answer:1. To find out how many lollipops remain after filling the maximum possible number of packages, we need to perform the division of 8362 lollipops by 12 lollipops per package and determine the remainder. 2. Let us divide (8362) by (12): [ 8362 div 12 = 696 text{ R } 10. ] 3. Here, (696) is the quotient, indicating that we can pack (696) full packages of lollipops. The remainder, (10), represents the lollipops that are left over after packing as many packages as possible. 4. This calculation means: [ 8362 - (12 times 696) = 8362 - 8352 = 10. ] 5. Therefore, (10) lollipops remain when (8362) lollipops are divided into packages of (12) each. Conclusion: [ boxed{(E)} ]

answer:To find the negation of the given proposition, we must understand what it means and then apply the logical rules for negation. The original proposition is: - exists xin R, such that x^{2}+3x+2 lt 0 The negation of an existential quantifier (exists) proposition is a universal quantifier (forall) proposition, and the inequality sign switches from < to geqslant. This is because the original statement asserts the existence of at least one x for which the inequality is true, and negating it asserts that for all x, the opposite of the inequality must hold. Therefore, the negation of the proposition is: - forall xin R, x^{2}+3x+2geqslant 0 Thus, the correct choice that represents the negation of the given proposition is: [boxed{B}]

answer:(1) Since C_n^8=C_n^9, we have n=17, thus T_{r+1}=C_{17}^rx^{(frac{17-r}{2}-frac{r}{3})}2^r, let frac{17-r}{2}-frac{r}{3}=1, solving this gives r=9, thus T_{10}=C_{17}^9x2^9, therefore, the coefficient of the linear term of x is C_{17}^9cdot2^9; (2) Let f(x)=(2x-1)^7, thus f(-1)=-a_0+a_1-a_2+ldots+a_7, f(1)=a_0+a_1+a_2+ldots+a_7, therefore, a_1+a_3+a_5+a_7= frac{f(1)+f(-1)}{2}= frac{1^7+(-3)^7}{2}=-1093. The final answers are: 1. The coefficient of the linear term of x is boxed{C_{17}^9cdot2^9}. 2. The value of a_1+a_3+a_5+a_7 is boxed{-1093}.

answer:Let the function y=2xf(x)-3=0, which leads to the equation f(x) = frac{3}{2x}, - When x in [1, 2), the function f(x) first increases and then decreases, reaching its maximum value of 1 at x = frac{3}{2}, and y = frac{3}{2x} also has y=1 at x = frac{3}{2}; - When x in [2, 2^2), f(x) = frac{1}{2}fleft(frac{1}{2}xright), and at x=3, the function f(x) reaches its maximum value frac{1}{2}, and y = frac{3}{2x} also has y= frac{1}{2} at x=3; - When x in [2^2, 2^3), f(x) = frac{1}{2}fleft(frac{1}{2}xright), and at x=6, the function f(x) reaches its maximum value frac{1}{4}, and y = frac{3}{2x} also has y= frac{1}{4} at x=6; - ...; - When x in [2^{10}, 2^{11}), f(x) = frac{1}{2}fleft(frac{1}{2}xright), and at x=1536, the function f(x) reaches its maximum value frac{1}{2^{10}}, and y = frac{3}{2x} also has y= frac{1}{2^{10}} at x=1536. Therefore, the number of zeros of the function y=2xf(x)-3 in the interval (1, 2017) is boxed{11}. By setting the function y=2xf(x)-3=0, we obtain the equation f(x) = frac{3}{2x}, thereby transforming the problem of finding the zeros of the function into finding the roots of the equation, which is further transformed into a problem of finding the intersection points of two functions, and then the answer is obtained by discussing each interval separately. This problem examines the relationship between the zeros of a function and the roots of an equation, as well as the application of the intersection points of functions, reflecting the mathematical transformation thought method and the mathematical thought method of classified discussion, making it a challenging problem.