answer:【Analysis】First satisfy the divisibility by 4 and 5, the unit digit is 0; then to be the smallest, the hundredth digit is 0, and then satisfy the divisibility by 3, the tenth digit is 2. This six-digit number is 325020.

answer:5 . a=3, b=-5.Given that x^{2}-x-1 can divide the polynomial a x^{5}+b x^{4}+1, there exists a polynomial f(x) such thata x^{5}+b x^{4}+1=left(x^{2}-x-1right) f(x) text {. }Let the two roots of the equation x^{2}-x-1=0 be x_{1} and x_{2}. Then we havebegin{array}{l}left{begin{array}{l}a x_{1}^{5}+b x_{1}^{4}+1=0, a x_{2}^{5}+b x_{2}^{4}+1=0 .end{array}right. because x_{1} x_{2}=-1 text {, } thereforeleft{begin{array}{l}a x_{1}^{5}+b x_{1}^{4}+left(x_{1} x_{2}right)^{4}=0, a x_{2}^{5}+b x_{2}^{4}+left(x_{1} x_{2}right)^{4}=0end{array}right. Rightarrowleft{begin{array}{l}a x_{1}+b+x_{2}^{4}=0, a x_{2}+b+x_{1}^{4}=0 .end{array}right. end{array}Subtracting, we get aleft(x_{1}-x_{2}right)=x_{1}^{4}-x_{2}^{4};Adding, we get aleft(x_{1}+x_{2}right)+2 b+x_{1}^{4}+x_{2}^{4}=0.text { Hence } begin{aligned}a & =left(x_{1}+x_{2}right)left(x_{1}^{2}+x_{2}^{2}right) & =left(x_{1}+x_{2}right)left[left(x_{1}+x_{2}right)^{2}-2 x_{1} x_{2}right] & =1 times[1-2 times(-1)]=3 .end{aligned}begin{array}{l}text { Also, } because x_{1}^{2}+x_{2}^{2}=left(x_{1}+x_{2}right)^{2}-2 x_{1} x_{2}=3, x_{1}^{4}+x_{2}^{4}=left(x_{1}^{2}+x_{2}^{2}right)^{2}-2left(x_{1} x_{2}right)^{2}=9-2=7 . therefore 3+2 b+7=0 . b=-5 .end{array}

answer:12. 2 cos n theta Let x=cos theta pm i sin theta, so x^{n}+frac{1}{x^{n}}=2 cos n theta.

answer:Consider triangles A C D and B C D.## SolutionUsing the Law of Cosines in triangles A C D and B C D, we find thatbegin{aligned}& A C^{2}=A D^{2}+C D^{2}-2 A D cdot C D cos 60^{circ}=100+16-40=76 & B D^{2}=B C^{2}+C D^{2}-2 B C cdot C D cos 120^{circ}=9+16+12=37end{aligned}Therefore, A C=2 sqrt{19}, B D=sqrt{37}.## Answer2 sqrt{19} ; sqrt{37}.

answer:I. solution. It is not difficult to determine the two numbers in question - let's denote these by x and y - and then the sum of their fourth powers. According to the problem,x+y=10, quad x y=4thus x and y are the two roots of thez^{2}-10 z+4=0quadratic equation, that is, +5+sqrt{21} and +5-sqrt{21} (we can consider either as x, the other as y). The sum of their fourth powers isbegin{aligned}& (5+sqrt{21})^{4}+(5-sqrt{21})^{4}=left((5+sqrt{21})^{2}right)^{2}+left((5-sqrt{21})^{2}right)^{2}= & =(46+10 sqrt{21})^{2}+(46-10 sqrt{21})^{2}=2left(46^{2}+10^{2} cdot 21right)=8432end{aligned}II. solution. The problem can also be solved more simply without calculating the two numbers. The sum of the fourth powers of two numbers can generally be expressed using the sum of the two numbers - let's denote it by p - and their product - let's denote it by q:begin{gathered}x^{4}+y^{4}=left(x^{2}+y^{2}right)^{2}-2 x^{2} y^{2}=left((x+y)^{2}-2 x yright)^{2}-2(x y)^{2}= =left(p^{2}-2 qright)^{2}-2 q^{2}=p^{4}-4 p^{2} q+2 q^{2}end{gathered}Substituting the data p=10 and q=4:x^{4}+y^{4}=8432Remarks. 1. The sum of the fourth powers of the two numbers can be written in several ways using their sum and product, for example,begin{aligned}(x+y)^{4} & =x^{4}+y^{4}+4 x^{3} y+4 x y^{3}+6 x^{2} y^{2}=x^{4}+y^{4}+4 x yleft(x^{2}+y^{2}right)+ & +6(x y)^{2}=x^{4}+y^{4}+4 x yleft((x+y)^{2}-2 x yright)+6(x y)^{2}end{aligned}From this, expressing x^{4}+y^{4} and rearranging, we again obtain the above expression.2. The polynomial x^{4}+y^{4} does not change if x and y are swapped, the same holds for the polynomials x+y and x y. Generally, a polynomial in the variables x_{1}, x_{2}, ldots, x_{n} is called a symmetric polynomial if the polynomial does not change (except possibly for the order of the terms) when the variables are replaced by the same variables in any order. If k leq n, the sum of all k-factor products that can be formed from the variables x_{1}, x_{2}, ldots, x_{n} is called the ( k-th degree) elementary symmetric polynomial of the variables. According to a famous theorem of algebra, every symmetric polynomial can be expressed as a polynomial in the elementary symmetric polynomials of its variables. This expression (in condensed form) is unique up to the order of the terms.This is the expression we produced in the II. solution, and it is not coincidental that the transformation in the 1st remark also leads to the same result.

