Web Puzzles

100 Puzzle Solution

• 111 - 11 = 100
• (3 * 33) + (3 / 3) = 100
• (5 * 5 * 5) - (5 * 5) = 100
• (5 + 5 + 5 + 5) * 5 = 100
• (11 - 1) ^ ( 1 + 1) = 100 [Thanks to Steven Renich for that one!]
• ((2 * 2 * 2) + 2) ^ 2 = 100 [Thanks to David Cohen for this other one!]
• ((99 * 9) + 9) / 9 = 100 [Thanks to Taylor Lowry for this other one!]
• ((22 - 2) / 2) ^ 2 = 100 [Thanks to Karen D. Miller for this other one!]
• 6! / 6 - 6! / (6 * 6) = 100 [Thanks to Karen D. Miller for this other one!]
• 5! - 5 - 5 - 5 - 5 = 100 [Thanks to Karen D. Miller for this other one!]
• 5! - (5 + 5 + 5 + 5) = 100 [Thanks to Jim St. Clair for this other one!]
• 5 * 5 * (5 - 5 / 5) = 100 [Thanks to Rishi Mohan Sanwal for this one!]
• 4! + 4! + 4! + 4! + 4 = 100 [Thanks Karen D. Miller and Saurabh Gupta!]
• 99 + 9 ^ (9 - 9) = 100 [Thanks Gopalakrishnan Thirumurthy for this one!]
• 5! - 5 * (5 - 5 / 5) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• (4! + 4 ^ (4 - 4)) * 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• 4! * 4 + 4 - 4 + 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• (4! * 4) + (4 * 4 / 4) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• 5 * 5 * (5 - (5 - 5)!) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
• ((9 - 9) */ 9)! + 99 = 0! + 99 = 1 + 99 = 100 [Thanks to Bala Neerumalla!]
• (3 - 3)! + 33 * 3 = 100 [Thanks Bala Neerumalla for this other one!]
• (5 + 5) ^ ((5 + 5) / 5) = 100 [Thanks Bala Neerumalla for this one!]
• ((2 ^ 2) * 2 + 2) ^ 2 = 100 [Thanks Bala Neerumalla for this one!]
• (4! + ((4 - 4) */ 4)!) * 4 = 100 [Thanks Bala Neerumalla for this one!]

Forum member Bealzbob reminds us that, whenever we have a subtraction like A - B, we can rewrite it as an addition with a negative number, like A + (-B). In the case of the first solution, 111 - 11 = 111 + (-11).

Glen Parnell points out that, in any solution in which the digit 2 is used, it could be replaced by 2!, whenever the 2 is not only a digit but also a number, as 2 = 2! (but of course, 22 is not equal to 22!).
Bala Neerumalla has suggested some alternative solutions using trigonometric functions Sin() and Cos():

• Sin(99 - 9) + 99 = Sin(90) + 99 = 1 + 99 = 100
• Cos((9 - 9) */ 9) + 99 = Cos(0) + 99 = 1 + 99 = 100
• Cos(3 - 3) + 33 * 3 = 100
• (4! + (Sin(4-4)*4)!)*4 = 100

Gopalakrishnan Thirumurthy has intelligently played around with cubic root functions. If we were to represent this function with a name like, say, CR(), the following alternative solutions could be achieved:

• 88 + 8 + CR(8) + CR(8) = 100
• (8 + CR(8)) ^ (CR(8) * (8 / 8)) = 100
• ((CR(8) * CR(8) * CR(8)) + CR(8)) ^ CR(8) = 100
• ((8 * (8 / 8)) + CR(8)) ^ CR(8) = 100

However! If the cubic root function is represented as it should be, ie ³Ö, then there is the extra "3" which invalidates the solution. These alternative solutions using CR(), have therefore been given as possible examples that will not be accepted in the future. Unless, of course, the "3" in ³Ö is actually one of the five digits used in a solution. Same goes with any other root or power. This also applies to any usage of the symbol "%", as it really involves the number 100.
Bala Neerumalla smartly noticed that he can produce number 100 by using a different number base than the normal base 10 (b10). The following example is in base 5 (b5):

• 444 - 44 = (4 * 5^2 + 4 * 5^1 + 4 * 5^0) - (4 * 5^1 + 4 * 5^0) = 100

However! The above equation mixes up two different number bases: the left hand side is b5, while the right hand side is b10 (100 b5 is equal to 25 b10). This is the reason why we will not be posting anymore solutions that mix different number bases. We will, however, accept solutions that use number bases different from 10, as long as they are consistent on both sides of the equation. Therefore, we will accept:

• 111 - 11 = 100 [Wolfgang Solfrank pointed out that this works in any number base, thanks!]

