🧠:Okay, let's see. The problem is to write the number 1.11 ten times and the number 1.01 eleven times. Then cross out some numbers so that the sum of the remaining ones is exactly 20.19. Hmm, interesting. Let me try to figure this out step by step.First, let me understand the total sum of all the numbers before any are crossed out. If there are ten 1.11s and eleven 1.01s, the total would be:10 * 1.11 + 11 * 1.01.Calculating that: 10 * 1.11 is 11.10, and 11 * 1.01. Let's compute 11*1.01. 10*1.01 is 10.10, plus 1.01 is 11.11. So total sum is 11.10 + 11.11 = 22.21. So the total sum is 22.21. We need to cross out numbers such that the remaining sum is 20.19. Therefore, the amount we need to cross out is 22.21 - 20.19 = 2.02. So we need to remove numbers that add up to 2.02.Now, the numbers we can cross out are either 1.11 or 1.01. So the problem reduces to finding a combination of some 1.11s and 1.01s that sum to 2.02. Let me denote the number of 1.11s crossed out as x and the number of 1.01s crossed out as y. Then, we have the equation:1.11x + 1.01y = 2.02.We need to find non-negative integers x and y such that this equation holds, with x ≤ 10 and y ≤ 11 because we can't cross out more numbers than are present.Let me try to solve this equation. Let's multiply both sides by 100 to eliminate decimals:111x + 101y = 202.Now, we need integer solutions for x and y. Let me see possible values of x and y that satisfy this equation.First, note that 111x must be less than or equal to 202, so x can be 0, 1, or 2 because 111*2=222 which is already more than 202. So x can be 0, 1, or 2.Let me check x=0:111*0 + 101y = 202 ⇒ 101y = 202 ⇒ y = 202/101 = 2. So y=2. Since 2 ≤11, this is possible. So one solution is x=0, y=2. Cross out two 1.01s. Then the total removed would be 2*1.01=2.02, which matches.Another possibility, x=1:111*1 + 101y = 202 ⇒ 111 +101y =202 ⇒101y=91 ⇒ y=91/101≈0.9009, which is not an integer. So y is not an integer here. So this is not possible.x=2:111*2=222. Then 222 +101y=202 ⇒101y=202-222= -20. Not possible since y can't be negative. So x=2 is invalid.Therefore, the only solution is x=0, y=2. So cross out two 1.01s. Let me verify.Original total is 22.21. Removing 2*1.01=2.02. So remaining sum is 22.21 - 2.02=20.19. Exactly what's needed. So that works.Wait, but the problem says "cross out one or more numbers". So at least one number must be crossed out. In this solution, we cross out two numbers (both 1.01s). So that's okay. So the answer would be to cross out two 1.01s. Let me check again.Original numbers: 10 of 1.11 and 11 of 1.01. If we remove two 1.01s, then remaining numbers are 10*1.11 + 9*1.01. Let's compute that:10*1.11=11.10, 9*1.01=9.09. Total sum is 11.10+9.09=20.19. Correct.But wait, let me make sure that there are indeed eleven 1.01s. If we cross out two, we have 9 left, which is allowed. Yes. So this seems to work.But let me check if there are other possible solutions. Because sometimes there might be multiple ways. Let me double-check.Suppose we try x=1, but as before, y would have to be (202 -111)/101=91/101≈0.9009, which is not an integer. So no solution here.Another approach: Maybe if we consider that 1.11 and 1.01 have some common factors. Let me see. 1.11 is 111 cents, 1.01 is 101 cents. The equation is 111x +101y=202. Are 111 and 101 coprime? Let's check GCD(111,101). 111 divided by 101 is 1 with remainder 10. Then GCD(101,10). 101 divided by 10 is 10 with remainder 1. Then GCD(10,1)=1. So GCD is 1. Therefore, there exists solutions, but we need to find non-negative solutions. Since the GCD is 1, and 1 divides 202, solutions exist. But in this case, the only solution in non-negative integers is x=0,y=2.Alternatively, perhaps using negative numbers, but since we can't cross out negative numbers, that's invalid.Therefore, the only solution is crossing out two 1.01s. Therefore, the answer is to cross out two instances of 1.01. But let me make sure again. Let me verify.Total numbers: 10*1.11 + 11*1.01=11.10+11.11=22.21. Subtract 2*1.01=2.02, gives 20.19. Correct.Alternatively, is there a way to cross out a combination of 1.11 and 1.01? For example, maybe cross out one 1.11 and one 1.01. Let's see what that would sum to.1.11 +1.01=2.12. But we need to remove 2.02. 