answer:1. **Initial Setup:** Let ( a_1, a_2, ldots, a_n ) be any ( n ) distinct integers. Take ( N ) to be a number divisible by all integers ( 2, 3, ldots, n ). For our purpose, ( N = n! ) (the factorial of ( n )) suffices. 2. **Adjusting Integers:** Define ( b_i = a_i N ) for each ( i ). Consequently, ( b_1, b_2, ldots, b_n ) are distinct integers since ( a_1, a_2, ldots, a_n ) are distinct and ( N ) is the same multiplier for all ( a_i ). 3. **Average of Subsets and Primes:** Observe that the average of any subset of ( {b_1, b_2, ldots, b_n} ) is an integer because the ( b_i )s are simply multiples of ( N ). Let ( {c_1, c_2, ldots, c_m} ) be the set of all distinct averages of subsets of ( {b_1, b_2, ldots, b_n} ). Note that ( m leq 2^n - 1 ) since every non-empty subset provides a distinct average. 4. **Prime Identification:** Let ( p_1, p_2, ldots, p_m ) be distinct primes. Each ( c_i ) (the average) is a ratio of sums of ( b_i )'s divided by subset size. If ( p ) is a prime dividing one or more of the ( c_i ), identify the highest power of ( p ) dividing ( c_i ). 5. **Chinese Remainder Theorem Application:** By the Chinese Remainder Theorem, construct ( K ) such that ( K equiv -d_i pmod{p_i} ) for each ( i ). 6. **Constructing ( N_p ):** Define ( N_p = p^K ) where ( K ) is tailored by the aforementioned construction such that the highest power of ( p ) dividing ( N_p c_i ) is ( p )-th power. 7. **Iterative Process for All Primes:** Repeat the Chinese Remainder Theorem process for each prime ( p ) dividing any ( c_i ). The product ( M ) of all such ( N_p ) will provide: [ M = N_{p_1} cdot N_{p_2} cdot ldots cdot N_{p_m} ] yielding that ( h_i = M c_i ) is a power of some prime ( p_i ). 8. **Final Set Construction:** Therefore, the required set ( X ) is ({M b_1, M b_2, ldots, M b_n}). These integers are built such that the average of elements in any subset of ( X ) is a ( p_i )-th power for some prime ( p_i ). **Conclusion:** Thus, we have shown the construction achieves the goal. For any integer ( n ), we can find a set ( X ) of ( n ) distinct integers greater than 1 such that the average of the elements of any subset of ( X ) is a square, cube, or higher power: [ boxed{} ]

answer:Given two distinct integers ( a ) and ( b ), we denote their operations as follows: 1. Addition: ( a + b ) 2. Subtraction: ( a - b ) 3. Multiplication: ( ab ) 4. Division: ( frac{a}{b} ) According to the problem, the sum of these four results is (-100): [ (a+b) + (a-b) + ab + frac{a}{b} = -100 ] Furthermore, when the addition result is omitted, the sum of the remaining three operations is also (-100): [ (a-b) + ab + frac{a}{b} = -100 ] We have the following equations based on the given conditions: 1. ( (a+b) + (a-b) + ab + frac{a}{b} = -100 ) 2. ( (a-b) + ab + frac{a}{b} = -100 ) Subtracting the second equation from the first, we get: [ (a+b) = 0 ] This implies: [ a + b = 0 ] [ a = -b ] Substituting ( a = -b ) into the second equation: [ (a-b) + ab + frac{a}{b} = -100 ] [ (-b - b) + (-b cdot b) + frac{-b}{b} = -100 ] [ -2b - b^2 - 1 = -100 ] Simplifying, we obtain: [ - (b+1)^2 = -100 ] Removing the negative sign from both sides: [ (b+1)^2 = 100 ] Taking the square root of both sides: [ b + 1 = pm 10 ] This gives us two possible solutions: [ b + 1 = 10 quad text{or} quad b + 1 = -10 ] [ b = 9 quad text{or} quad b = -11 ] Correspondingly, substituting back for ( a ): [ a = -b ] [ a = -9 quad text{if} quad b = 9 ] [ a = 11 quad text{if} quad b = -11 ] Therefore, the pairs of integers that Bára could have written are ( -9 ) and ( 9 ) or ( 11 ) and ( -11 ). Conclusion: [ boxed{-9 text{ and } 9 text{ or } 11 text{ and } -11} ]

answer:We have the 2x2 matrix: [ begin{pmatrix} 7 & -2 -3 & 5 end{pmatrix} ] The determinant of a 2x2 matrix is calculated using the formula: [ text{det}(A) = ad - bc ] where (a = 7), (b = -2), (c = -3), and (d = 5). Plugging in these values, we get: [ text{det}(A) = (7)(5) - (-2)(-3) = 35 - 6 = 29 ] The determinant of the matrix is (boxed{29}).

answer:To solve this problem, we need to understand how positive and negative numbers represent direction. 1. Moving north is considered a positive direction, so moving north 8m is denoted as +8m. 2. Conversely, moving south is considered the opposite direction to north. Therefore, to represent moving south 5m, we use a negative sign to denote the opposite direction, resulting in -5m. Thus, moving south 5m is correctly denoted as -5m. Therefore, the correct choice is boxed{text{B: -5m}}.