Here are some more examples of equations using different number bases:

• b16 4 ^ 4 +- 4 * (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
• b16 (2 * 2 * 2 * 2) ^ 2 = 100 [Thanks to Glen Parnell for this one!]
• b8 4 * 4 * 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
• b8 (4 ^ 4) / 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
• b8 4 * 4 * 4 * (4 / 4) = 100 [Thanks to Glen Parnell for this one!]
• b8 (2 ^ (2 * 2)) * 2 * 2 = 100 [Thanks to Glen Parnell for this one!]
• b2 11 - 1 + 1 + 1 = 100 [Thanks to Glen Parnell for this one!]

Furthermore, Glen Parnell noticed that there exists a general rule that applies to equations using any number base:

• bX+1 ((X - X) */ X)! + XX = 100

Here are some examples of it:

• b16 ((F - F) */ F)! + FF = 100
• b12 ((B - B) */ B)! + BB = 100
• b10 ((9 - 9) */ 9)! + 99 = 100 [As already shown by Bala Neerumalla]
• b8 ((7 - 7) */ 7)! + 77 = 100
• b4 ((3 - 3) */ 3)! + 33 = 100
• b2 ((1 - 1) */ 1)! + 11 = 100

Jeff Shall cleverly discovered that he can produce 100 using the Roman numeral 'L', which corresponds to number 50:

• ((L / L) + (L / L)) * L = ((50 / 50) + (50 / 50)) * 50 = 100

Unashamedly based on the above solution by Jeff Shall, we found that we can also use the Roman numeral 'C', which corresponds to number 100:

• (C / C) * (C / C) * C = (100 / 100) * (100 / 100) * 100 = 100
• (C / C) - (C / C) + C = (100 / 100) - (100 / 100) + 100 = 100
• CCC - CC = 300 - 200 = 100
• CC - CC + C = 200 - 200 + 100 = 100

Gopalakrishnan Thirumurthy has discovered a lot of new combinations with Roman numerals, including numeral 'X', which is number 10:

• (L + L) * L ^ (L - L) = (50 + 50) * 50 ^ (50 - 50) = 100
• (L + L) / L ^ (L - L) = (50 + 50) / 50 ^ (50 - 50) = 100
• XX * X - X * X = 20 * 10 - 10 * 10 = 100
• (X + (X / X)) * X - X = (10 + (10 / 10)) * 10 - 10= 100
• X * X * (X ^ (X - X)) = 10 * 10 * (10 ^ (10 - 10) = 100
• X * X / (X ^ (X - X)) = 10 * 10 / (10 ^ (10 - 10) = 100
• C * C * C / C * C = 100 * 100 * 100 / 100 * 100 = 100
• (C - C) + (C - C) + C = (100 - 100) + (100 - 100) + 100 = 100
• L * (L - L) + L + L = 50 * (50 - 50) + 50 + 50 = 100
• C * (CC / C) - C = 100 * (200 / 100) - 100 = 100
• C * (C / C) + (C - C) = 100 * (100 / 100) + (100 - 100) = 100

Gopalakrishnan Thirumurthy has also noticed that all the solutions involving five times the number 5, could be rewritten by substituting it with Roman numeral 'V'. This is also true if, instead of number 5 and numeral 'V', we were talking about number 1 and Roman numeral 'I'.
Now, how about some solution using Roman numerals 'D' (500), or 'M' (1000)? If you find any alternative solutions, get in touch with us by email!

Easy!

A Law-Abiding Citizen Puzzle Solution

The electrician packed the tube diagonally into a flat-ish squared package, with sides of less than one metre. More precisely, the sides were about 0.85 metres long, because [squareroot(1.2² / 2) = 0.84852...]

