Web Puzzles

Job Interview Puzzle Solution

"You set light to both ends of the rope 1 and just one end of rope 2. It will take half an hour for the two burning ends of rope 1 to meet. Then you set light to the remaining end of rope 2. The time it will take for rope 2 to finish burning will be a further 15 minutes. Hence all together, both ropes burned in this manner will take 45 minutes to burn." Tom leaned back in his chair and folded his arms.

That squinty eye stared at him some more. "Very well," stated Patrick after some time. "We will let you know about the success of your application shortly." He stood and shook Tom's hand.
As Tom left the room, he grinned again. "Elderberries," he muttered to himself.

Napoleon's Star Puzzle Solution

To be able to place nine coins, it is necessary to make the 3rd point of each step equal to the start point of the previous step. For example:

a-g; i-a; c-i; f-c; e-f; h-e; b-h; j-b; d-j.


 With such a simple solution, it's hard to believe that Napoleon stayed in charge for so long.

Southern Cross Puzzle Solution

The missing number is 18. The numbers in the bottom row are the square of the numbers in the top row, but with their digits reversed.

4	5	6	7	8	9
61	52	63	94	46	18

Bread And Water Puzzle Solution

Each slice of bread was divided into 3 equal pieces, making a total of 24 pieces. These were then divided between the three men, 8 pieces each.

Mohammed had 5 slices, and so contributed 15 pieces. He ate 8 pieces himself, so 7 were eaten by Alek.

Azad had 3 slices, which contributed 9 pieces. Azad ate 8, leaving 1 piece for Alek.

Therefore, Mohammed gave 7 pieces of bread away, and Azad only gave 1 piece. So Mohammed deserves 7 gold coins and Azad only 1. If he were smarter or less greedy, Azad should have accepted Mohammed's initial offer of 5 - 3.

Canopus Puzzle Solution

A nice and simple sum does the trick.…

110,000	+
11,000	+
1,100	+
11	=
__________
122,111

Deneb Puzzle Solution

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

Visit At The Kremlin Puzzle Solution

There are 20 buttons. This is because there are 10 combinations of Heroes in pairs of 2, and this is multiplied by 2 because these combinations have to be mapped to two different persons (the President of the Praesidium or the Chief of KGB). The magnetic cards, as held by the eight persons, and marked with an X, are distributed as follows:

Person	Buttons
Secretary of Party	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X
President of Praesidium	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X	-	-	-	-	-
Chief of KGB	-	-	-	-	-	X	X	X	X	X	X	X	X	X	X	X	X	X	X	X
1st Hero	X	X	X	X	-	-	-	-	-	-	-	X	X	X	X	X	X	X	X	-
2nd Hero	X	X	X	-	X	X	X	X	-	-	-	-	-	-	X	X	X	X	-	X
3rd Hero	X	X	-	X	X	-	-	X	-	X	X	-	-	X	-	X	X	-	X	X
4th Hero	X	-	X	X	X	-	X	-	X	-	X	-	X	-	-	X	-	X	X	X
5th Hero	-	X	X	X	X	X	-	-	X	X	-	X	-	-	-	-	X	X	X	X

Frogs Puzzle Solution

The only frog is Amanda. The table below shows, on the left, all possible combinations of frogs (X stands for frog, - stands for non-frog), while on the right pane there is the control of the truthfullness (T for true, F for false) of each given statement, for each combination of frog population. The only combination where every girl tells one true statement and one false one, is the emphasised row. Hence, there's only one frog, and that's Amanda.

Possible Frogs				Statement
A	B	C	D	A		B		C		D
-	-	-	-	F	F	T	F	F	F	F	T
-	-	-	X	F	F	T	F	T	F	F	F
-	-	X	-	F	F	T	F	T	T	F	T
-	-	X	X	F	F	T	T	F	T	T	F
-	X	-	-	F	F	F	F	T	F	F	T
-	X	-	X	F	F	F	F	F	F	T	F
-	X	X	-	F	F	F	T	F	T	T	T
-	X	X	X	F	T	F	T	F	T	F	F
X	-	-	-	T	F	T	F	T	F	F	T
X	-	-	X	T	F	T	F	F	F	T	F
X	-	X	-	T	F	T	T	F	T	T	T
X	-	X	X	T	T	T	T	F	T	F	F
X	X	-	-	T	F	F	F	F	F	T	T
X	X	-	X	T	T	F	F	F	F	F	F
X	X	X	-	T	T	F	T	F	T	F	T
X	X	X	X	T	T	F	T	F	T	F	F

Swapping Art Puzzle Solution

The swap is unfair. If we abbreviate with an E the drawings by Ernst, an M for Magritte's sketches, and a B for Bacon's paintings, we can rewrite the equations as stated by the Italian dealer as:

1. EEEE = MMMMM (both dealers agreed on this)
2. MME = BB (again, both dealers agreed on this too)
3. MMMBB = EEEEM (the French dealer wasn't sure of this)

Now, if for the 3rd equation we substitute BB with MME (from equation 2), we'll have a 4th equation: MMMMME = EEEEM. But the 1st equation states that MMMMM = EEEE, so it is possible to cancel out, in equation 4, MMMMM on the left and EEEE on the right. This would leave us with the equation E = M, which does not fit with equation 1, where E is obviously greater than M.


The problem gives us 3 equations:
    1. 4x = 5y
    2. 2y + x = 2z
    3. 3y + 2z = 4x + y

Equation 2. can be rewritten as:
    4. x = -2y + 2z

Equation 3. can be rewritten as:
    5. 4x = 2y + 2z

Adding equations 4. and 5. gives us:
    6. 5x = 4z

Combining 1. and 6., we get:
    7. 4x = 5y = (16/5)z

If we convert the offer (3.) in terms of y, we get:
    8. 3y + 2(25/16)y = 5y + y
    ie 3y + (25/8)y = 6y
    ie 6.125y = 6y

The last equation is not absolutely true, which proves the swap is unfair.

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.