6. From the 11 integers 1 to 11, if any 6 numbers are taken, then the correct conclusions among the following are ( ) in number.
(1) There must be two numbers that are coprime;
(2) There must be one number that is a multiple of another number;
(3) There must be one number whose double is a multiple of another number.
(A) 3
(B) 2
(C) 1
(D) 0
Answer: B.
For statements like "for any... there must be", the "Pigeonhole Principle" should be considered. If it is necessary to prove the conclusion is correct, then pigeonholes should be constructed, and the number of pigeonholes should be less than 6! Look at the question:

The first sentence says that there must be two numbers that are coprime. If this is correct, then it is necessary to construct pigeonholes less than 6, and each group of pigeonholes must contain numbers that are pairwise coprime. It is easy to think that any two adjacent numbers are coprime, so they can be constructed as $(1,2,3)(4,5)(6,7)(8,9)(10,11)$, and the construction method is not unique, there are many other construction methods such as: 【(1,3,4)(2,9) $(8,11)(5,6)(7,10) 】$;

The second sentence is easily refuted by counterexamples, $(6, 7, 8, 9, 10, 11)$, the largest six numbers have no multiple relationship, similarly, there are: $(4, 5, 6, 7, 9, 11$;

The third sentence, based on the counterexample of the second sentence, it can be seen that the third sentence is valid, so it needs to be rigorously proven, still by constructing pigeonholes, how to construct, can be constructed according to the 5 prime factors in 1 to 11 to construct 5 pigeonholes, 1 can be placed in any pigeonhole, $(1,2,4,8)(3,6,9)(5,10)(7)(11)$ these five pigeonholes, to arbitrarily take 6 numbers, there must be two numbers in the same pigeonhole, which must satisfy that one number is a multiple of the other number's 2 times. So two sentences are correct, the final answer is B.