answer：First, I need to determine the value of one stamp. Since 6 stamps are worth 15, I can divide 15 by 6 to find the value per stamp.Next, with the value of one stamp known, I can multiply this value by the total number of stamps, which is 18, to find the total value of Ella's entire stamp collection.Finally, by performing these calculations, I can determine the total value of her stamp collection.

answer：First, I'll define the current ages of A and B as ( A ) and ( B ) respectively.From the problem, I know that in 10 years, A's age will be twice what B's age was 10 years ago. This gives me the equation:[ A + 10 = 2(B - 10) ]Additionally, I'm told that A is currently 8 years older than B, which provides the second equation:[ A = B + 8 ]Now, I'll substitute the expression for ( A ) from the second equation into the first equation:[ (B + 8) + 10 = 2(B - 10) ]Simplifying this equation:[ B + 18 = 2B - 20 ]Subtracting ( B ) from both sides:[ 18 = B - 20 ]Finally, adding 20 to both sides to solve for ( B ):[ B = 38 ]Therefore, B's current age is 38 years old.

answer：Alright, so I've got this problem about wheat harvesting in a township. It mentions two types of harvesters, type A and type B. Each type A harvester can harvest 80 mu per day, and each type B can harvest 120 mu per day. The township introduced a total of 20 harvesters, and together, they can harvest 2080 mu in one day. First, I need to figure out how many type A and type B harvesters they introduced. Okay, so let's break this down. If I let the number of type A harvesters be 'a' and the number of type B harvesters be 'b', then I can set up two equations based on the information given.The first equation is straightforward: the total number of harvesters is 20. So, a + b = 20.The second equation relates to the total amount of wheat harvested in a day, which is 2080 mu. Since each type A harvester does 80 mu and each type B does 120 mu, the total would be 80a + 120b = 2080.Now, I have a system of two equations:1. a + b = 202. 80a + 120b = 2080I need to solve this system to find the values of 'a' and 'b'. One way to solve this is by substitution or elimination. Maybe elimination is easier here. If I multiply the first equation by 80, I get:80a + 80b = 1600Now, subtract this from the second equation:(80a + 120b) - (80a + 80b) = 2080 - 1600Which simplifies to:40b = 480So, b = 480 / 40 = 12Now that I have b = 12, I can plug this back into the first equation:a + 12 = 20So, a = 8Alright, so there are 8 type A harvesters and 12 type B harvesters.Now, moving on to the second part. Due to weather changes, they need to harvest more than 2900 mu in one day. They plan to introduce 8 more harvesters of these two types. I need to find how many different introduction plans there are.Let me denote the number of additional type A harvesters as 'x'. Then, the number of additional type B harvesters would be '8 - x'.The total harvesting capacity with the additional harvesters should be greater than 2900 mu. So, the equation would be:80(8 + x) + 120(12 + (8 - x)) > 2900Let me simplify this:80(8 + x) = 640 + 80x120(20 - x) = 2400 - 120xAdding these together:640 + 80x + 2400 - 120x > 2900Combine like terms:3040 - 40x > 2900Subtract 3040 from both sides:-40x > -140Divide both sides by -40, remembering to flip the inequality sign:x < 3.5Since x has to be an integer (you can't have half a harvester), the possible values for x are 0, 1, 2, 3. But wait, if x is 0, that means introducing 0 type A and 8 type B. Is that allowed? The problem says "introduce 8 more harvesters of these two types," so I think x can be 0 as well.But let's check the total harvesting capacity for x = 0:80(8 + 0) + 120(12 + 8) = 640 + 2400 = 3040 > 2900, which is acceptable.Similarly, for x = 1:80(9) + 120(19) = 720 + 2280 = 3000 > 2900x = 2:80(10) + 120(18) = 800 + 2160 = 2960 > 2900x = 3:80(11) + 120(17) = 880 + 2040 = 2920 > 2900x = 4:80(12) + 120(16) = 960 + 1920 = 2880 < 2900, which doesn't satisfy the condition.So, x can be 0, 1, 2, or 3. That gives us four different introduction plans.Wait, but in the initial calculation, I got x < 3.5, which would mean x can be up to 3. So, x = 0, 1, 2, 3.But in the problem statement, it says "introduce 8 more harvesters of these two types." So, introducing 0 type A and 8 type B is acceptable, as well as introducing all 8 as type A. However, introducing all 8 as type A would mean x = 8, but that would make the total harvesting capacity:80(16) + 120(12) = 1280 + 1440 = 2720 < 2900, which doesn't meet the requirement.So, x can only be 0, 1, 2, or 3 to meet the requirement of harvesting more than 2900 mu.But wait, when x = 0, introducing 8 type B harvesters, the total becomes 3040, which is more than 2900.When x = 1, introducing 1 type A and 7 type B, total is 3000.x = 2: 2 type A and 6 type B, total 2960.x = 3: 3 type A and 5 type B, total 2920.x = 4: 4 type A and 4 type B, total 2880, which is less than 2900.So, only x = 0, 1, 2, 3 are valid.But in the initial inequality, x < 3.5, so x can be 0, 1, 2, 3.Therefore, there are four different introduction plans.But wait, in the first part, we had 8 type A and 12 type B. If we introduce x type A, the total type A becomes 8 + x, and type B becomes 12 + (8 - x) = 20 - x.So, the total number of type A and type B harvesters after introduction would be:Type A: 8 + xType B: 20 - xBut the total number of harvesters after introduction is 20 + 8 = 28.So, the possible values of x are 0, 1, 2, 3, 4, 5, 6, 7, 8.But only x = 0, 1, 2, 3 satisfy the harvesting capacity requirement.So, the different introduction plans are:1. x = 0: 0 type A, 8 type B2. x = 1: 1 type A, 7 type B3. x = 2: 2 type A, 6 type B4. x = 3: 3 type A, 5 type BSo, four different plans.But wait, the problem says "how many different introduction plans are there? Please list them all."So, the answer is four plans, with x = 0, 1, 2, 3.But in the initial calculation, I thought x < 3.5, so x can be up to 3, which is correct.Therefore, the number of different introduction plans is four, and they are:1. 0 type A and 8 type B2. 1 type A and 7 type B3. 2 type A and 6 type B4. 3 type A and 5 type BSo, that's the solution.