Tamerlano's Trap II Puzzle Solution

Moschin asked the guard: "If I had asked you which door leads to freedom, which door would you have pointed me to?"

If the guard was truthful, he would have shown the right door. If he lies, he would have again shown the right door, because he would have given the merchant the opposite answer to what he would have given if he was asked a direct question.

Tamerlano's Trap Puzzle Solution

The merchant asked one of the soldiers, it didn't matter which one: "If I had asked your colleague which door leads to freedom, which of the two doors would he have pointed me to?"

If the interrogated soldier was the one that tells the truth, he would have pointed him to the door that leads to death, because that's the door that the liar would have showed. But even if the same question was asked the liar, the same door (the one that leads to death) would have been the one pointed at, ie the door opposite the one that would have been shown by the truthful soldier.

Once obtained the answer, the merchant went to the door he was NOT pointed at, and enjoyed his freedom.

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

Hydra Puzzle Solution

If 6 boys and 4 girls spend 4 hours to complete the jigsaw puzzle, then 3 boys and 2 girls will need 8 hours, which is also the amount of time needed by the team of 2 boys and 5 girls. The reader can see that the input of 1 extra boy on the first team equals the input of 3 extra girls on the other team. The conclusion deduced is that the input from 1 boy is worth as much as the input from 3 girls.

To be more precise, it is possible to demonstate that each boy can put together, in one hour, 75 pieces of the puzzle, compared to the 25 for each girl. This is because, if we say that 6 boys and 4 girls (ie team 1) are equal to 22 girls ((6 * 3) + 4), and since that team puts together 550 pieces per hour (2200 / 4), then the workforce of each team member equals 25 pieces/hour (550 / 22).

Sorry ladies about the "non politically correct" nature of this puzzle, but I'm just translating. I was actually thinking of using Martians and Venusians instead, but then I thought "Who cares!" - Mickey.

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.

Top Secret Puzzle Solution

The answers given were the the number of letters in the question. When asked "Nine?" the secret agent should have answered "Four".

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

A Struggle For Survival Puzzle Solution

The answer is nine planets. The geoanthropologist states that one of the habitable planets is eighth toward the sun, so there must be at least 8 planets. The navigator mentions that there are 3 planets between the 6th planet and the 8th towards the sun. The only number that can satisfy both statements is the number nine. You will then realise that the navigator was talking about Earth.

The Greatest Show On Earth Puzzle Solution

There are 69 bipeds and 43 four-legged animals. If all animals were four-legged, the officer would have counted 448 legs, not 310. Obviously 138 legs are missing, hence 69 subjects are bipeds.

It's worth pointing out that, within the bipeds, there are 13 stilt birds, which include flamingos and storks. The cook must have counted them as they were standing on one leg only, so whilst he is not crazy, he almost drove the customs officer mad.

A Thinking Man Puzzle Solution

The symbol ¿ means the difference between the number made up of the all digits of the operation, and the mirror of this last number. i.e,
5 ¿ 2 = 52 - 25 = 27
6 ¿ 3 = 63 - 36 = 27
8 ¿ 4 = 84 - 48 = 36

An alternative solution for this symbol (submitted by Alfa Chan... many thanks!) is simply the difference between the two numbers multiplied by 9. i.e,
5 ¿ 2 = (5 - 2) × 9 = 27
6 ¿ 3 = (6 - 3) × 9 = 27
8 ¿ 4 = (8 - 4) × 9 = 36

Another alternative solution submitted by Mickey Kawick... thanks!! We have x ¿ y; If x is odd, then the result is 5x + y, otherwise it's 5x - y. i.e,
5 ¿ 2 = 5 × 5 + 2 = 27 (5 is odd, so we add the 2)
6 ¿ 3 = 6 × 5 - 3 = 27 (6 is even, so we subtract the 3)
8 ¿ 4 = 8 × 5 - 4 = 36 (8 is even, so we subtract the 4)

