answer:According to the question, each of the three positions (hundreds, tens, and ones) has 4 choices, since we can use any of the digits 1, 2, 3, and 4 without restrictions.- For the hundreds place, we can choose any one of the 4 digits.- Similarly, the tens place can also be filled with any one of the 4 digits.- The ones place also has 4 options to choose from.Now, we will apply the multiplication principle of combinatorics, which states that if one event can occur in `m` ways and a second can occur independently of the first in `n` ways, then the number of ways the two events can occur in sequence is `m × n`.Thus, the total number of three-digit numbers that can be formed is given by the product of the choices for each of the three places:4 times 4 times 4 = 64Therefore, there are boxed{64} three-digit numbers that can be composed using the digits 1, 2, 3, and 4.

answer:■ Answer. All N such that t^{2} leq N<t^{2}+t for some positive integer t.ब Solution 1 (local). If t^{2} leq N<t^{2}+t then the sequence s_{n}=t n+1 satisfies both conditions. It remains to show that no other values of N work.Define a_{n}:=s_{n}-s_{n-1}, and let p be the minimal period of left{a_{n}right}. For each k in mathbb{Z}_{geq 0}, let f(k) be the integer such thats_{f(k)}-s_{k} leq N<s_{f(k)+1}-s_{k} .Note that fleft(s_{n-1}right)=s_{n} for all n. Since left{a_{n}right} is periodic with period p, f(k+p)=f(k)+p for all k, so k mapsto f(k)-k is periodic with period p. We also note that f is nondecreasing: if k<k^{prime} but fleft(k^{prime}right)<f(k) thenN<s_{fleft(k^{prime}right)+1}-s_{k^{prime}}<s_{f(k)}-s_{k} leq N,which is absurd. We now claim thatmax _{k}(f(k)-k)<p+min _{k}(f(k)-k) .Indeed, if fleft(k^{prime}right)-k^{prime} geq p+f(k)-k then we can shift k and k^{prime} so that 0 leq k-k^{prime}<p, and it follows that k leq k^{prime} leq fleft(k^{prime}right)<f(k), violating the fact that f is nondecreasing. Therefore max _{k}(f(k)-k)<p+min _{k}(f(k)-k), so f(k)-k is uniquely determined by its value modulo p. In particular, since a_{n}=fleft(s_{n-1}right)-s_{n-1}, a_{n} is also uniquely determined by its value modulo p, so left{a_{n} bmod pright} also has minimal period p.Now work in mathbb{Z} / p mathbb{Z} and consider the sequence s_{0}, fleft(s_{0}right), fleft(fleft(s_{0}right)right), ldots. This sequence must be eventually periodic; suppose it has minimal period p^{prime}, which must be at most p. Then, sincef^{n}left(s_{0}right)-f^{n-1}left(s_{0}right)=s_{n}-s_{n-1}=a_{n},and left{a_{n} bmod pright} has minimal period p, we must have p^{prime}=p. Therefore the directed graph G on mathbb{Z} / p mathbb{Z} given by the edges k rightarrow f(k) is simply a p-cycle, which implies that the map k mapsto f(k) is a bijection on mathbb{Z} / p mathbb{Z}. Therefore, f(k+1) neq f(k) for all k (unless p=1, but in this case the following holds anyways), hencef(k)<f(k+1)<cdots<f(k+p)=f(k)+p .This implies f(k+1)=f(k)+1 for all k, so f(k)-k is constant, therefore a_{n}= fleft(s_{n-1}right)-s_{n-1} is also constant. Let a_{n} equiv t. It follows that t^{2} leq N<t^{2}+t as we wanted.『 Solution 2 (global). Define left{a_{n}right} and f as in the previous solution. We first show that s_{i} not equiv s_{j}(bmod p) for all i<j<i+p. Suppose the contrary, i.e. that s_{i} equiv s_{j} (bmod p) for some i, j with i<j<i+p. Then a_{s_{i}+k}=a_{s_{j}+k} for all k geq 0, therefore s_{s_{i}+k}-s_{s_{i}}=s_{s_{j}+k}-s_{s_{j}} for all k geq 0, thereforea_{i+1}=fleft(s_{i}right)-s_{i}=fleft(s_{j}right)-s_{j}=a_{j+1} quad text { and } quad s_{i+1}=fleft(s_{i}right) equiv fleft(s_{j}right)=s_{j+1} quad(bmod p) .Continuing this inductively, we obtain a_{i+k}=a_{j+k} for all k, so left{a_{n}right} has period j-i<p, which is a contradiction. Therefore s_{i} not equiv s_{j}(bmod p) for all i<j<i+p, and this implies that left{s_{i}, ldots, s_{i+p-1}right} forms a complete residue system modulo p for all i. Consequently we must have s_{i+p} equiv s_{i}(bmod p) for all i.Let T=s_{p}-s_{0}=a_{1}+cdots+a_{p}. Since left{a_{n}right} is periodic with period p, and {i+1, ldots, i+k p} contains exactly k values of each residue class modulo p,s_{i+k p}-s_{i}=a_{i+1}+cdots+a_{i+k p}=k Tfor all i, k. Since p mid T, it follows that s_{s_{p}}-s_{s_{0}}=frac{T}{p} cdot T=frac{T^{2}}{p}. Summing up the inequalitiess_{s_{n}}-s_{s_{n-1}} leq N<s_{s_{n}+1}-s_{s_{n-1}}=s_{s_{n}}-s_{s_{n-1}}+a_{s_{n}+1}for n in{1, ldots, p} then impliesfrac{T^{2}}{p}=s_{s_{p}}-s_{s_{0}} leq N p<frac{T^{2}}{p}+a_{s_{1}+1}+a_{s_{2}+1}+cdots+a_{s_{p}+1}=frac{T^{2}}{p}+T,where the last equality holds because left{s_{1}+1, ldots, s_{p}+1right} is a complete residue system modulo p. Dividing this by p yields t^{2} leq N<t^{2}+t for t:=frac{T}{p} in mathbb{Z}^{+}.Remark (Author comments). There are some similarities between this problem and IMO 2009/3, mainly that they both involve terms of the form s_{s_{n}} and s_{s_{n}+1} and the sequence s_{0}, s_{1}, ldots turns out to be an arithmetic progression. Other than this, I don't think knowing about IMO 2009/3 will be that useful on this problem, since in this problem the fact that left{s_{n+1}-s_{n}right} is periodic is fundamentally important.The motivation for this problem comes from the following scenario: assume we have boxes that can hold some things of total size leq N, and a sequence of things of size a_{1}, a_{2}, ldots (where a_{i}:=s_{i+1}-s_{i} ). We then greedily pack the things in a sequence of boxes, 'closing' each box when it cannot fit the next thing. The number of things we put in each box gives a sequence b_{1}, b_{2}, ldots. This problem asks when we can have left{a_{n}right}=left{b_{n}right}, assuming that we start with a sequence left{a_{n}right} that is periodic.(Extra motivation: I first thought about this scenario when I was pasting some text repeatedly into the Notes app and noticed that the word at the end of lines are also (eventually) periodic.)