answer：Okay, so I have this problem about a rectangular box with prime number dimensions a, b, and c, where a ≤ b ≤ c. The volume and twice the surface area are numerically equal, and the sum a + b + c is a prime number. I need to find how many ordered triples (a, b, c) satisfy these conditions. The answer choices are from 0 to 4, so I need to figure out if there are none, one, two, three, or four such triples.First, let me write down the given conditions mathematically. The volume of the box is abc, and the surface area is 2(ab + bc + ac). The problem states that the volume is equal to twice the surface area, so:abc = 2(ab + bc + ac)Also, the sum a + b + c must be a prime number.Since a, b, and c are prime numbers and a ≤ b ≤ c, I can start by considering the smallest primes and see if they satisfy the conditions. The primes to consider are 2, 3, 5, 7, 11, etc.Let me first try to manipulate the equation abc = 2(ab + bc + ac). Maybe I can factor something out or simplify it.Dividing both sides by abc (assuming a, b, c are non-zero, which they are since they are primes):1 = 2(ab + bc + ac) / abcSimplify the right side:1 = 2(ab + bc + ac) / abc = 2(1/c + 1/a + 1/b)So,1 = 2(1/a + 1/b + 1/c)Therefore,1/2 = 1/a + 1/b + 1/cSo, the sum of the reciprocals of the primes a, b, c must be equal to 1/2.Since a ≤ b ≤ c, and all are primes, let's try small primes for a.Case 1: a = 2Then, 1/2 = 1/2 + 1/b + 1/cWait, that would mean 1/2 = 1/2 + 1/b + 1/c, which implies 1/b + 1/c = 0. But since b and c are positive primes, their reciprocals are positive, so this can't be. So, maybe I made a mistake in the algebra.Wait, let's go back.We had:abc = 2(ab + bc + ac)Divide both sides by abc:1 = 2(ab + bc + ac)/abc = 2(1/c + 1/a + 1/b)So,1 = 2(1/a + 1/b + 1/c)Therefore,1/2 = 1/a + 1/b + 1/cSo, that's correct.So, if a = 2, then:1/2 = 1/2 + 1/b + 1/cWhich implies 1/b + 1/c = 0, which is impossible because b and c are primes, so their reciprocals are positive. Therefore, a cannot be 2? Wait, that can't be right because 2 is the smallest prime.Wait, maybe I made a mistake in the algebra. Let me check again.Starting from abc = 2(ab + bc + ac)Divide both sides by abc:1 = 2(ab + bc + ac)/abc = 2(1/c + 1/a + 1/b)So, 1 = 2(1/a + 1/b + 1/c)So, 1/2 = 1/a + 1/b + 1/cSo, if a = 2, then 1/2 = 1/2 + 1/b + 1/cWhich implies 1/b + 1/c = 0, which is impossible. So, does that mean a cannot be 2? But 2 is the smallest prime, so if a can't be 2, then the next prime is 3.Case 2: a = 3Then, 1/2 = 1/3 + 1/b + 1/cSo, 1/b + 1/c = 1/2 - 1/3 = 1/6So, 1/b + 1/c = 1/6Since a ≤ b ≤ c, and a = 3, so b ≥ 3, c ≥ b.Let me denote b as the next prime after 3, which is 5.So, b = 5, then 1/c = 1/6 - 1/5 = (5 - 6)/30 = -1/30, which is negative. That's impossible because c is a positive prime.So, b can't be 5. Next, b = 7.1/c = 1/6 - 1/7 = (7 - 6)/42 = 1/42So, c = 42, but 42 is not a prime. So, that's invalid.Next, b = 11.1/c = 1/6 - 1/11 = (11 - 6)/66 = 5/66So, c = 66/5 = 13.2, which is not an integer, so invalid.Wait, maybe I need to find b and c such that 1/b + 1/c = 1/6.Let me write this as:1/c = 1/6 - 1/b = (b - 6)/(6b)So, c = 6b / (b - 6)Since c must be an integer, 6b must be divisible by (b - 6). Let me denote d = b - 6, so b = d + 6.Then, c = 6(d + 6)/d = 6 + 36/dSo, d must be a divisor of 36.Since b is a prime greater than or equal to a = 3, and b = d + 6, d must be a positive integer such that d + 6 is prime.Also, since c must be greater than or equal to b, c = 6 + 36/d ≥ b = d + 6So, 6 + 36/d ≥ d + 6Subtract 6 from both sides:36/d ≥ dMultiply both sides by d (positive):36 ≥ d^2So, d ≤ 6So, d is a positive divisor of 36, d ≤ 6.Divisors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.But d ≤ 6, so d can be 1, 2, 3, 4, 6.Now, let's check each d:d = 1:b = 1 + 6 = 7 (prime)c = 6 + 36/1 = 42 (not prime)So, invalid.d = 2:b = 2 + 6 = 8 (not prime)Invalid.d = 3:b = 3 + 6 = 9 (not prime)Invalid.d = 4:b = 4 + 6 = 