answer:1. Let's define the function ( f(m, k) ) to represent the number of students who have scored ( k ) points after ( m ) rounds. The tournament conditions ensure that students with different scores only meet if the difference is 1 point. 2. Notice that if ( f(m, k) ) is even for any ( k in {0, ldots, m} ), the participants with different scores will not meet in the ((m + 1))-th round (otherwise, there would be at least two such encounters). Additionally, if a student with ( n ) points meets one with ( n + 1 ) points, then their resulting scores will be ( n ) and ( n + 2 ) respectively. 3. First, we need to prove two claims: 1. **Claim 1**: In a tournament with ( 2^n ) participants, after ( m ) rounds, the number of players with ( k ) points is given by: [ f(m, k) = 2^{n-m} cdot binom{m}{k} ] where ( m leq n ) and ( k in {0, ldots, m} ). Use induction on ( m ): - **Base Case**: If ( m = 0 ), then ( k = 0 ), and ( f(0, 0) = 2^n ). - **Inductive Step**: Assume for some ( m < n ) that the required equality holds. In each group of participants with the same number of points, there are even numbers of participants, so they can be paired off evenly. After ((m+1)) rounds, groups of sportsmen will halve in size, leading to: [ f(m + 1, 0) = frac{1}{2} f(m, 0) = frac{1}{2} cdot 2^{n - m} = 2^{n - (m + 1)} cdot binom{m + 1}{0} ] For ( k in {1, ldots, m} ): [ f(m + 1, k) = frac{1}{2} cdot f(m, k) + frac{1}{2} cdot f(m, k - 1) = 2^{n - m - 1} left( binom{m}{k} + binom{m}{k - 1} right) = 2^{n - (m + 1)} cdot binom{m + 1}{k} ] This completes the induction. 2. **Claim 2**: In a tournament with ( 2^n - 2 ) participants and fewer than ( n ) rounds, the number ( f(2m, k) ) is even for any ( k ) from 0 to ( n ); ( f(2m-1, k) ) is odd only for ( k = m - 1 ) or ( k = m ). Use induction on ( n ): - **Base Case**: When ( n = 2 ), it's evident. - **Inductive Step**: Assume for some ( n ), the statement holds. In a tournament with ( 2^{n + 1} ) participants, after the first round, groups of ( 2^n - 1 ) participants will have scored either 0 or 1 points. Applying the first claim shows how to proceed. 4. Introduce two fictitious participants ( A ) and ( B ). Assume they play each other in an odd round, with ( A ) winning and ( B ) losing. This does not affect other participants. In any even ( 2m )-th round, ( A ) and ( B ) maintain scores as per defined pairing rules, which doesn't affect the overall distribution. 5. Apply the established function for ( n = 8 ). The tournament will end after 8 rounds, and the number of participants attaining exactly 5 points is: [ binom{8}{5} = frac{8 cdot 7 cdot 6}{3 cdot 2 cdot 1} = 56 ] It is crucial to note that the fictitious participants ( A ) and ( B ) accrue 4 points each, hence not included. # Conclusion: [ boxed{56} ]

answer:Blockchain is a publicly shared online record that encrypts data to guarantee transaction security and thwart fraudulent activity.

answer:To find the length of the line segment from (0,0) to (1,1) in the given non-standard metric, we'll employ the tools of Riemannian geometry, particularly the metric tensor. The given metric tensor is: g = [1 1] [1 1] The expression for the length, ds, of an infinitesimal segment in this space is given by the quadratic form associated with the metric tensor: ds^2 = g_ij dx^i dx^j = g_11 dx^1 dx^1 + g_12 dx^1 dx^2 + g_21 dx^2 dx^1 + g_22 dx^2 dx^2 = (dx^1)^2 + 2 dx^1 dx^2 + (dx^2)^2 where we've used the given metric tensor components and the fact that g_12 = g_21. For the line segment from (0,0) to (1,1), we can parameterize the curve by letting x^1 = t and x^2 = t, with t varying from 0 to 1. Then, dx^1 = dt and dx^2 = dt. Plugging these into the expression for ds^2, we have: ds^2 = (dt)^2 + 2(dt)(dt) + (dt)^2 = 4(dt)^2 Taking the square root of both sides, we get: ds = 2 dt To find the total length of the line segment, we integrate this expression from t = 0 to t = 1: Length = ∫[0,1] 2 dt = 2 ∫[0,1] dt = 2 [t] from 0 to 1 = 2 So, the length of the line segment from (0,0) to (1,1) in this non-standard metric is 2. It's worth noting that, in the standard Euclidean metric, this line segment would have length √2. The difference here highlights how the non-standard metric changes our notion of distance and geometry in this space.

answer:To find the angle between the line segment from (0,0) to (1,1) and the x-axis in this non-standard metric, we can use the fact that the inner product (or dot product) between two vectors is related to the metric tensor. Specifically, the inner product of two vectors u and v is given by: u · v = g_ij u^i v^j Let's consider the vector u = (1, 0) along the x-axis and the vector v = (1, 1) along the line segment. We want to find the angle θ between these two vectors. First, we compute the inner product u · v using the metric tensor: u · v = g_ij u^i v^j = g_11 u^1 v^1 + g_12 u^1 v^2 + g_21 u^2 v^1 + g_22 u^2 v^2 = (1)(1) + (1)(1) + (0) + (0) = 2 Next, we compute the norms (or lengths) of the vectors u and v: ||u||^2 = g_ij u^i u^j = g_11 u^1 u^1 + g_12 u^1 u^2 + g_21 u^2 u^1 + g_22 u^2 u^2 = (1)(1) + 0 + 0 + 0 = 1 ||v||^2 = g_ij v^i v^j = g_11 v^1 v^1 + g_12 v^1 v^2 + g_21 v^2 v^1 + g_22 v^2 v^2 = (1)(1) + 2(1)(1) + (1)(1) = 4 Now, we can use the formula for the cosine of the angle between two vectors: cos(θ) = (u · v) / (||u|| ||v||) Plugging in the values, we get: cos(θ) = 2 / (√1 √4) = 2 / 2 = 1 Since cos(θ) = 1, we have θ = 0. However, this result might seem counterintuitive, as the line segment from (0,0) to (1,1) is not along the x-axis. The key insight here is that the non-standard metric changes our notion of angles and orthogonality. In this metric, the vectors u = (1, 0) and v = (1, 1) are indeed parallel in the sense that their inner product is maximal, which is reflected in the cosine of the angle between them being 1. This highlights the importance of considering the specific metric when working with geometric concepts like angles and lengths.