2.12 is more than 2.02. So that's too much. Alternatively, cross out one 1.11 and subtract some 1.01s. But then we need 1.11 +1.01y=2.02 ⇒1.01y=0.91 ⇒y≈0.9009, which is not an integer. So no.Alternatively, cross out two 1.11s. That would be 2.22. But we need to remove 2.02. So that's 0.20 too much. Not helpful.Alternatively, cross out one 1.01. That removes 1.01, so remaining sum is 22.21-1.01=21.20, which is more than 20.19. Not enough.So indeed, only removing two 1.01s gives the correct amount. Therefore, the solution is to cross out two 1.01s. Let me confirm once more.Yes, 11*1.01=11.11. If we cross out two, 9*1.01=9.09. 10*1.11=11.10. 11.10+9.09=20.19. Perfect.So the answer is to cross out two of the 1.01 numbers. Therefore, the final answer is to cross out two instances of 1.01.

✅:To solve the problem, we first calculate the total sum of all numbers: 10 × 1.11 + 11 × 1.01 = 11.10 + 11.11 = 22.21. The goal is to remove numbers summing to 22.21 − 20.19 = 2.02. We set up the equation: 1.11x + 1.01y = 2.02 (where ( x ) and ( y ) are the counts of numbers removed). Multiplying by 100 to eliminate decimals: 111x + 101y = 202. Testing possible integer solutions: - ( x = 0 ): ( 101y = 202 ) ⇒ ( y = 2 ). - ( x = 1 ) or ( x = 2 ) yield non-integer ( y ). Thus, the only valid solution is ( x = 0 ), ( y = 2 ). This means removing two 1.01s. Verification: Remaining sum: 10 × 1.11 + (11 − 2) × 1.01 = 11.10 + 9.09 = 20.19. Answer: Cross out two instances of 1.01.

🔑:1. First, let's write down the numbers as given in the problem: - Ten times the number (1.11) - Eleven times the number (1.01)2. Calculate the total sum of all these numbers: [ text{Sum} = 10 times 1.11 + 11 times 1.01 ] - Compute (10 times 1.11): [ 10 times 1.11 = 11.10 ] - Compute (11 times 1.01): [ 11 times 1.01 = 11.11 ] - Add these values together: [ 11.10 + 11.11 = 22.21 ]3. Determine how much this total exceeds the required sum (20.19): [ 22.21 - 20.19 = 2.02 ]4. Since we need to remove the excess (2.02) from the total: - We identify that to achieve this, we need to remove two instances of the number (1.01), because removing (2 times 1.01 = 2.02) will achieve the required adjustment.5. Now, rewrite the sequence of numbers and indicate which numbers to remove: [ 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01 ] - Remove (1.01) and (1.01):6. Verify that the remaining numbers sum to (20.19): - Remaining numbers: [ 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01 ] - Calculate their sum: [ 10 times 1.11 = 11.10 ] [ 9 times 1.01 = 9.09 ] [ 11.10 + 9.09 = 20.19 ]Conclusion:[boxed{1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.11, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01, 1.01}]

🧠:Okay, let's try to figure out this problem step by step. So, there are five baskets with apples of two varieties. Each basket has only one variety. The baskets have 20, 30, 40, 60, and 90 apples. Then, one basket was completely sold, and after that, the number of apples of the first variety became twice as many as the second variety. We need to find out which basket could have been sold. If there are multiple possibilities, list them all.First, let me restate the problem in my own words to make sure I understand. There are five baskets, each with a single variety of apples. The quantities are 20, 30, 40, 60, 90. After selling one entire basket, the remaining apples of the first variety are double the second variety. So, we