10 (not prime)Invalid.d = 6:b = 6 + 6 = 12 (not prime)Invalid.So, none of these d values give b as a prime. Therefore, when a = 3, there are no solutions.Case 3: a = 5Then, 1/2 = 1/5 + 1/b + 1/cSo, 1/b + 1/c = 1/2 - 1/5 = 3/10So, 1/b + 1/c = 3/10Again, since a = 5, b ≥ 5, c ≥ b.Let me try b = 5:1/c = 3/10 - 1/5 = 3/10 - 2/10 = 1/10So, c = 10, which is not prime.Next, b = 7:1/c = 3/10 - 1/7 = (21 - 10)/70 = 11/70c = 70/11 ≈ 6.36, not integer.b = 11:1/c = 3/10 - 1/11 = (33 - 10)/110 = 23/110c = 110/23 ≈ 4.78, not integer.b = 13:1/c = 3/10 - 1/13 = (39 - 10)/130 = 29/130c = 130/29 ≈ 4.48, not integer.Wait, maybe I should approach this differently.Let me write 1/b + 1/c = 3/10So, 1/c = 3/10 - 1/b = (3b - 10)/(10b)So, c = 10b / (3b - 10)c must be an integer, so 10b must be divisible by (3b - 10). Let me denote k = 3b - 10, so 3b = k + 10, so b = (k + 10)/3Then, c = 10b / k = 10*(k + 10)/(3k) = (10k + 100)/(3k) = 10/3 + 100/(3k)Since c must be an integer, 100/(3k) must be a rational number, but since k divides 10b, and b is prime, k must be a factor of 10b.But this seems complicated. Maybe instead, let's try to find integer solutions for c.c = 10b / (3b - 10)We need 3b - 10 to divide 10b.Let me denote d = 3b - 10, so d divides 10b.But d = 3b - 10, so 3b = d + 10, so b = (d + 10)/3Since b must be an integer, d + 10 must be divisible by 3, so d ≡ -10 ≡ 2 mod 3.Also, since b ≥ 5, d = 3b - 10 ≥ 3*5 - 10 = 15 - 10 = 5.So, d ≥ 5, and d ≡ 2 mod 3.Also, c = 10b / d = 10*(d + 10)/(3d) must be an integer.So, 10*(d + 10) must be divisible by 3d.Let me write c = (10d + 100)/(3d) = 10/3 + 100/(3d)For c to be integer, 100/(3d) must be a rational number, but since d is an integer, 3d must divide 100.So, 3d divides 100, meaning d divides 100/3, but d must be an integer, so 3d divides 100.Therefore, d must be a divisor of 100/3, but since d is integer, 3d must divide 100.So, 3d | 100, meaning d must be a divisor of 100/3, but 100/3 is not integer, so this approach might not be helpful.Alternatively, let's consider that c must be an integer, so 10b must be divisible by (3b - 10). Let me try small primes for b starting from 5.b = 5:c = 10*5 / (15 - 10) = 50 / 5 = 10 (not prime)b = 7:c = 70 / (21 - 10) = 70 / 11 ≈ 6.36 (not integer)b = 11:c = 110 / (33 - 10) = 110 / 23 ≈ 4.78 (not integer)b = 13:c = 130 / (39 - 10) = 130 / 29 ≈ 4.48 (not integer)b = 17:c = 170 / (51 - 10) = 170 / 41 ≈ 4.15 (not integer)b = 19:c = 190 / (57 - 10) = 190 / 47 ≈ 4.04 (not integer)b = 23:c = 230 / (69 - 10) = 230 / 59 ≈ 3.89 (not integer)It seems like for a = 5, there are no solutions where c is a prime.Case 4: a = 7Then, 1/2 = 1/7 + 1/b + 1/cSo, 1/b + 1/c = 1/2 - 1/7 = 5/14So, 1/b + 1/c = 5/14Again, a = 7, so b ≥ 7, c ≥ b.Let me try b = 7:1/c = 5/14 - 1/7 = 5/14 - 2/14 = 3/14c = 14/3 ≈ 4.67 (not integer)b = 11:1/c = 5/14 - 1/11 = (55 - 14)/154 = 41/154c = 154/41 ≈ 3.756 (not integer)b = 13:1/c = 5/14 - 1/13 = (65 - 14)/182 = 51/182c = 182/51 ≈ 3.568 (not integer)b = 17:1/c = 5/14 - 1/17 = (85 - 14)/238 = 71/238c = 238/71 ≈ 3.352 (not integer)b = 19:1/c = 5/14 - 1/19 = (95 - 14)/266 = 81/266c = 266/81 ≈ 3.284 (not integer)This is getting too small, and c must be at least b, which is at least 7, so c is at least 7. But the values are getting smaller, so it's unlikely to find a solution here.Case 5: a = 11Then, 1/2 = 1/11 + 1/b + 1/cSo, 1/b + 1/c = 1/2 - 1/11 = 9/22So, 1/b + 1/c = 9/22With a = 11, b ≥ 11, c ≥ b.Let me try b = 11:1/c = 9/22 - 1/11 = 9/22 - 2/22 = 7/22c = 22/7 ≈ 3.14 (not integer)b = 13:1/c = 9/22 - 1/13 = (117 - 22)/286 = 95/286c = 286/95 ≈ 3.01 (not integer)b = 17:1/c = 9/22 - 1/17 = (153 - 22)/374 = 131/374c = 374/131 ≈ 2.855 (not integer)Again, c is getting smaller, which is not possible since c ≥ b ≥ 11.It seems like as a increases, the possible values for b and c become smaller, which contradicts the requirement that c ≥ b ≥ a.Therefore, it's likely that there are no solutions for