answer:1. Group terms into pairs: (1-2), (3-4), dots, (99-100), leaving 101 unpaired. 2. Each pair sums to -1, and we have 50 pairs, so sum of pairs is -1 cdot 50 = -50. 3. Add the unpaired number 101 to -50: -50 + 101 = 51. Thus, the sum is boxed{51}.

answer:Given the set ( S subseteq {1, 2, ldots, 2023} ) and the condition that the sum of the squares of any two elements in ( S ) is not a multiple of 9, we aim to determine the maximum possible size of ( S ), denoted ( |S| ). 1. **Identifying quadratic residues mod 9**: - For an integer ( x ): - ( x equiv 0 pmod{9} implies x^2 equiv 0 pmod{9} ) - ( x equiv pm 1 pmod{9} implies x^2 equiv 1 pmod{9} ) - ( x equiv pm 2 pmod{9} implies x^2 equiv 4 pmod{9} ) - ( x equiv pm 3 pmod{9} implies x^2 equiv 0 pmod{9} ) - ( x equiv pm 4 pmod{9} implies x^2 equiv 7 pmod{9} ) 2. **Exclusion of elements**: - To satisfy the condition that the sum of the squares of any two elements in ( S ) is not a multiple of 9, elements whose squares are congruent to 0 mod 9 (i.e., multiples of 3) must be carefully controlled. 3. **Counting multiples of 3 in ({1, 2, ldots, 2023})**: - The multiples of 3 within this range are ( 3, 6, 9, ldots, 2022 ). - Number of terms in the arithmetic sequence ( a_n = 3n leq 2023 ): [ 3n leq 2023 implies n leq leftlfloor frac{2023}{3} rightrfloor = 674 ] - Thus, there are 674 multiples of 3 in ({1, 2, ldots, 2023}). 4. **Determining the maximal size**: - We can include at most one element for which ( x^2 equiv 0 pmod{9} ) in ( S ). - If we exclude all but one of the multiples of 3, then we exclude 673 elements (since there are 674 multiples). - Hence, the maximum number of elements we can include is: [ 2023 - 673 = 1350 ] 5. **Conclusion**: - Therefore, the maximum possible size of ( |S| ) is ( 1350 ). [ boxed{1350} ]

answer:Note that when n is odd, a_n + a_{n+1} = -(3n - 2) + (3(n+1) - 2) = 3. Thus, a_1 + a_2 + ... + a_{20} = (a_1 + a_2) + (a_3 + a_4) + ... + (a_{19} + a_{20}) = 3 times 10 = boxed{30}. Therefore, the answer is A. It is clear that when n is odd, a_n + a_{n+1} = -(3n - 2) + (3(n+1) - 2) = 3. Consequently, the solution is obtained. This problem tests the application of the pairwise summation method.

answer:The matrix mathbf{E} for a dilation centered at the origin with scale factor 12 is given by: [mathbf{E} = begin{pmatrix} 12 & 0 0 & 12 end{pmatrix}.] The determinant of a 2 times 2 matrix (begin{pmatrix} a & b c & d end{pmatrix}) is calculated as (ad - bc). Applying this formula to matrix mathbf{E}: [det mathbf{E} = (12 times 12) - (0 times 0) = 144 - 0 = 144.] Thus, (det mathbf{E} = boxed{144}.)