need to figure out which basket, when removed, leaves the remaining apples in a 2:1 ratio between the two varieties.Since each basket is only one variety, the total number of each variety before selling is the sum of the baskets of that variety. But we don't know which baskets are which variety. That's part of the problem. So, we need to consider possible ways the baskets could be divided into two varieties, such that when one basket is removed, the ratio becomes 2:1.But wait, the problem doesn't tell us how the baskets are divided into the two varieties initially. So we have to figure out possible distributions of the baskets into the two varieties, such that after removing one basket, the remaining apples satisfy the ratio. Hmm. That seems a bit tricky because we don't know the initial distribution. But maybe we can model it.Let me denote the two varieties as Variety A and Variety B. Let’s say that some baskets are Variety A and others are Variety B. The key is that after selling one basket (which is either A or B), the remaining number of A apples is twice the remaining number of B apples.Let’s denote S_A as the sum of all apples in Variety A baskets, and S_B as the sum of all apples in Variety B baskets. Then, after selling a basket, either S_A or S_B will decrease by the number of apples in that basket. Then, the new ratio (S_A' / S_B') or (S_A / S_B') must be 2/1.So, depending on which basket was sold (A or B), we can set up equations.But since we don't know which baskets are A or B, maybe we need to consider all possible assignments of baskets to A or B, and check which ones satisfy the ratio after removing a basket. But since there are 5 baskets, each can be either A or B, leading to 2^5 = 32 possible assignments. That's a lot. However, maybe we can find a smarter way.Alternatively, perhaps instead of considering all possible assignments, we can think in terms of total apples. Let's denote that before selling, the total number of apples is 20 + 30 + 40 + 60 + 90 = 240 apples. After selling a basket with x apples, the total becomes 240 - x. According to the problem, after selling, the number of Variety A is twice Variety B, so total remaining apples would be 2y + y = 3y, where y is the number of Variety B apples. Therefore, 240 - x must be divisible by 3. So, 240 - x ≡ 0 mod 3 ⇒ x ≡ 240 mod 3. Since 240 divided by 3 is 80, so 240 ≡ 0 mod 3. Therefore, x ≡ 0 mod 3. So the basket sold must have a number of apples divisible by 3.Looking at the baskets: 20, 30, 40, 60, 90. Which of these are divisible by 3? Let's check:20 ÷ 3 = 6.666… Not divisible.30 ÷ 3 = 10. Divisible.40 ÷ 3 ≈ 13.333… Not divisible.60 ÷ 3 = 20. Divisible.90 ÷ 3 = 30. Divisible.So possible candidates for the sold basket are 30, 60, or 90. So the sold basket must be one of those. Wait, but that's only considering the divisibility by 3. But maybe there's more to it. Let me verify.If after selling, the total apples are 3y, where y is Variety B, so 3y = 240 - x. Therefore, 240 - x must be divisible by 3. Therefore, x must be congruent to 240 modulo 3. Since 240 is 0 mod 3, x must be 0 mod 3. So x is 30, 60, or 90. Therefore, the sold basket must be 30, 60, or 90. So that's a key insight.But wait, the problem says "the number of apples of the first variety became twice as many as the number of apples of the second variety." So it's possible that before selling, the ratio wasn't 2:1, but after selling a basket, it became 2:1. So the key is that the act of removing a basket adjusts the counts to achieve that ratio.But the initial ratio is not given. So perhaps the initial counts are such that when you subtract the basket sold (which could be of either variety), the remaining counts are in 2:1.But how do we know which variety the sold basket was? If we remove a Variety A basket, then S_A becomes S_A - x, and S_B remains the