must have had to be 29; but since he was given number 25, he was able to know the correct combination requested: 20*9*5. When Zaxre saw that Wadzru had the answer, she was also able to find the solution, because she had also been able to follow the selection process. For her, it was only possible to give the answer on her 4th turn, because if Wadzru could have been able to come up with the answer on his 3rd turn (when the numbers appering just once in columns A+B and A+C were more than one), then Zaxre would not have been able to know which of the combinations contained the number given to her opponent.

If instead, the question was first asked to Zaxre, who was given number B, then after her first "don't know" it would have been possible to discard from the table (the one with all 32 combinations) combinations 1, 11, 12. After her opponent also said "don't know", the combinations to be discarded would have been 2, 3, 4, 5, 6, 7, 8, 9, 10, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23 and 26, leaving the table looking like this:

Combination #	Factors	A+B	A+C	B
19	25 * 18 * 2	43	27	18
24	25 * 12 * 3	37	28	12
25	20 * 15 * 3	35	23	15
27	25 * 9 * 4	34	29	9
28	30 * 6 * 5	36	35	6
29	20 * 9 * 5	29	25	9
30	18 * 10 * 5	28	23	10
31	15 * 12 * 5	27	20	12
32	15 * 10 * 6	25	21	10

For the second turn, the "don't know" answer of the contestant answering first (Zaxre in this scenario) would have automatically discarded combinations 19, 25 and 28, and the "don't know" from her opponent would have further discarded another one: 31. For the third turn, the remaining possible combinations are: 24, 27, 29, 30 and 32. If Zaxre could not answer yet, then combination 24 would be discarded, while Wadzru's "don't know" would discard combination 30. The situation would now be:

Combination #	Factors	A+B	A+C	B
27	25 * 9 * 4	34	29	9
29	20 * 9 * 5	29	25	9
32	15 * 10 * 6	25	21	10

If Zaxre would now be able to know the solution, she would have been given, as B, number 10 (combination # 32), resulting with the product 15*10*6. But her opponent would also be able to answer then, thanks to Zaxre being able to find the solution. This scenario, like the previous one with Wadzru starting, where both opponents are able to find the solution at the same time, can only happen on the fourth turn. It's worth knowing that a fourth "don't know" from the candidates would have led to the problem being left unsolved.

Killing Time Puzzle Solution

The logical conclusion must necessarily be: I avoid kangaroos. Based on the 10 statements, it is possible to deduce the following conclusions:

Deduction #	From statements #	Deduction
11	1 and 5	All animals in this house kill mice
12	8 and 11	All animals in this house are carnivorous
13	4 and 12	All animals in this house move at night
14	6 and 13	All animals I own move at night
15	10 and 14	All animals I own love watching the moon
16	2 and 15	All animals I own are tamable
17	7 and 16	I do not own any kangaroo
18	9 and 17	I hate kangaroos
final	3 and 18	I avoid kangaroos

Faulty Batch Puzzle Solution

The secretary placed on the scale 1 coin from the first batch, 2 from the second, and so on until he put 8 from the eighth batch.

If all coins weighed 10 grams each, then the weight displayed on the scale should have been 360 grams ((1 + 2 + ... + 8) × 10). But, since one batch of coins weighs less, the difference between 360 grams and the weight displayed on the scale should point us to the faulty batch. For example, if the faulty batch was the fifth one, then the total weight displayed on the scale would be 355 grams. Or if it was the seventh batch, the weight would have been 353 grams, ie 7 grams less than the theoretical total weight of 360 grams.

An 'optimisation' on this solution is to omit the 8 coins from the eighth batch. In this case, the maximum weight of the coins would be 280 grams, and if it equals 280, then the eighth batch is the faulty one. Thanks to Denis Borris for this observation.

By using the same logic, one could omit the coins from any one of the other batches, instead of the eighth one. For example, if we omit the fourth batch, we'll be left with a theoretical 320 grams and, if it is indeed the total weight, then we will know that the fourth batch was the faulty one. Thanks to Glen Parnell for noticing this.

Mizar Puzzle Solution:

11 seconds. The 5 seconds needed to signal six o'clock are the 5 silent intermissions between rings. At twelve, the 12 rings are interleaved by 11 silent intermissions, which need 11 seconds to be executed.