a ≥ 2, 3, 5, 7, 11, etc.Wait, but earlier when a = 2, we had an issue where 1/b + 1/c = 0, which is impossible. So, maybe there's no solution where a is 2, 3, 5, 7, 11, etc.But let me double-check the initial equation.We had abc = 2(ab + bc + ac)Let me try small primes for a, b, c and see if any combination satisfies this.Let's try a = 2, b = 3, c = 5:Volume = 2*3*5 = 30Surface area = 2(2*3 + 3*5 + 2*5) = 2(6 + 15 + 10) = 2*31 = 62Twice the surface area would be 124, which is not equal to 30.So, that doesn't work.Next, a = 2, b = 3, c = 7:Volume = 2*3*7 = 42Surface area = 2(6 + 21 + 14) = 2*41 = 82Twice the surface area = 164 ≠ 42Nope.a = 2, b = 5, c = 5:Volume = 2*5*5 = 50Surface area = 2(10 + 25 + 10) = 2*45 = 90Twice surface area = 180 ≠ 50Nope.a = 2, b = 3, c = 11:Volume = 2*3*11 = 66Surface area = 2(6 + 33 + 22) = 2*61 = 122Twice surface area = 244 ≠ 66Nope.a = 2, b = 5, c = 7:Volume = 2*5*7 = 70Surface area = 2(10 + 35 + 14) = 2*59 = 118Twice surface area = 236 ≠ 70Nope.a = 2, b = 7, c = 7:Volume = 2*7*7 = 98Surface area = 2(14 + 49 + 14) = 2*77 = 154Twice surface area = 308 ≠ 98Nope.a = 3, b = 3, c = 3:Volume = 27Surface area = 2(9 + 9 + 9) = 54Twice surface area = 108 ≠ 27Nope.a = 3, b = 3, c = 5:Volume = 45Surface area = 2(9 + 15 + 15) = 2*39 = 78Twice surface area = 156 ≠ 45Nope.a = 3, b = 5, c = 5:Volume = 75Surface area = 2(15 + 25 + 15) = 2*55 = 110Twice surface area = 220 ≠ 75Nope.a = 3, b = 5, c = 7:Volume = 105Surface area = 2(15 + 35 + 21) = 2*71 = 142Twice surface area = 284 ≠ 105Nope.a = 5, b = 5, c = 5:Volume = 125Surface area = 2(25 + 25 + 25) = 150Twice surface area = 300 ≠ 125Nope.a = 5, b = 5, c = 7:Volume = 175Surface area = 2(25 + 35 + 35) = 2*95 = 190Twice surface area = 380 ≠ 175Nope.a = 5, b = 7, c = 7:Volume = 245Surface area = 2(35 + 49 + 35) = 2*119 = 238Twice surface area = 476 ≠ 245Nope.a = 7, b = 7, c = 7:Volume = 343Surface area = 2(49 + 49 + 49) = 294Twice surface area = 588 ≠ 343Nope.It seems like none of these combinations work. Maybe I need to try larger primes, but as I saw earlier, when a increases, the required c becomes smaller, which contradicts c ≥ b ≥ a.Wait, maybe I missed something in the equation manipulation.We had abc = 2(ab + bc + ac)Let me try to factor this differently.abc - 2ab - 2bc - 2ac = 0Factor:ab(c - 2) - 2c(b + a) = 0Hmm, not sure if that helps.Alternatively, let's factor out c:c(ab - 2a - 2b) = 2abSo,c = 2ab / (ab - 2a - 2b)Since c must be a prime, the denominator (ab - 2a - 2b) must divide 2ab.Let me write this as:c = 2ab / (ab - 2a - 2b) = 2ab / [a(b - 2) - 2b]Hmm, not sure.Alternatively, let's write it as:c = 2ab / (ab - 2a - 2b) = 2ab / [ab - 2(a + b)]Let me factor numerator and denominator:c = 2ab / [ab - 2(a + b)] = 2ab / [ab - 2a - 2b]Let me factor out a from the denominator:c = 2ab / [a(b - 2) - 2b]Hmm, still not helpful.Wait, maybe I can write it as:c = 2ab / (ab - 2a - 2b) = 2ab / [ab - 2(a + b)] = 2ab / [ab - 2a - 2b]Let me factor out b from the denominator:c = 2ab / [b(a - 2) - 2a]Hmm, not helpful.Alternatively, let me write it as:c = 2ab / (ab - 2a - 2b) = 2ab / [ab - 2(a + b)] = 2ab / [ab - 2a - 2b]Let me add and subtract 4 in the denominator:c = 2ab / [ab - 2a - 2b + 4 - 4] = 2ab / [(a - 2)(b - 2) - 4]Hmm, not sure.Alternatively, let me consider that c must be a prime, so the denominator must be a factor of 2ab.Let me denote d = ab - 2a - 2bSo, d divides 2ab, and c = 2ab / dSince c is prime, 2ab must be equal to d * c, where c is prime.Therefore, d must be either 1, 2, a, b, 2a, 2b, or ab, but considering that d = ab - 2a - 2b.Let me check possible values of d.Case 1: d = 1Then, ab - 2a - 2b = 1So, ab - 2a - 2b = 1Let me rearrange:ab - 2a - 2b = 1Add 4 to both sides:ab - 2a - 2b + 4 = 5Factor:(a - 2)(b - 2) = 5Since 5 is prime, the possible factor pairs are (1,5) and (5,1)So,a - 2 = 1, b - 2 = 5 => a = 3, b = 7Or,a - 2 = 5, b - 2 = 1 => a = 7, b = 3But since a ≤ b, a = 3, b = 7So, a = 3, b = 7Then, c = 2ab / d = 2*3*7 / 1 = 42But 42 is not a prime, so invalid.Case 2: d = 2Then, ab - 2a - 2b = 2So,ab - 2a - 2b = 2Add 4 to both sides:ab - 2a - 2b + 4 = 6Factor:(a - 2)(b - 2) = 6Possible factor pairs of 6: (1,6), (2,3), (3,2), (6,1)Since a ≤ b, consider:a - 2 = 1, b - 2 = 6 => a = 3, b = 8 (not prime)a - 2 = 2, b - 2 = 3 => a = 4 (not prime), b = 5a - 2 = 3, b - 2 = 2 => a = 5, b = 4 (not prime)a - 2 = 6, b - 2 = 1 => a = 8 (not prime), b = 3So, only possible prime b is when a = 4, b = 5, but a = 4 is not prime. So, no solution here.Case 3: d = aThen, ab - 2a - 2b = aSo,ab - 2a - 2b = aSubtract a:ab - 3a - 2b = 0Factor:a(b - 3) - 2b = 0a(b - 3) = 2bSo,a = 2b / (b - 3)Since a must be a prime, let's try b > 3 (since denominator must be positive)b = 5:a = 10 / 2 = 5 (prime)So, a = 5, b = 5Then, c = 2ab / d = 2*5*5 / 5 = 10But c = 10 is not prime.b = 7:a = 14 / 4 = 3.5 (not integer)b = 11:a = 22 / 8 = 2.75 (not integer)b = 13:a = 26 / 10 = 2.6 (not integer)So, only b = 5 gives a = 5, but c = 10 is not prime.Case 4: d = bThen, ab - 2a - 2b = bSo,ab - 2a - 2b = bSubtract b:ab - 2a - 3b = 0Factor:a(b - 2) - 3b = 0a(b - 2) = 3bSo,a = 3b / (b - 2)Since a must be a prime, let's try b > 2b = 3:a = 9 / 1 = 9 (not prime)b = 5:a = 15 / 3 = 5 (prime)So, a = 5, b = 5Then, c = 2ab / d = 2*5*5 / 5 = 10 (not prime)b = 7:a = 21 / 5 = 4.2 (not integer)b = 11:a = 33 / 9 = 3.666 (not integer)b = 13:a = 39 / 11 ≈ 3.545 (not integer)So, only b = 5 gives a = 5, but c = 10 is not prime.Case 5: d = 2aThen, ab - 2a - 2b = 2aSo,ab - 2a - 2b = 2aSubtract 2a:ab - 4a - 2b = 0Factor:a(b - 4) - 2b = 0a(b - 4) = 2bSo,a = 2b / (b - 4)Since a must be a prime, let's try b > 4b = 5:a = 10 / 1 = 10 (not prime)b = 7:a = 14 / 3 ≈ 4.666 (not integer)b = 11:a = 22 / 7 ≈ 3.142 (not integer)b = 13:a = 26 / 9 ≈ 2.888 (not integer)So, no solution here.Case 6: d = 2bThen, ab - 2a - 2b = 2bSo,ab - 2a - 2b = 2bSubtract 2b:ab - 2a - 4b = 0Factor:a(b - 2) - 4b = 0a(b - 2) = 4bSo,a = 4b / (b - 2)Since a must be a prime, let's try b > 2b = 3:a = 12 / 1 = 12 (not prime)b = 5:a = 20 / 3 ≈ 6.666 (not integer)b = 7:a = 28 / 5 = 5.6 (not integer)b = 11:a = 44 / 9 ≈ 4.888 (not integer)b = 13:a = 52 / 11 ≈ 4.727 (not integer)So, no solution here.Case 7: d = abThen, ab - 2a - 2b = abSubtract ab:-2a - 2b = 0 => a + b = 0, which is impossible since a and b are positive primes.So, no solution here.Therefore, the only possible case where d divides 2ab is when d = 1, which gave us a = 3, b = 7, c = 42, but c is not prime. All other cases either result in non-integer a or c, or c is not prime.Therefore, there are no ordered triples (a, b, c) that satisfy the given conditions.

answer：Okay, so I have this geometry problem here about a parallelogram ABCD. Inside triangle BCD, angle DCB is 75 degrees, and triangle BCD is isosceles with BC equal to CD. I need to find the measure of angle A. Hmm, let me try to visualize this.First, I remember that in a parallelogram, opposite sides are equal and opposite angles are equal. So, angle A should be equal to angle C, and angle B should be equal to angle D. Also, consecutive angles in a parallelogram add up to 180 degrees because they are supplementary.Now, focusing on triangle BCD. It's given that BC equals CD, so triangle BCD is isosceles with BC = CD. That means the angles opposite these sides should be equal. In triangle BCD, the sides BC and CD are equal, so the angles opposite them, which are angles at B and D, should be equal. Let me denote these angles as angle B and angle D in the triangle.Wait, but in the parallelogram, angle C is angle DCB, which is given as 75 degrees. So, in triangle BCD, angle C is 75 degrees, and the other two angles at B and D are equal. Let me call these angles x each. So, in triangle BCD, the sum of angles should be 180 degrees. Therefore, I can write the equation:75 degrees + x + x = 180 degreesSimplifying that, I get:2x = 180 - 752x = 105x = 52.5 degreesSo, angles at B and D in triangle BCD are each 52.5 degrees. But wait, in the parallelogram ABCD, angle B is not just the angle in triangle BCD. Because in the parallelogram, angle B is actually the angle at vertex B, which is adjacent to side BC and AB. Similarly, angle D is at vertex D, adjacent to side CD and AD.Wait, maybe I'm confusing the angles here. Let me clarify. In the parallelogram, angle at vertex C is angle DCB, which is 75 degrees. Since opposite angles in a parallelogram are equal, angle A should also be 75 degrees. But hold on, is that correct?Wait, no. Because in the parallelogram, angle A is opposite to angle C, so angle A equals angle C. But angle C is given as 75 degrees, so angle A should also be 75 degrees. But wait, that seems too straightforward. Let me double-check.In triangle BCD, angle C is 75 degrees, and angles at B and D are each 52.5 degrees. But in the parallelogram, angle B is equal to angle D, right? Because in a parallelogram, opposite angles are equal. So, angle B equals angle D, and angle A equals angle C.But in triangle BCD, angles at B and D are 52.5 degrees each. Does that mean that in the parallelogram, angles at B and D are 52.5 degrees? Wait, that can't be because in a parallelogram, consecutive angles are supplementary. So, if angle C is 75 degrees, then angle B should be 180 - 75 = 105 degrees. But according to triangle BCD, angle B is 52.5 degrees. That's a contradiction.Hmm, so I must have made a mistake somewhere. Let me go back. Maybe I misapplied the properties of the parallelogram. Let me draw a diagram in my mind. Parallelogram ABCD, with AB parallel to CD and AD parallel to BC. So, angle at C is angle DCB, which is 75 degrees. Triangle BCD is inside the parallelogram, with BC = CD.Wait, so triangle BCD is formed by points B, C, and D. In the parallelogram, BC is equal to AD, and CD is equal to AB. But it's given that BC = CD, so that means AD = AB. So, sides AD and AB are equal, making triangle ABD isosceles? Wait, no, triangle ABD is not necessarily a triangle unless we connect the diagonals.Wait, maybe I should consider the diagonals. In a parallelogram, diagonals bisect each other. But I'm not sure if that helps here. Let me think again about triangle BCD.In triangle BCD, sides BC and CD are equal, so it's an isosceles triangle with BC = CD. Therefore, angles opposite these sides are equal. So, angle at B (which is angle CBD) and angle at D (which is angle CDB) are equal. Let me denote these angles as x each.Given that angle at C is 75 degrees, the sum of angles in triangle BCD is:75 + x + x = 1802x = 105x = 52.5 degreesSo, angles at B and D in triangle BCD are each 52.5 degrees. Now, in the parallelogram, angle at B is angle ABC, which is adjacent to side AB and BC. Similarly, angle at D is angle ADC, adjacent to side AD and DC.Wait, so in the parallelogram, angle at B is angle ABC, which is equal to angle ADC (angle D). But in triangle BCD, angle at B is angle CBD, which is 52.5 degrees. Is angle ABC equal to angle CBD?Wait, no. Because angle ABC is the angle at vertex B in the parallelogram, which is adjacent to sides AB and BC. Whereas angle CBD is the angle at vertex B in triangle BCD, which is adjacent to sides CB and BD (the diagonal). So, they are different angles.Therefore, I cannot directly say that angle ABC is equal to angle CBD. Hmm, this complicates things. Maybe I need to find the measure of angle ABC using the information from triangle BCD.Let me denote the diagonal BD. In triangle BCD, we have sides BC = CD, and angles at B and D are each 52.5 degrees. So, diagonal BD can be found using the Law of Sines or Law of Cosines if needed, but maybe I don't need its length.Wait, in the parallelogram, diagonals bisect each other, so the diagonals AC and BD intersect at point O, which is the midpoint of both diagonals. But I'm not sure if that helps