answer:26. Let us first examine the case that all the inequalities in the problem are actually equalities. Then (a_{n-2}=a_{n-1}+a_{n}, a_{n-3}=2 a_{n-1}+a_{n}, ldots, a_{0} = F_{n} a_{n-1} + F_{n-1} a_{n} = 1), where (F_{n}) is the (n)th Fibonacci number. Then it is easy to see (from (F_{1} + F_{2} + cdots + F_{k} = F_{k+2})) that (a_{0} + cdots + a_{n} = (F_{n+2} - 1) a_{n-1} + F_{n+1} a_{n} = frac{F_{n+2} - 1}{F_{n}} + left(F_{n+1} - frac{F_{n-1} (F_{n+2} - 1)}{F_{n}}right) a_{n}). Since (frac{F_{n-1} (F_{n+2} - 1)}{F_{n}} leq F_{n+1}), it follows that (a_{0} + a_{1} + cdots + a_{n} geq frac{F_{n+2} - 1}{F_{n}}), with equality holding if and only if (a_{n} = 0) and (a_{n-1} = frac{1}{F_{n}}). We denote by (M_{n}) the required minimum in the general case. We shall prove by induction that (M_{n} = frac{F_{n+2} - 1}{F_{n}}). For (M_{1} = 1) and (M_{2} = 2) it is easy to show that the formula holds; hence the inductive basis is true. Suppose that (n > 2). The sequences (1, frac{a_{2}}{a_{1}}, ldots, frac{a_{n}}{a_{1}}) and (1, frac{a_{3}}{a_{2}}, ldots, frac{a_{n}}{a_{2}}) also satisfy the conditions of the problem. Hence we have [ a_{0} + cdots + a_{n} = a_{0} + a_{1} left(1 + frac{a_{2}}{a_{1}} + cdots + frac{a_{n}}{a_{1}}right) geq 1 + a_{1} M_{n-1} ]and [ a_{0} + cdots + a_{n} = a_{0} + a_{1} + a_{2} left(1 + frac{a_{3}}{a_{2}} + cdots + frac{a_{n}}{a_{2}}right) geq 1 + a_{1} + a_{2} M_{n-2} ]Multiplying the first inequality by (M_{n-2} - 1) and the second one by (M_{n-1}), adding the inequalities and using that (a_{1} + a_{2} geq 1), we obtain ((M_{n-1} + M_{n-2} + 1)(a_{0} + cdots + a_{n}) geq M_{n-1} M_{n-2} + M_{n-1} + M_{n-2} + 1), so [ M_{n} geq frac{M_{n-1} M_{n-2} + M_{n-1} + M_{n-2} + 1}{M_{n-1} + M_{n-2} + 1} ]Since (M_{n-1} = frac{F_{n+1} - 1}{F_{n-1}}) and (M_{n-2} = frac{F_{n} - 1}{F_{n-2}}), the above inequality easily yields (M_{n} geq frac{F_{n+2} - 1}{F_{n}}). However, we have shown above that equality can occur; hence (frac{F_{n+2} - 1}{F_{n}}) is indeed the required minimum.