same. Then, (S_A - x) = 2 * S_B.Alternatively, if we remove a Variety B basket, then S_B becomes S_B - x, and S_A remains the same. Then, S_A = 2 * (S_B - x).Therefore, there are two cases:Case 1: The sold basket is Variety A. Then:S_A - x = 2 * S_BBut we know that S_A + S_B = 240 - x (before selling). Wait, no. Wait, before selling, S_A + S_B = 240. Then, after selling a basket of Variety A, S_A becomes S_A - x, and S_B remains. So total after selling is 240 - x. Then:(S_A - x) = 2 * S_BBut S_A + S_B = 240So substituting S_A = 240 - S_B into the first equation:240 - S_B - x = 2 * S_B240 - x = 3 * S_BTherefore, S_B = (240 - x)/3But S_B must be equal to the sum of the remaining Variety B baskets. Similarly, S_A = 240 - S_B.But since the baskets are all assigned to A or B, S_B must be the sum of some combination of the baskets (excluding the sold one if it's A). But this might not directly help.Wait, but in this case, if we sold a Variety A basket, then S_B remains the same. So S_B is the sum of the original B baskets. But we need to have S_B = (240 - x)/3.Similarly, in Case 2: Sold a Variety B basket. Then:S_A = 2 * (S_B - x)And S_A + S_B = 240So substituting S_A = 2*(S_B - x) into S_A + S_B = 240:2*(S_B - x) + S_B = 2403*S_B - 2x = 2403*S_B = 240 + 2xS_B = (240 + 2x)/3So S_B must be an integer, so (240 + 2x) must be divisible by 3. Since 240 is divisible by 3, 2x must also be ≡ 0 mod 3. Therefore, 2x ≡ 0 mod 3 ⇒ x ≡ 0 mod 3. Which is the same as before. Therefore, in both cases, x must be divisible by 3. Therefore, sold basket must be 30, 60, or 90.So regardless of whether the sold basket is A or B, x must be divisible by 3. Therefore, possible x: 30, 60, 90.Therefore, the possible sold baskets are 30, 60, 90. But need to check if each is possible.But let's check each possible x (30, 60, 90) and see if there exists a distribution of the baskets into A and B such that after removing x, the ratio is 2:1.Let's start with x = 30.Case 1: Sold basket is 30 apples, which is Variety A.Then:S_A - 30 = 2 * S_BBut original S_A + S_B = 240So substituting S_A = 240 - S_B into first equation:240 - S_B - 30 = 2*S_B210 - S_B = 2*S_B210 = 3*S_B ⇒ S_B = 70So S_B = 70. So the sum of Variety B baskets is 70. But the original baskets are 20, 30, 40, 60, 90. But one of them is sold (30, which is Variety A). So the remaining baskets are 20, 40, 60, 90. These need to be partitioned into A and B such that the sum of B is 70. But wait, the sold basket is 30 (A), so the remaining baskets are 20,40,60,90. So the original S_A included 30, which is now sold. So original S_A = 30 + sum of other A baskets. But S_A = 240 - S_B = 240 - 70 = 170. Therefore, original S_A = 170. Since the sold basket is 30 (A), then the remaining A baskets sum to 170 - 30 = 140. So the remaining A baskets sum to 140, and the B baskets sum to 70.But the remaining baskets are 20,40,60,90. We need to assign these to A and B such that sum of B is 70 and sum of A is 140.Looking for subsets of {20,40,60,90} that sum to 70. Let's check:Possible combinations:20 + 40 + 10? Not possible.20 + 50? No. 40 + 30? No. Wait, available numbers are 20,40,60,90.Possible subsets:20 + 40 = 6020 + 60 = 8020 + 90 = 11040 + 60 = 10040 + 90 = 13060 + 90 = 150Single numbers: 20, 40, 60, 90. 20, 40, 60, 90. None of these sum to 70. So no possible combination. So S_B cannot be 70. Therefore, if x=30 is sold as Variety A, there's no solution.Case 2: Sold basket is 30 apples, which is Variety B.Then:S_A = 2*(S_B - 30)But S_A + S_B = 240Substitute S_A:2*(S_B - 30) + S_B = 2402S_B - 60 + S_B = 2403S_B = 300 ⇒ S_B = 100Therefore, original S_B = 100. Then S_A = 240 - 100 = 140. But the sold basket is 30 (B), so the remaining B apples are 100 - 30 = 70. Then remaining apples: S_A = 140, remaining S_B = 70, which is 2:1 ratio. So this works.But we need to check if there's a