answer:First, solve for (a) and (b): [ a = -sqrt{frac{9}{27}} = -frac{sqrt{9}}{sqrt{27}} = -frac{3}{3sqrt{3}} = -frac{1}{sqrt{3}} ] [ b = sqrt{frac{(3+sqrt{8})^2}{12}} = frac{3+sqrt{8}}{sqrt{12}} = frac{3+sqrt{8}}{2sqrt{3}} = frac{3 + sqrt{8}}{2sqrt{3}} ] Evaluate ((a + b)^4): [ a + b = -frac{1}{sqrt{3}} + frac{3 + sqrt{8}}{2sqrt{3}} = frac{-2 + 3 + sqrt{8}}{2sqrt{3}} = frac{1 + sqrt{8}}{2sqrt{3}} = frac{1 + 2sqrt{2}}{2sqrt{3}} ] [ (a + b)^4 = left(frac{1 + 2sqrt{2}}{2sqrt{3}}right)^4 = left(frac{1 + 2sqrt{2}}{2sqrt{3}}right)^4 = frac{(1 + 2sqrt{2})^4}{(2sqrt{3})^4} = frac{81 + 108sqrt{2}}{144} ] Continue simplification: [ frac{81 + 108sqrt{2}}{144} = frac{27 + 36sqrt{2}}{48} = frac{27 + 36sqrt{2}}{48} = frac{9 + 12sqrt{2}}{16} ] Thus, (x = 9), (y = 2), and (z = 16). Therefore, (x + y + z = 9 + 2 + 16 = boxed{27}).

answer:1. **Claim**: No such partition exists. 2. **Proof by Contradiction**: Assume there exists an integer ( n > 1 ) such that the set of positive integers can be partitioned into ( n ) nonempty subsets ( A_1, A_2, ldots, A_n ) with the given property. 3. **Case 1**: At least one set in the partition contains both odd and even elements. - **Lemma 1**: All sets ( A_i ) have at least one odd element and at least one even element. - Suppose ( A_1 ) has both an even and an odd element. For ( i geq 1 ), take ( n-1 ) integers from each set besides ( A_i ). Their sum plus an element from ( A_i ) must lie in ( A_i ). By changing the one integer chosen from ( A_1 ), we can make this integer in ( A_i ) take on either parity, so ( A_i ) must have at least one odd and one even integer. - **Lemma 2**: There exists ( a ) such that ( a ) and ( a+1 ) are in different sets. - This is true because otherwise one set would contain all positive integers, which contradicts the partition into ( n ) nonempty subsets. - **Lemma 3**: Let ( d = 2(a_2 + a_3 + cdots + a_{n-1}) ), where ( a_i in A_i ). Then if ( a_1 in A_1 ), ( a_1 + d in A_1 ). - To prove this, consider ( a_1 + a_2 + cdots + a_{n-1} in A_n ). Then ( a_2 + cdots + a_{n-1} + (a_1 + a_2 + cdots + a_{n-1}) = a_1 + d in A_1 ). - Arrange the subsets in order ( A_1, A_2, ldots, A_n ) with ( a_i in A_i ) and ( a_1' in A_1 ) such that ( a_1 ) is even and ( a_1' ) is odd. Let ( a_n = a_{n-1} + 1 ). - Let ( d_1 = 2(a_2 + cdots + a_{n-2} + a_{n-1}) ) and ( d_2 = 2(a_2 + cdots + a_{n-2} + a_n) ). Then ( d_2 = d_1 + 2 ) and ( gcd(d_1, d_2) = 1 ). - It is well known that there exists some ( N ) such that any even integer greater than ( N ) can be expressed as ( Xd_1 + Yd_2 ) for some nonnegative integers ( X, Y ). Since ( a_1 + Xd_1 + Yd_2 in A_1 ) and ( a_1' + Xd_1 + Yd_2 in A_1 ) by Lemma 3, eventually all integers are in ( A_1 ). This is a contradiction because if we take some high enough integer ( z in A_1 ), there will be a higher integer ( z + a_2 + a_3 + cdots + a_{n-1} in A_n ). 4. **Case 2**: All sets in the partition contain either only odd integers or only even integers. - We can apply the same argument as in Case 1. For all even (or all odd) numbers, all even (or odd) integers starting at some point will be in ( A_1 ). This is also clearly impossible. If we take ( z in A_1 ) where ( z ) is high enough, then ( z + a_2 + a_3 + cdots + a_{n-1} in A_n ). If ( z + a_2 + a_3 + cdots + a_{n-1} ) is the same parity as ( z ), then this is a contradiction. Otherwise, we can show that ( A_n ) contains all odd numbers greater than a certain number. Then all integers after some point are in either ( A_1 ) or ( A_n ), so we can apply the same argument and reach a contradiction. Therefore, no such partition exists. (blacksquare)

answer:If Alan collected 48 shells and he collected four times as many shells as Ben, then Ben collected 48 / 4 = 12 shells. Since Ben collected a third of what Laurie did, Laurie collected 3 times as many shells as Ben. Therefore, Laurie collected 12 * 3 = boxed{36} shells.

answer:1. **Binary Interpretation**: Each x_n transforms by doubling x_{n-1}. The change from subtraction of 1 to subtraction of 2 effectively shifts the condition to cut off when x_n reaches or exceeds 2, but since 0 leq x_0 < 1, this doesn't change the operation, and we only observe the wrap around to 0 when exceeding 1. 2. **Modifying x_n Generation**: - If 2x_{n-1} < 1, then x_n = 2x_{n-1}. - If 1 leq 2x_{n-1} < 2, then x_n = 2x_{n-1} - 1 (subtracting 1, not 2 as the condition isn't met). 3. **Pattern Analysis**: Similar to the original, we look for a cyclic pattern that allows x_0 = x_7. We want the binary sequence to repeat after seven shifts: a_i = a_{i+7}. 4. **Counting Valid Initials**: - Each of a_1, a_2, dots, a_7 can still be either 0 or 1. - The sequence should not be all ones because x_0 would not be able to return to itself after 7 shifts (similar argument as in the original). 5. **Calculation**: - There are 2^7 = 128 possible combinations. - By excluding the all-ones case, we have 128 - 1 = 127. (127) Conclusion: There are 127 possible values for x_0 such that x_0 = x_7. The final answer is boxed{C}.