answer:To prove the equivalence of the two conditions, we will show that each condition implies the other. # Step 1: Prove (1) implies (2) 1. **Assume Condition (1):** There exists a polynomial ( P(x) ) with degree (leq frac{p-1}{2}) such that ( P(i) equiv a_i pmod{p} ) for all ( 1 leq i leq p ). 2. **Define ( Q(x) ):** Let ( Q(x) = P(x+d) - P(x) ). Since ( P(x) ) has degree (leq frac{p-1}{2}), ( Q(x) ) has degree (leq frac{p-3}{2} ). 3. **Square ( Q(x) ):** Consider ( Q(x)^2 ). The degree of ( Q(x)^2 ) is (leq p-3). 4. **Sum of ( Q(x)^2 ):** By Lemma 1(ii), since the degree of ( Q(x)^2 ) is less than ( p-1 ), the sum of ( Q(x)^2 ) over all ( x in mathbb{F}_p ) is zero: [ sum_{i=1}^p Q(i)^2 = 0. ] 5. **Substitute ( Q(i) ):** Since ( Q(i) = P(i+d) - P(i) equiv a_{i+d} - a_i pmod{p} ), we have: [ sum_{i=1}^p (a_{i+d} - a_i)^2 equiv 0 pmod{p}. ] Thus, Condition (1) implies Condition (2). # Step 2: Prove (2) implies (1) 1. **Assume Condition (2):** For any natural ( d leq frac{p-1}{2} ), [ sum_{i=1}^p (a_{i+d} - a_i)^2 equiv 0 pmod{p}. ] 2. **Define ( g(x) ):** Let ( g(x) = a_{x+1} - a_x ). By the given condition, we have: [ sum_{i=1}^p g(i)^2 equiv 0 pmod{p}. ] 3. **Finite Differences:** Define the finite difference operator (Delta) such that (Delta(f)(x) = f(x+1) - f(x)). For higher-order differences, (Delta^n(f)(x) = Delta(Delta^{n-1}(f))(x)). 4. **Degree Reduction:** By Lemma 2(i), (deg(Delta^i(f)) = deg(f) - i). If (deg(f) = m), then (deg(Delta^m(f)) = 0). 5. **Leading Coefficient:** By Lemma 2(ii), the leading coefficient of (Delta^i(f)) is ((m-i+1) cdots m). 6. **Sum of Differences:** By the given condition and Lemma 2(iv), there exists a polynomial ( f ) such that (Delta(f) = g) and (deg(f) = deg(g) + 1). 7. **Inductive Step:** By induction, we show that: [ sum_{i=1}^p g(i) i^j equiv 0 pmod{p} text{ for each } 1 leq j leq m. ] 8. **Contradiction:** If ( m geq frac{p-1}{2} ), then ( g(x)x^m ) will have a nonzero coefficient of the term ( x^{p-1} ), which contradicts the given condition. 9. **Conclusion:** Hence, ( m leq frac{p-3}{2} ). Therefore, there exists a polynomial ( f ) with degree at most (frac{p-1}{2}) such that (Delta(f) = g). Thus, ( f ) and the function ( h(i) = a_i ) differ by a constant. Therefore, Condition (2) implies Condition (1). (blacksquare)

answer:1. **Identify A and C**: Given A = 3B and C = B - 30. 2. **Express A in terms of C**: Substitute B = C + 30 into the equation for A: [ A = 3(C + 30) ] Simplify: [ A = 3C + 90 ] 3. **Calculate the percentage increase**: The percentage increase from C to A is: [ text{Percentage increase} = 100 left(frac{A - C}{C}right) ] Substitute A = 3C + 90 into the equation: [ text{Percentage increase} = 100 left(frac{3C + 90 - C}{C}right) ] Simplify: [ text{Percentage increase} = 100 left(frac{2C + 90}{C}right) ] Factor out terms: [ text{Percentage increase} = 200 + 100 left(frac{90}{C}right) ] Conclusion: Thus, the expression for the percentage increase is 200 + 100left(frac{90}{C}right), which requires knowing the value of C to compute exactly, but this form correctly expresses the percentage increase in terms of C. The answer is boxed as [ 200 + frac{9000{C}} ] The final answer is **A)** boxed{200 + frac{9000}{C}}

answer:To solve this problem, we need to first determine the number of 10-pound and 20-pound boxes in the original shipment. Let's denote the number of 10-pound boxes as x and the number of 20-pound boxes as y. We have two equations based on the given information: 1. The total number of boxes is 30: x + y = 30 2. The average weight of the boxes is 18 pounds, so the total weight of all boxes is 18 * 30 = 540 pounds: 10x + 20y = 540 Now we can solve these two equations to find the values of x and y. From equation 1, we can express y in terms of x: y = 30 - x Substitute y in equation 2 with the expression from equation 1: 10x + 20(30 - x) = 540 10x + 600 - 20x = 540 -10x = 540 - 600 -10x = -60 x = 6 Now that we have the value of x, we can find y: y = 30 - x y = 30 - 6 y = 24 So, there are 6 boxes that weigh 10 pounds and 24 boxes that weigh 20 pounds. To reduce the average weight of the boxes to a certain amount, we need to know the target average weight. Without that information, we cannot determine the exact number of boxed{20-pound} boxes that must be removed. Please provide the target average weight so we can continue with the calculation.

answer:1. **Calculate work time before lunch:** Maria starts at 8:00 A.M. and breaks for lunch at 1:00 P.M. The time worked before lunch is: [ 1:00 - 8:00 = 5 text{ hours} ] 2. **Account for the lunch break:** The lunch break lasts 1 hour. Therefore, she resumes work at: [ 1:00 + 1:00 = 2:00 text{ P.M.} ] 3. **Calculate remaining work hours:** Maria needs to work a total of 8 hours each day. She has already worked 5 hours before lunch. The remaining time needed is: [ 8 text{ hours} - 5 text{ hours} = 3 text{ hours} ] 4. **Determine the end of the workday:** Adding the remaining work time to the time she resumed after lunch gives: [ 2:00 text{ P.M.} + 3 text{ hours} = 5:00 text{ P.M.} ] Maria's workday ends at 5:00 text{ P.M.} The final answer is boxed{C) 5:00 P.M.}