answer:# Problem: **a)** In a three-digit number, the first digit from the left was struck out, then the resulting two-digit number was multiplied by 7, and the original three-digit number was obtained. Find such a number. **b)** In a three-digit number, the middle digit was struck out, and the resulting number was 6 times less than the original. Find such a three-digit number. Part (a) Let's denote the digits of the three-digit number as (x, y, z). Therefore, the number can be represented as (100x + 10y + z). 1. When the first digit (x) is struck out, the resulting two-digit number is (10y + z). 2. According to the given condition: [ 7(10y + z) = 100x + 10y + z ] 3. Rearrange and simplify this equation: [ 100x + 10y + z = 7(10y + z) ] [ 100x + 10y + z = 70y + 7z ] [ 100x = 70y + 7z - 10y - z ] [ 100x = 60y + 6z ] [ 50x = 3(10y + z) ] 4. Observe that (x) must divide 3, so possible values are (x = 3) (since (x) being 0 is impossible for a 3-digit number). 5. Substitute (x = 3) into the equation: [ 50 times 3 = 3(10y + z) ] [ 150 = 30y + 3z ] [ 50 = 10y + z ] 6. Solve for (10y + z): [ 10y + z = 50 ] This gives us the two-digit number (50), therefore the original three-digit number is: [ 100 cdot 3 + 50 = 350 ] [ boxed{350} ] # Part (b): 1. Let's denote the digits of the three-digit number as (x, y, z). Thus, the number can be represented as (100x + 10y + z). 2. When the middle digit (y) is struck out, the resulting number is (10x + z). 3. According to the given condition: [ 100x + 10y + z = 6(10x + z) ] 4. Rearrange and simplify this equation: [ 100x + 10y + z = 60x + 6z ] [ 100x - 60x + 10y + z - 6z = 0 ] [ 40x + 10y - 5z = 0 ] [ 8x + 2y = z ] 5. Test values for (x, y, z) that satisfy this equation. Let (x = 1): [ 8(1) + 2y = z ] [ 8 + 2y = z ] 6. Set (y = 0): [ 2(0) = 0 quad text{then} quad z = 8 ] Thus, the satisfactory values are (x = 1), (y = 0), and (z = 8), giving us the three-digit number: [ 100 cdot 1 + 10 cdot 0 + 8 = 108 ] [ boxed{108} ]

answer:A: Since alpha parallel beta, and a subset beta, it follows that a parallel alpha. Therefore, option A is correct. B: Since b subset alpha, and a parallel b, if a subset alpha, then a cannot be parallel to alpha. Therefore, option B is incorrect. C: Since a parallel beta, and alpha parallel beta, if a subset alpha, then the conclusion does not hold. Therefore, option C is incorrect. D: Since b parallel alpha, and a parallel b, if a subset alpha, then the conclusion does not hold. Therefore, option D is incorrect. Therefore, the correct answer is boxed{text{A}}.

answer:Since the equation of the latus rectum of the parabola y^{2}=2px is x=-2, we have frac{p}{2}=2, which implies p=4. Therefore, the answer is boxed{4}. By using the equation of the latus rectum of the parabola y^{2}=2px, we can find the value of frac{p}{2} and subsequently solve for p. This problem tests the understanding of parabolic properties and basic calculation skills.

answer:Let's denote the common multiple of the ratio as x. So, the shares of p, q, and r will be 3x, 7x, and 12x respectively. According to the problem, the difference between the shares of p and q is Rs. 3200. So we can write: 7x - 3x = 3200 4x = 3200 x = 3200 / 4 x = 800 Now that we have the value of x, we can find the difference between q and r's share: 12x - 7x = 5x 5x = 5 * 800 5x = 4000 So, the difference between q and r's share is Rs. boxed{4000} .