here.Alternatively, maybe I can consider triangle ABC. In parallelogram ABCD, triangle ABC shares side BC with triangle BCD. But I'm not sure if that helps.Wait, perhaps I should consider the entire parallelogram and the fact that consecutive angles are supplementary. So, if I can find angle B, then angle A would be 180 - angle B.But how do I find angle B? Let's see. In triangle BCD, angle at B is 52.5 degrees, which is angle CBD. But angle ABC in the parallelogram is adjacent to angle CBD. So, perhaps angle ABC is equal to 180 - angle CBD?Wait, no. Because in the parallelogram, angle ABC is at vertex B, between sides AB and BC. Whereas angle CBD is at vertex B, between sides CB and BD. So, they are adjacent but not supplementary.Wait, maybe I can find angle ABC by considering triangle ABC. But I don't have much information about triangle ABC.Alternatively, maybe I can use the fact that in the parallelogram, the sum of angles at B and C is 180 degrees. So, angle B + angle C = 180. Since angle C is 75 degrees, angle B should be 105 degrees. Therefore, angle A, being opposite to angle C, is also 75 degrees.Wait, but earlier I thought that angle B in the parallelogram is 105 degrees, but in triangle BCD, angle at B is 52.5 degrees. How does that reconcile?Ah, I think I see the confusion. In the parallelogram, angle at B is angle ABC, which is 105 degrees, while in triangle BCD, angle at B is angle CBD, which is 52.5 degrees. These are two different angles at vertex B, but they are related.So, in the parallelogram, angle ABC is 105 degrees, and in triangle BCD, angle CBD is 52.5 degrees. Since these two angles are adjacent at vertex B, their sum should be equal to angle ABC. Wait, no, because angle ABC is the angle between AB and BC, while angle CBD is the angle between BC and BD.Wait, actually, in the parallelogram, at vertex B, there are two angles: one is angle ABC (105 degrees) and the other is angle ABD, which is part of triangle ABD. But I'm getting confused.Maybe I should approach this differently. Since ABCD is a parallelogram, sides AB and CD are equal and parallel, and sides AD and BC are equal and parallel. Given that BC = CD, which are sides of the parallelogram, that would mean that sides AD and BC are equal, and sides AB and CD are equal. But if BC = CD, then AB = CD = BC, so AB = BC. Therefore, sides AB and BC are equal, making triangle ABC isosceles with AB = BC.Wait, that's an important point. If AB = BC, then triangle ABC is isosceles with AB = BC, so angles at A and C are equal. But angle at C is 75 degrees, so angle at A would also be 75 degrees. Therefore, angle A is 75 degrees.Wait, but earlier I thought angle B was 105 degrees, which would make angle A 75 degrees as well, since consecutive angles in a parallelogram are supplementary. So, both approaches lead to angle A being 75 degrees.But let me verify this again. Since ABCD is a parallelogram, opposite sides are equal, so AB = CD and AD = BC. Given that BC = CD, then AB = CD = BC, so AB = BC. Therefore, triangle ABC has sides AB = BC, making it isosceles with base AC. Therefore, angles at A and C are equal. Since angle at C is 75 degrees, angle at A is also 75 degrees.Yes, that makes sense. So, angle A is 75 degrees.Wait, but earlier I was confused about the angles in triangle BCD. Let me make sure that doesn't affect the result. In triangle BCD, angles at B and D are 52.5 degrees each, and angle at C is 75 degrees. In the parallelogram, angle at B is 105 degrees, which is different from the 52.5 degrees in triangle BCD. But that's okay because those are different angles at vertex B.So, to summarize, since ABCD is a parallelogram and BC = CD, which leads to AB = BC, making triangle ABC isosceles with AB = BC. Therefore, angles at A and C are equal, and since angle C is 75 degrees, angle A is also 75 degrees.I think that's the correct approach. So, the measure of angle A is 75 degrees.