The symbol ¤ means the difference between the two numbers multiplied by the larger of the two numbers. i.e,
5 ¤ 2 = (5 - 2) * 5 = 3 * 5 = 15
6 ¤ 4 = (6 - 4) * 6 = 2 * 6 = 12
3 ¤ 8 = (8 - 3) * 8 = 5 * 8 = 40

An alternative solution for this symbol, as submitted by Jeff Schall (many thanks!), is the difference between the square of the bigger of the two numbers and their product. i.e,
5 ¤ 2 = (5 ^ 2) - (5 * 2) = 25 - 10 = 15
6 ¤ 4 = (6 ^ 2) - (6 * 4) = 36 - 24 = 12
3 ¤ 8 = (8 ^ 2) - (3 * 8) = 64 - 24 = 40

Three Divine Comedians Puzzle Solution

Dante figured that first of all, he had to find out which of the three monsters was the lunatic one. Let's call the three monsters A, B, C.

In order to do so, he asked one of them (let's say monster A for this example) a question like: "If I asked a question to monster B, would I stand a greater chance of obtaining a truth than if I asked the same question to monster C?"

The possible combinations (where + means the monster that tells the truth, - means the monster that lies, x means the lunatic monster) are:

Answer given	Monster A	Monster B	Monster C
No	+	-	x
Yes	+	x	-
No	-	+	x
Yes	-	x	+
Yes/No	x	+	-
Yes/No	x	-	+

If the answer was a "Yes", then it was safe to say that x couldn't have been monster C; on the other hand, if the answer was a "No", then x couldn't possibly have been Monster B.

Once that Dante obtained this information, he used his second question to find out whether the monster that is definately NOT the lunatic one was + or -, and the easiest way of achieved this was by asking a question with an obvious answer such as, "Is 2 an even number?" After finding out whether the monster is the liar or the sincere one, the third question was used to resolve the other two monsters; a question like "Is monster A the lunatic one" did the trick.

Barrels 'O' Fun Puzzle Solution

There are multiple ways of solving this. One way is given below, and it's probably the fastest one. Each set of 3 numbers separated by hyphens is the amount of wine (in litres) in the 3 barrels after each "pouring operation". The 3 barrels are always in the same order: 12, 7, and 5 litres.

• 12-0-0
• 5-7-0
• 5-2-5
• 10-2-0
• 10-0-2
• 3-7-2
• 3-4-5
• 8-4-0
• 8-0-4
• 1-7-4
• 1-6-5
• 6-6-0.

Three Camels Puzzle Solution

The Great Sage added his own camel to the other seventeen. He then gave 9 camels (one half of 18) to the eldest of the three, 6 camels (one third) to Ali, and 2 camels (one 9th) to Ismail. Then took his own camel back and sat in front of the tent, thanking God for His generosity.

Another way to explain the solution, perhaps in a more mathematical manner, is the following, courtesy of Gopalakrishnan Thirumurthy.
The three sons are assigned their shares: 1/2, 1/3, and 1/9. The sum of their shares is 1/2 + 1/3 + 1/9 = 17/18. Out of 18 camels, 17 of them are left by their father. So,

• #1 gets 1/2 of 18 = 9
• #2 gets 1/3 of 18 = 6
• #3 gets 1/9 of 18 = 2

9 + 6 + 2 = 17.

Cheers To Statistics Puzzle Solution

Three artist would do the trick. This is because twenty-four artists would paint twenty-four models in two and a half hours. Since the available time increases eight-fold (2.5 * 8 = 20), it is possible to reduce the number of painters by the same number of times (24 / 8 = 3).

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.

Faulty Batches Puzzle Solution

They had to weigh 1 coin from the 1st batch, 2 from the 2nd, 4 from the 3rd, 8 from the 4th, and 16 from the 5th one.

If all coins weighed 10 grams as they should, the scale would display 310 grams ((1 + 2 + 4 + 8 + 16) * 10). However, since one batch has 9 grams coins, and another 11 grams coins, then the total weight of this combination of coins will be:

Total Weight	Number of 9g coins	Number of 11g coins
311	1	2
313	1	4
317	1	8
325	1	16
312	2	4
316	2	8
324	2	16
314	4	8
322	4	16
318	8	16
309	2	1
307	4	1
303	8	1
295	16	1
308	4	2
304	8	2
296	16	2
306	8	4
298	16	4
302	16	8

After seeing the solution to this puzzle, it is clear that it would be a lot easier to simply use the scales up to 5 times rather than go through all this, but where is the fun in that?