answer:AnalysisThis question examines the truth of propositions and the determination of the truth of compound propositions formed by simple logical connectives. According to the problem, p is true and q is false, thus leading to the results of each option.SolutionSince the line 6x+2y-1=0 is parallel to the line y=5-3x,Therefore, proposition p is a true proposition,Therefore, neg p is a false proposition,Since the line 6x+2y-1=0 is not perpendicular to the line 2x+6y-4=0,Therefore, proposition q is a false proposition,Therefore, (neg p) wedge q is a false proposition.Hence, the correct choice is boxed{text{B}}.

answer: Step-by-Step Solution# Part 1: Finding the Analytical Expression of the Quadratic FunctionGiven the quadratic function y=x^{2}+bx+c.- Condition 1: The graph intersects the y-axis at point A(0,-3). This gives us the equation 0^2 + b cdot 0 + c = -3, leading to c = -3.- Condition 2: The sum of the squares of the x-intercepts is 15. If the x-intercepts are x_1 and x_2, then x_1^2 + x_2^2 = 15. We know that (x_1 + x_2)^2 = x_1^2 + x_2^2 + 2x_1x_2. Given that x_1x_2 = c = -3, we can substitute to obtain: [ (x_1 + x_2)^2 = 15 + 2(-3) = 9 ] This simplifies to x_1 + x_2 = pm 3. Since x_1 + x_2 = -b, we find b = mp 3.- Quadratic Function: Substituting b and c into the original equation, the quadratic function becomes y = x^2 pm 3x - 3. So, the analytical expression of the quadratic function is boxed{y = x^2 pm 3x - 3}.# Part 2: Solution Set of the Quadratic InequalityGiven b 1}. - When 0 1, and the solution set is (1, frac{3}{a}). - When a = 3, there is no solution since frac{3}{a} = 1. - When a > 3, the root frac{3}{a} 1}}.- For 0 3: boxed{(frac{3}{a},1)}.

answer:1. A Prompt: Left side =sin A cos A+sin A cos B+cos A sin B+sin B cos Bbegin{array}{l}=frac{1}{2}(sin 2 A+sin 2 B)+sin (A+B) =sin (A+B) cos (A-B)+sin (A+B)end{array}Right side =2 sin left[180^{circ}-(A+B)right]=2 sin (A+B)Thus, sin (A+B) cos (A-B)+sin (A+B)=2 sin (A+B)which simplifies to sin (A+B)[cos (A-B)-1]=0Since sin (A+B)>0it follows that cos (A-B)=1, so A=BTaking A=B=30^{circ}, C=120^{circ} and substituting into the condition, we find that it satisfies the condition.Therefore, triangle A B C is an isosceles triangle, but not necessarily a right triangle.

answer:Given that U = {x in mathbb{N} | 0 < x leq 6} = {1, 2, 3, 4, 5, 6}, and N = {2, 3, 4},Therefore, the complement of N in U, denoted as mathbb{C}_U N, is {1, 5, 6}. Also, given that M = {1, 4, 5},Then, M cap (mathbb{C}_U N) = {1, 5}.Hence, the correct option is boxed{text{B}}.