possible assignment of baskets to B such that their sum is 100. The original baskets (before selling) are 20,30,40,60,90. If S_B = 100, then the B baskets must sum to 100. The sold basket is 30 (B), so original B baskets include 30 and others summing to 70. So the other B baskets must sum to 70.Looking for subsets of {20,40,60,90} that sum to 70. As before, subsets of {20,40,60,90} summing to 70.Again, check possible combinations:20 + 40 + 10? No.40 + 30? Not in the set. 20 + 50? No. Wait, same as before. The available numbers are 20,40,60,90.Looking for subsets:20 + 40 = 6020 + 60 = 8020 + 90 = 11040 + 60 = 10040 + 90 = 13060 + 90 = 150Single numbers: 20,40,60,90. None of these sum to 70. So again, no possible combination. Therefore, it's impossible to have original S_B = 100 by having B baskets as 30 plus others summing to 70. Therefore, x=30 cannot be the sold basket, whether it's Variety A or B. Hmm, that's confusing. Wait, but we had S_B = 100, but the baskets are 30 (sold) and others. So original B baskets are 30 and some others. The sum needs to be 100, so other B baskets must sum to 70. But as before, no such subsets. Therefore, x=30 is not possible.Wait, but this is conflicting with the earlier reasoning. If x=30, and it's B, then S_B original is 100. But 100 - 30 = 70. But we need the other B baskets to sum to 70. But there are no subsets in {20,40,60,90} that sum to 70. Therefore, x=30 is invalid.So maybe x=30 is not possible. Let's try x=60.Case 1: Sold basket is 60 (Variety A).Then:S_A - 60 = 2*S_BOriginal total S_A + S_B = 240.Substitute S_A = 240 - S_B:240 - S_B - 60 = 2*S_B180 - S_B = 2*S_B180 = 3*S_B ⇒ S_B = 60So S_B = 60. Original S_A = 240 - 60 = 180. After selling 60 (A), remaining S_A = 120, S_B = 60. So ratio 2:1. Good.Now, need to check if S_B = 60 is possible. Original B baskets sum to 60. The sold basket is 60 (A), so the original B baskets are among 20,30,40,90. Need to see if any combination of these sums to 60.Check subsets of {20,30,40,90} summing to 60:20 + 40 = 60Yes. So B baskets could be 20 and 40. Then S_B = 60. The A baskets would be 30,60,90. Sum S_A = 30 + 60 + 90 = 180. Then, selling the 60 (A) basket, remaining A = 180 - 60 = 120, B = 60. 120:60 = 2:1. Perfect. So this works.Therefore, x=60 is possible if the sold basket is Variety A (60) and the B baskets are 20 and 40.Case 2: Sold basket is 60 (Variety B).Then:S_A = 2*(S_B - 60)Original S_A + S_B = 240.Substitute S_A = 2*(S_B - 60):2*(S_B - 60) + S_B = 2402S_B - 120 + S_B = 2403S_B = 360 ⇒ S_B = 120Therefore, original S_B = 120. Sold basket is 60 (B), remaining S_B = 60. Then S_A = 240 - 120 = 120. So after selling, S_A = 120, S_B = 60. Ratio 2:1. Good.Need to check if S_B can be 120. Original B baskets include 60 and others summing to 60 (since sold is 60). Original B baskets: 60 and others. The other baskets are 20,30,40,90. Need subsets of these summing to 60.Check subsets of {20,30,40,90} summing to 60:20 + 40 = 6030 + 30 = 60 (but there's only one 30)20 + 30 + 10 = 60 (no)So only 20 + 40 = 60. Therefore, original B baskets could be 60, 20, 40. Sum S_B = 60 + 20 + 40 = 120. Then A baskets are 30 and 90. Sum S_A = 30 + 90 = 120. After selling 60 (B), remaining B = 60, A = 120. Ratio 2:1. So this works.Therefore, x=60 is possible whether it's sold as A or B. Wait, but in the first case (sold as A), B baskets are 20 and 40. In the second case (sold as B), B baskets are 20, 40, 60. So both scenarios are possible. Therefore, x=60 is a valid solution.Now, moving to x=90.Case 1: Sold basket is 90 (Variety A).Then:S_A - 90 = 2*S_BOriginal S_A + S_B = 240.Substitute S_A = 240 - S_B:240 - S_B - 90 = 2*S_B150 - S_B = 2*S_B150 = 3*S_B ⇒ S_B = 50So original S_B = 50. Then original S_A = 240 - 50 = 190. After selling 90 (A), remaining S_A = 100, S_B = 50. Ratio 2:1. Good.Need to check if S_B = 50 is possible. Original