answer:1. Consider two consecutive odd numbers. The sum of these numbers is always divisible by 4. For example, if the odd numbers are ( (2k-1) ) and ( (2k+1) ), then their sum is: [ (2k-1) + (2k+1) = 4k ] This shows that the sum of two consecutive odd numbers is divisible by 4. 2. Extending this argument, the sum of any even number of consecutive odd numbers is also divisible by 4. Let's denote these numbers as (a, a+2, a+4, ldots, a+2(n-1)). The sum is: [ sum_{i=0}^{n-1} (a + 2i) = an + 2sum_{i=0}^{n-1} i = an + 2 cdot frac{(n-1)n}{2} = an + (n-1)n ] Since (n) is even, let's write (n = 2m): [ an + n(n-1) = a cdot 2m + 2m(2m-1) = 2am + 4m^2 - 2m ] Clearly, this sum is divisible by 4. 3. Consider now sums of an odd number of consecutive odd integers. The sum (a, a+2, a+4, ldots, a+2(n-1)) where (n) is odd can be written as: [ sum_{i=0}^{n-1} (a + 2i) = an + 2sum_{i=0}^{n-1} i = an + 2 cdot frac{(n-1)n}{2} = an + (n-1)n ] For (n) odd, let's write (n = 2m + 1): [ an + n(n-1) = a cdot (2m+1) + (2m+1)(2m) = a(2m+1) + 4m^2 + 2m ] Here, (a(2m + 1)) is divisible by (2m + 1), and the sum (4m^2 + 2m) is even. 4. Therefore, all composite odd numbers are sums of odd prime factors and can form sums of odd consecutive numbers. 5. To prove sufficiency: - If (a) is a number divisible by 4, then: [ a = 4k = (2k-1) + (2k+1) ] - Odd composite numbers can be written as the product of two odd numbers (p) and (q): [ a = p cdot q ] If we consider (q) odd integers centered around (p), the sum evaluates to: [ sum_{i=0}^{q-1} left(p - (q-1) + 2iright) = pq ] 6. Given (a = x^2 - y^2 = (x-y)(x+y)), for non-adjacent squares, it is clear (1 < x-y < x+y) and both terms share the same parity. 7. If we rearrange and utilize same, (a) thus expressed can always be decomposed into non-adjacent squares differences. Thus, the numbers that can be written as the sum of two or more consecutive positive odd numbers are precisely the multiples of 4 and the odd composite numbers. [boxed{text{Multiples of 4 and odd composite numbers}}]

answer:1. **Identify the initial and final exchange rates:** According to the problem, the exchange rate of the dollar on January 1, 2014, was 32.6587 rubles, and on December 31, 2014, it was 56.2584 rubles. [ text{Initial rate} = 32.6587 , text{rubles} ] [ text{Final rate} = 56.2584 , text{rubles} ] 2. **Calculate the difference in exchange rates:** To find out how much the exchange rate changed, subtract the initial exchange rate from the final exchange rate: [ Delta text{rate} = text{Final rate} - text{Initial rate} ] [ Delta text{rate} = 56.2584 - 32.6587 ] 3. **Perform the subtraction:** [ 56.2584 - 32.6587 = 23.5997 ] 4. **Round the result to the nearest whole number:** According to the problem statement, the answer needs to be rounded to the nearest integer. The number 23.5997 rounds to 24. 5. **Present the final result:** [ boxed{24} ]

answer:The contrapositive of a proposition involves swapping the hypothesis and the conclusion and negating both. Therefore, the answer is: Given a, b, c, d in mathbb{R}, if a+c neq b+d, then a neq b or c neq d. The contrapositive of a proposition involves swapping the hypothesis and the conclusion and negating both, which leads to the answer. This question mainly tests the understanding of the four types of propositions, focusing on distinguishing between the hypothesis and the conclusion, and then applying the definition. Thus, the final answer is boxed{text{Given } a, b, c, d in mathbb{R}, text{ if } a+c neq b+d, text{ then } a neq b text{ or } c neq d.}

answer:1. **Identify right-angled triangles:** We start by noting that the triangles APH_A and AH_AB have right angles at P and H_A respectively since H_A is the foot of the altitude from A to BC and P is the projection of H_A onto AB. 2. **Determine equal angles:** Both triangles APH_A and AH_AB share the angle widehat{BAH_A} . Therefore, they are similar because they have a right angle and another angle in common. Hence, triangle APH_A sim triangle AH_AB. 3. **Establish similarity ratios:** Using the similarity of the triangles, we obtain the ratio of the corresponding sides: [ frac{AP}{AH_A} = frac{AH_A}{AB} ] 4. **Cross-multiply to find a relationship:** By cross-multiplying the terms, we get: [ AP cdot AB = AH_A^2 ] 5. **Repeat process for another side:** Similarly, consider the triangles AQH_A and AH_AC where Q is the projection of H_A on AC. These triangles are also similar, triangle AQH_A sim triangle AH_AC, which leads us to: [ frac{AQ}{AH_A} = frac{AH_A}{AC} ] and by cross-multiplying, [ AQ cdot AC = AH_A^2 ] 6. **Equate the expressions obtained:** From the two equalities derived: [ AP cdot AB = AH_A^2 quad text{and} quad AQ cdot AC = AH_A^2 ] we see that: [ AP cdot AB = AQ cdot AC ] 7. **Conclude by Power of a Point theorem:** By the Power of a Point theorem, this equation asserts that point A has the same power relative to a circle passing through points B, P, Q, and C. This implies that B, P, Q, and C are concyclic, i.e., they all lie on the same circle. # Conclusion: Thus, the points B, P, Q, and C are indeed on the same circle. [ boxed{} ]