answer:1. **Set the problem constraints**: Distribute 8 distinct pieces of candy among three bags (red, blue, white), with red and blue bags needing at least two pieces each. 2. **Setup the distribution**: Let r be the number of candies in the red bag where 2 leq r leq 6. The remaining candies, 8-r, are to be distributed between the blue and white bags. 3. **Count ways to distribute candies to the red bag**: For each value of r, choose r candies for the red bag in binom{8}{r} ways. 4. **Distribute the remaining candies**: The remaining 8-r candies to be distributed between the blue and white bags. Since the blue bag needs at least two candies, distribute 8-r candies ensuring at least 2 go to the blue bag. 5. **Calculate total arrangements**: [ sum_{r=2}^{6} binom{8}{r} cdot 2^{8-r} ] 6. **Calculate the sum**: [ sum_{r=2}^{6} binom{8}{r} cdot 2^{8-r} = 2^8 - binom{8}{1} cdot 2^7 + binom{8}{0} cdot 2^8 - 2 = 256 - 128 + 1 - 2 = 127 ] 7. **Conclusion**: The number of valid distributions of candies is 127. The final answer is boxed{C}

answer:**(a)** To show that such functions ( f(n, m) ) exist, consider the following function: [ f(n, m) = n quad text{for all} ; (n, m) in mathbb{Z} times mathbb{Z} ] We need to verify that this function satisfies the given identity: [ f(n, m) equiv frac{1}{4} left( f(n-1, m) + f(n+1, m) + f(n, m-1) + f(n, m+1) right) ] Substituting ( f(n, m) = n ) into both sides of the identity, we get: [ n = frac{1}{4} left( (n-1) + (n+1) + n + n right) ] Simplifying the right-hand side: [ frac{1}{4} left( (n-1) + (n+1) + n + n right) = frac{1}{4} left( 4n right) = n ] Thus, the function ( f(n, m) = n ) satisfies the given identity. Hence, such functions exist. **(b)** We need to prove that for any integer ( k in mathbb{Z} ), each such function ( f(n, m) ) can take values both greater than and less than ( k ). Proof by contradiction: - Assume the statement is false. That is, there exists some integer ( k ) such that all values of a function ( f(n, m) ) satisfying the condition do not exceed ( k ). Let ( l ) be the largest value of ( f(n, m) ): [ l = max{ f(n, m) ; | ; (n, m) in mathbb{Z} } ] Suppose ( l = f(n_0, m_0) ) for some particular pair ( (n_0, m_0) ). For ( f(n, m) ) to satisfy the given identity at ( (n_0, m_0) ), we must have: [ f(n_0, m_0) = frac{1}{4} left( f(n_0-1, m_0) + f(n_0+1, m_0) + f(n_0, m_0-1) + f(n_0, m_0+1) right) ] Since ( l ) is the maximum value, we have: [ f(n_0-1, m_0) leq l, quad f(n_0+1, m_0) leq l, quad f(n_0, m_0-1) leq l, quad f(n_0, m_0+1) leq l ] Thus, [ l geq frac{1}{4} left( f(n_0-1, m_0) + f(n_0+1, m_0) + f(n_0, m_0-1) + f(n_0, m_0+1) right) Rightarrow l geq l ] Equality must hold, implying that: [ f(n_0-1, m_0) = l quad text{and} quad f(n_0+1, m_0) = l quad text{and} quad f(n_0, m_0-1) = l quad text{and} quad f(n_0, m_0+1) = l ] By induction, we can extend this equality to infer: [ f(n, m) = l quad text{for all} ; (n, m) in mathbb{Z} times mathbb{Z} ] This would imply that ( f(n, m) ) is a constant function, which contradicts the initial assumption that ( f(n, m) ) is non-constant. Thus, ( f(n, m) ) must take values both greater than and less than any integer ( k ). The conclusion follows: (blacksquare)

answer:Let's compute the angle between the vectors overrightarrow{a} and overrightarrow{b} step by step. 1. First, we find the magnitudes of the vectors: The magnitude of overrightarrow{a} is |overrightarrow{a}| = sqrt{1^2 + 1^2} = sqrt{2}. The magnitude of overrightarrow{b} is |overrightarrow{b}| = sqrt{2^2 + 0^2} = 2. 2. The dot product of overrightarrow{a} and overrightarrow{b} is overrightarrow{a} cdot overrightarrow{b} = 1 cdot 2 + 1 cdot 0 = 2. 3. Using the formula for the dot product, overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}| |overrightarrow{b}| coslangle overrightarrow{a}, overrightarrow{b} rangle, we can find the cosine of the angle between the vectors: 2 = sqrt{2} cdot 2 cdot coslangle overrightarrow{a}, overrightarrow{b} rangle. 4. Solving for coslangle overrightarrow{a}, overrightarrow{b} rangle gives us coslangle overrightarrow{a}, overrightarrow{b} rangle = frac{2}{sqrt{2} cdot 2} = frac{sqrt{2}}{2}. 5. Given that the cosine of the angle equals frac{sqrt{2}}{2}, we can determine the angle itself. Since this value corresponds to the standard 45-degree angle or frac{pi}{4} radians, we conclude that the angle between the vectors is langle overrightarrow{a}, overrightarrow{b} rangle = boxed{frac{pi}{4}}.

answer:Let's denote the actual distance traveled by the person as ( D ) km. If the person walks at 10 km/hr, the time taken to travel the actual distance ( D ) would be ( frac{D}{10} ) hours. If the person walks at 15 km/hr, the time taken to travel the actual distance ( D ) plus the additional 20 km would be ( frac{D + 20}{15} ) hours. Since the time taken to travel at both speeds is the same, we can set up the following equation: [ frac{D}{10} = frac{D + 20}{15} ] Now, we can solve for ( D ): [ 15D = 10(D + 20) ] [ 15D = 10D + 200 ] [ 15D - 10D = 200 ] [ 5D = 200 ] [ D = frac{200}{5} ] [ D = 40 ] So, the actual distance traveled by the person is boxed{40} km.