answer：Alright, so I've got this inequality problem here, and I need to figure out the solution set for another inequality based on some given information. Let me try to break it down step by step.First, the problem says that the solution to the inequality ((a - b)x + a + 2b > 0) is (x > frac{1}{2}). From this, I need to find the solution set for the inequality (ax < b).Okay, let's start by understanding what the first inequality tells us. The inequality is linear in terms of (x), so it's a straight line when graphed. The solution (x > frac{1}{2}) means that the inequality holds true for all values of (x) greater than (frac{1}{2}). This suggests that the coefficient of (x) must be positive because the inequality opens upwards.So, looking at the inequality ((a - b)x + a + 2b > 0), the coefficient of (x) is ((a - b)). For the inequality to hold for (x > frac{1}{2}), ((a - b)) must be positive. Therefore, we can conclude that (a - b > 0), which simplifies to (a > b).Now, to find the exact relationship between (a) and (b), let's solve the inequality ((a - b)x + a + 2b > 0) for (x). First, isolate the term with (x):[(a - b)x > -a - 2b]Now, divide both sides by ((a - b)). Since we've already established that (a - b > 0), the direction of the inequality sign remains the same:[x > frac{-a - 2b}{a - b}]According to the problem, this solution is (x > frac{1}{2}). Therefore, we can set up the equation:[frac{-a - 2b}{a - b} = frac{1}{2}]Now, let's solve this equation for the ratio (frac{b}{a}). First, cross-multiply to eliminate the fractions:[2(-a - 2b) = 1(a - b)]Simplify both sides:[-2a - 4b = a - b]Bring all terms to one side:[-2a - 4b - a + b = 0]Combine like terms:[-3a - 3b = 0]Factor out (-3):[-3(a + b) = 0]Divide both sides by (-3):[a + b = 0]So, (a = -b). This means that (b = -a).Now, we can express (b) in terms of (a). Let's substitute (b = -a) into the inequality (ax < b).Substituting (b) gives:[ax < -a]Now, since we know from earlier that (a > b), and (b = -a), we can deduce that (a > -a). Adding (a) to both sides:[2a > 0]Divide both sides by 2:[a > 0]So, (a) is positive. Knowing that (a) is positive, we can safely divide both sides of the inequality (ax < -a) by (a) without changing the direction of the inequality:[x < frac{-a}{a}]Simplify the fraction:[x < -1]Therefore, the solution set for the inequality (ax < b) is all real numbers less than (-1).Just to recap:1. We started with the given inequality and its solution.2. We determined that (a > b) and found the relationship between (a) and (b) by setting up and solving an equation.3. We substituted (b) in terms of (a) into the second inequality.4. We found that (a) is positive, allowing us to solve for (x) in the second inequality.5. We concluded that (x < -1) is the solution set.I think that covers all the steps. Let me just double-check to make sure I didn't make any mistakes in the algebra.Starting from:[frac{-a - 2b}{a - b} = frac{1}{2}]Cross-multiplying:[2(-a - 2b) = a - b]Expanding:[-2a - 4b = a - b]Bringing all terms to one side:[-2a - 4b - a + b = 0]Combining like terms:[-3a - 3b = 0]Factoring:[-3(a + b) = 0]Dividing:[a + b = 0 Rightarrow a = -b]Yes, that seems correct.Then substituting into (ax < b):[ax < -a]Since (a > 0), dividing both sides by (a):[x < -1]Yep, that checks out. So, I'm confident that the solution set is (x < -1).**Final Answer**The solution set for the inequality (ax < b) is boxed{{x | x < -1}}.