Orbiting Logic Puzzle Solution

During the next orbit, the sleepy but correct answer came forth from the astronaut: 2 of clubs, 5 of hearts, 7 of diamonds, 9 of spades, and 10 of hearts.

Following from (g) - no two cards are the same rank - and (b), the strongest combination (ie highest ranks) possible is 10, 9, 7, 6, 4, which adds up to 36. If we write down all combinations of 2 numbers that have a difference of 9, and a maximum sum of 36, we'll have:

• 2, 11
• 3, 12
• 4, 13
• 5, 14
• 6, 15
• 7, 16
• 8, 17
• 9, 18
• 10,19
• 11, 20
• 12, 21
• 13, 22

The number of odd ranks in the 5-card combination could be 1 or 3 (not 2 and not 4 because their sum would be an even number, and a difference of 9 between two even numbers is inexistent). If there was only 1 odd rank, then there would be 4 even ranks, and the weakest combination of evens would be 2, 4, 6, 8 which add up to 20: the difference, 9, would result in a single number, 11, which does not exist in the deck of cards (the Royals were excluded). Therefore we have 3 odd ranks. The sum of the weakest 3 odd ranks (excluding the Aces) is 15, ie 3 + 5 + 7. So, from the combinations above, we can exclude all combinations that contain an odd number less than 15. We are left with

• 6, 15
• 8, 17
• 10, 19
• 12, 21

The first one (6, 15) must be excluded because the 3 odd numbers (3, 5, 7) would be joined by the only 2 even numbers which would add up to 6 (2,4), and therefore we would have a sequence of 5 consecutive numbers, which doesn't match the constraint set by (b). We are left with 3 combinations:

• 8, 17 (and the 5 ranks would be 2, 3, 5, 6, 9)
• 10, 19 (and the 5 ranks would be 3, 4, 6, 7, 9)
• 12, 21 (and the 5 ranks would be 2, 5, 7, 9, 10)

But (d) states that the sum of red cards is twice of the sum of black cards, so the first 2 combinations must be excluded, because it's not possible to find 2 sets of numbers, one of which is twice the other; therefore the 5 ranks can only be 2, 5, 7, 9, 10. The red cards must be 5, 7, 10.

Following (e), rank 2 must be clubs and 10 must be hearts, so rank 9 must be spades; following (f), rank 7 has got to be diamonds and rank 5 must be hearts.

Three Skullcaps Puzzle Solution

At first, any explorer could have guessed the colour of his own skullcap only if the other two wore green skullcaps. Unfortunately, the first explorer admits to not being able to work it out, and is killed.

Then, with two people left it is possible for either explorer to know if he wears a red skullcap only if the other wears green. In that case, he could reason, "The other person wears green, so if I also have a green skullcap, then the first man would have deduced that he was wearing a red skullcap, since there are only two green caps. Therefore my skullcap is, without doubt, red." Of course, this is not the case. Stupidly, the second explorer admits he does not know and is killed as punishment.

After seeing that the other two could not deduce their colours, and believing in their deductive capabilities, the third prisoner was then sure he was wearing a red skullcap.

Ivanovic had to turn around two posters: the first one (Lenin) and the third one (the Red Star).

All that Nikita Proskoijev said was that all the posters representing Lenin show Hammer And Sickle on their opposite face. Therefore it is needed to check the back of the first poster and the front of the third one, to make sure that the Red Star wasn't linked with Lenin's face, cause if it was, then Proskoijev's statement would have been false.

Checking the front of the fourth poster (which is what Ivanovic did, that's why he's now a lumberjack near Jakutsk) is useless; if the 4th poster shows Lenin face as its front, it would just confirm what Proskoijev stated; but then, if the 4th poster showed Marx, this would not have falsified Proskoijev's statement, because he said that Lenin is linked to Hammer And Sickle, while he didn't state that Hammer And Sickle are linked to Lenin.

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.