You need 6 tests:
(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)
If at least one of these tests is positive, then you have found two working lanterns.
It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.

The sentence says that buffalo (animals) from Buffalo (city, US), which are buffaloed (intimidated) by Buffalo (city, US) buffalo (animals), themselves buffalo (intimidate) buffalo (animals) from Buffalo (city, US).
Source: WikiWorld Comics

The original puzzle has 12 unique solutions, up to rotation and symmetry. With the additional restriction imposed, there is only one solution.

The German owns the pet fish.
Since the Norwegian lives in the leftmost house (9) and the house next to him is blue (14), the second house must be blue. Since the green house is on the left of the white house (4), the person living in the center house drinks milk (8), and the green house’s owner drinks coffee (5), the fourth house must be green and the fifth one must be white. Since the Brit lives in the red house (1) and the Norwegian lives in the leftmost house (9), the leftmost house must be yellow and the center house must be red. Therefore, the colors of the houses are: YELLOW, BLUE, RED, GREEN, WHITE.
Since the Norwegian from the yellow house smokes Dunhill (7), the man from the blue house must keep a horse (11). The person smoking Blends cannot be in the red house, because this would imply that the person in the green house keeps cats and the Swede keeps dogs in the white house (2, 10). However, in this case the Dane must be drinking tea in the blue house (3) and the person smoking Blends does not have a neighbor drinking water (5), which is a contradiction (15). Also, the person smoking Blends cannot be in the green house, because this would imply that the person in the white house drinks water (15), the Dane lives in the blue house (3), and the German and the Swede live in the last two houses. Since the German smokes Prince (13) and the Swede keeps dogs (2), there is nobody who could smoke Bluemaster and drink beer (12). The person smoking Blends cannot be in the white house either, because this would imply that the person in the green house drinks water (15), when in fact he drinks coffee (5).
Therefore, the person smoking Blends must be in the blue house, and then the German and the Swede must live in the last two houses (2, 13). Since the person who smokes Bluemasters drinks beer (12), this must be the Swede with his dogs in the white house (2). The only option for the person who smokes Pall Mall and raising birds (6) is the red house. Then the Norwegian must keep cats (10) and the German is left with the pet fish in the green house.

The answer is 1024 ways, up to rotation around the table. To see this, note that the youngest honorable knight must sit right next to King Arthur – there are two possible places for him. Then, the second-youngest knight must sit right next to this group of two. Once again, there are two possible places for him. Continuing like this, we see that for all honorable knights, except for the oldest one, there are two possible spots on the table. Multiplying two to the power of ten out, we get 1024.

The simplest solution is given by:
2, 2, 4, 4, 9, 9;
1, 1, 6, 6, 8, 8;
3, 3, 5, 5, 7, 7.
Another solution is given by the so-called “Miwin’s dice”. They are as follows:
1, 1, 3, 5, 5, 6;
2, 3, 3, 4, 4, 5;
1, 2, 2, 4, 6, 6.

Only 3 times. Once you have two socks of the same color, they already form a matching pair.

The poorest among the rich is richer than the richest among the poor. That is because any rich person is by default richer than any poor person.

Let us denote the professors with A, B and C. The thought process of A would go like this: “If my face is not painted, then B will see that C is laughing at him and will realize immediately that he is being pranked. However, B was laughing for a while and therefore my I must being pranked as well.”. Then A will stop laughing and the same will happen with the other two professors B and C.

Open the box labeled “Apples and Bananas”. If you pick a banana from it, then the box labeled “Bananas” will contain apples, and therefore the box labeled “Apples” will contain apples and bananas. Similarly, if you pick an apple from it, then the box labeled “Apples” will contain bananas, and therefore the box labeled “Bananas” will contain apples and bananas”.