B baskets sum to 50. Sold basket is 90 (A), so remaining baskets are 20,30,40,60. Need subsets of these summing to 50.Possible subsets:20 + 30 = 50Yes. So B baskets could be 20 and 30. Sum S_B = 50. Then A baskets are 40,60,90. Sum S_A = 40 + 60 + 90 = 190. After selling 90 (A), remaining A = 100, B = 50. Ratio 2:1. Perfect. So this works.Case 2: Sold basket is 90 (Variety B).Then:S_A = 2*(S_B - 90)Original S_A + S_B = 240.Substitute S_A:2*(S_B - 90) + S_B = 2402S_B - 180 + S_B = 2403S_B = 420 ⇒ S_B = 140Original S_B = 140. Sold basket is 90 (B), remaining S_B = 50. Then S_A = 240 - 140 = 100. After selling, ratio 100:50 = 2:1. Good.Need to check if S_B = 140 is possible. Original B baskets include 90 and others summing to 50. Other baskets are 20,30,40,60. Need subsets of these summing to 50.Possible subsets:20 + 30 = 5040 + 10 = 50 (not possible)So 20 + 30 = 50. Therefore, original B baskets are 90, 20, 30. Sum S_B = 90 + 20 + 30 = 140. Then A baskets are 40,60. Sum S_A = 100. After selling 90 (B), remaining B = 50, A = 100. Ratio 2:1. This works.Therefore, x=90 is possible whether sold as A or B. In Case 1, B baskets are 20,30; in Case 2, B baskets are 20,30,90. Both valid.So summarizing:- x=30: Not possible because no subsets sum to required amounts.- x=60: Possible as both A and B.- x=90: Possible as both A and B.Therefore, the sold basket could be 60 or 90. Wait, but we need to check x=60 and x=90. However, when we checked x=60, both cases (sold as A or B) worked. Similarly for x=90. So both 60 and 90 are possible.But wait, the answer options are 30,60,90. But we saw that x=30 doesn't work. So the correct answers are 60 and 90.Wait, but let's recheck. For x=60, both cases work. For x=90, both cases work. So both 60 and 90 are possible. Let me verify once more.For x=60:Case 1: Sold as A.B baskets: 20 + 40 = 60. A baskets: 30,60,90. After selling 60, A becomes 120, B remains 60. 120:60=2:1.Case 2: Sold as B.B baskets: 20 + 40 + 60 = 120. A baskets: 30,90. After selling 60 (B), B becomes 60, A remains 120. 120:60=2:1.For x=90:Case 1: Sold as A.B baskets: 20 + 30 = 50. A baskets:40,60,90. After selling 90, A becomes 100, B remains 50. 100:50=2:1.Case 2: Sold as B.B baskets:20 +30 +90=140. A baskets:40,60. After selling 90, B becomes 50, A remains 100. 100:50=2:1.Therefore, both 60 and 90 are possible.But wait, what about x=30? We initially thought x=30 might not work, but let's check again.For x=30:Case 1: Sold as A.S_B should be 70. Need B baskets to sum to 70 from remaining baskets 20,40,60,90. No subsets sum to 70.Case 2: Sold as B.S_B original is 100. Need B baskets to be 30 + subset of 20,40,60,90 summing to 70. Again, no subsets. So x=30 is invalid.Therefore, only 60 and 90 are possible.Wait, but the problem says "the contents of one of the baskets were completely sold". It doesn't specify whether the sold basket was of the first or second variety, just that after selling, the first variety became twice the second. So depending on which basket was sold (either variety), it's possible.Therefore, the answer should be 60 and 90. But let me check the problem statement again."The number of apples of the first variety became twice as many as the number of apples of the second variety."The first variety is mentioned, but we don't know which variety was first. Wait, actually, in the problem statement, "the first basket contains 20 apples", etc. But the "first variety" and "second variety" are just two types, not related to the order of baskets. So the problem is saying that after selling a basket, one variety (which is now referred to as the first variety) is twice the other. So maybe the first variety is whichever one is larger after selling. Wait, no. The problem says "the number of apples of the first variety became twice as many as the number of apples of the second variety". So