answer:To solve this problem, we will work backwards from the final state where the peasant has no money left. We will determine the amount of money he had before each payment until we reach the initial amount. 1. **Final State:** After the peasant paid the third merchant, he had no money left. Let the amount of money he had before paying the third merchant be x_3. According to the problem, the peasant paid the third merchant half of his remaining money plus 1 ruble: [ left(frac{x_3}{2} + 1 right) = x_3 Rightarrow frac{x_3}{2} + 1 = 0 Rightarrow frac{x_3}{2} = -1 Rightarrow x_3 = 2 ] Thus, before coming to the third merchant, the peasant had 2 rubles left. 2. **Before the Second Merchant:** Let x_2 be the amount of money the peasant had before paying the second merchant. According to the problem, he paid the second merchant half of his remaining money plus 2 rubles: [ left(frac{x_2}{2} + 2 right) = 2 Rightarrow frac{x_2}{2} + 2 = 2 Rightarrow frac{x_2}{2} = 0 Rightarrow x_2=4 ] Thus, before coming to the second merchant, the peasant had 4 rubles left. 3. **Before the First Merchant:** Let x_1 be the amount of money the peasant had before paying the first merchant. According to the problem, he paid the first merchant half of his remaining money plus 1 ruble: [ left(frac{x_1}{2} + 1 right) = 4 Rightarrow frac{x_1}{2} + 1 = 4 Rightarrow frac{x_1}{2} = 3 Rightarrow x_1 = 6 ] Thus, before coming to the first merchant, the peasant had 6 rubles left. Therefore, initially, the peasant had: boxed{18 text{ rubles}}

answer:① Since 2011 div 3 = 670 with a remainder of 1, it follows that 2011in[1], so conclusion ① is correct; ② If ain[1], bin[2] then a=3k+1, b=3m+2, a+b=3(k+m)+3=3(k+m+1), which means a+bin[0], so conclusion ② is correct; ③ Since the set of integers can be divided into three categories based on the remainder when divided by 3, it follows that Z=[0]cup[1]cup[2], so conclusion ③ is correct; ④ Since integers a, b belong to the same "heap", the remainder of a, b when divided by 3 is the same, thus the remainder of a-b when divided by 3 is 0. When a, b both belong to [0], then a-bin[0], so conclusion ④ is incorrect. Therefore, the number of correct conclusions is 3. Hence, the correct choice is boxed{C}.

answer:To determine the maximum length ( n ) of an arithmetic progression of natural numbers ( a_1, a_2, ldots, a_n ) with a difference of 2, such that ( a_k^2 + 1 ) is a prime number for all ( k = 1, 2, ldots, n ), we follow these steps. 1. Consider the general form of the terms in the arithmetic progression. Given the difference of 2, the terms can be written as: [ a_1, a_1 + 2, a_1 + 4, ldots, a_1 + 2(n - 1) ] 2. We need ( a_k^2 + 1 ) to be prime for each ( k ). First, check if there are any constraints for ( a_k ) by exploring specific values. 3. Notice that if ( a = 2 ), then the sequence ( 2, 4, 6, ldots ) satisfies the condition for the first few terms. 4. Check ( a_k = 2k ) since the common difference is 2: - For ( k = 1 ), ( a_1 = 2 ), thus ( a_1^2 + 1 = 2^2 + 1 = 5 ) (prime). - For ( k = 2 ), ( a_2 = 4 ), thus ( a_2^2 + 1 = 4^2 + 1 = 17 ) (prime). - For ( k = 3 ), ( a_3 = 6 ), thus ( a_3^2 + 1 = 6^2 + 1 = 37 ) (prime). - For ( k = 4 ), ( a_4 = 8 ), but ( a_4^2 + 1 = 8^2 + 1 = 65 ) (not a prime number). Therefore, the sequence stops satisfying the condition at ( a_4 = 8 ). 5. Let’s analyze a general term ( a_k = 2k ) again. We need to check if having more than three terms is possible while ensuring ( a_k^2 + 1 ) remains prime: [ text{If } a = 5m pm 2, text{ then } a^2 + 1 text{ is divisible by 5.} ] This can be shown by: [ begin{aligned} &text{For } a = 5m + 2, quad a^2 + 1 = (5m + 2)^2 + 1 = 25m^2 + 20m + 4 + 1 = 25m^2 + 20m + 5 equiv 0 pmod{5}. &text{For } a = 5m - 2, quad a^2 + 1 = (5m - 2)^2 + 1 = 25m^2 - 20m + 4 + 1 = 25m^2 - 20m + 5 equiv 0 pmod{5}. end{aligned} ] Since ( a^2 + 1 ) is divisible by 5, it cannot be a prime unless ( a = 2 ) or the sequence does not follow ( 5m pm 2 ). 6. Thus, for ( n geq 3 ), the sequence ( {2, 4, 6} ) works: [ a_1 = 2, a_2 = 4, a_3 = 6 quad text{give primes } 5, 17, 37. ] # Conclusion: Hence, the maximum length ( n ) of the desired arithmetic progression is: [ boxed{3} ]

answer:If there were 20 books left on a floor after removing the first and the last book, it means there were originally 20 + 2 = 22 books on each floor of a bookshelf. Since each bookshelf has 6 floors, the number of books on one bookshelf would be 22 books per floor * 6 floors = 132 books. Given that there are 28 identical bookshelves in the library, the total number of books on all the shelves would be 132 books per bookshelf * 28 bookshelves = boxed{3696} books.