the first variety is the one that ends up being twice as much. So it's not that they are labeled beforehand, but after selling, one is twice the other, and they are referred to as first and second based on that ratio. So it could be either way, but the key is that the ratio is 2:1.Therefore, our analysis is correct. Both 60 and 90 are possible. Wait, but according to our analysis, when we sold 60 or 90, both cases work regardless of whether they are A or B. Therefore, both 60 and 90 are valid answers.But let me check if there's another possible basket. For example, x=40. Even though 40 is not divisible by 3, but perhaps...Wait, earlier reasoning said x must be divisible by 3, but let's confirm.Total apples after selling: 240 - x.Since after selling, the ratio is 2:1, so total apples must be 3y. Therefore, 240 - x must be divisible by 3. Therefore, x ≡ 240 mod 3. 240 ÷3=80, so x ≡0 mod3. Therefore, x must be 30,60,90. So x=40 is invalid. So the sold basket must be 30,60,90. But we saw that 30 is invalid. So only 60 and 90.Therefore, the answer is 60 and 90.But wait, let me check x=90 again. If x=90 is sold as A:Original S_A = 190, S_B=50. After selling 90, S_A=100, S_B=50. 100:50=2:1.Original B baskets sum to 50: 20+30.Original A baskets:40,60,90. Yes.If x=90 sold as B:Original S_B=140 (20+30+90), S_A=100 (40+60). After selling 90, S_B=50, S_A=100. Ratio 2:1.Yes, both valid.Similarly for x=60 sold as A or B. So both 60 and 90 are possible. Hence, the answer is 60 and 90. So the possible sold baskets are 60 and 90.But wait, the problem says "the number of apples of the first variety became twice as many as the number of apples of the second variety". It doesn't specify which is which. So whether A becomes twice B or B becomes twice A, but since it's referred to as "first variety" and "second variety", I think the problem assumes that the first variety is the one that ends up being twice the second. So regardless of the original labels, after selling, one is twice the other, and they are referred to as first and second accordingly. Therefore, our analysis holds.Therefore, the possible sold baskets are 60 and 90.But wait, let's check x=60:If we sell 60 as A:A becomes 120, B remains 60.If we sell 60 as B:A becomes 120, B becomes 60.So in both cases, A is 120 and B is 60. But since the problem refers to "first variety" and "second variety", which could be A or B. But since after selling, one is twice the other, and the problem states "the first variety became twice as many as the second", regardless of which was which before. So in both cases, the variety that is twice is considered the first. So both scenarios satisfy the problem statement.Similarly for x=90:If sold as A:A becomes 100, B remains 50.If sold as B:A remains 100, B becomes 50.Either way, the first variety (the one that's twice) is 100 vs 50.Therefore, both x=60 and x=90 are valid. Hence, the answers are 60 and 90.But wait, let me check the initial problem again. The baskets are:1st basket: 202nd:303rd:404th:605th:90The problem says "the contents of one of the baskets were completely sold". So the answer is the possible number of apples in the sold basket. The baskets have 20,30,40,60,90 apples. So possible answers are 60 and 90. The sold basket could be the one with 60 apples or the one with 90 apples. So the answer is 60 and 90.But let me check once more for x=30. Even though x=30 is divisible by 3, we saw that there's no possible assignment. So x=30 is invalid. So only 60 and 90.Therefore, the final answer is 60 and 90. But let me check again for x=60 and x=90.Yes, both 60 and 90 have valid assignments. Therefore, the answer is 60 and 90.

✅:After systematically analyzing the problem, we determine that the sold basket must have a quantity divisible by 3 to satisfy the 2:1 ratio condition. This narrows the candidates to 30, 60, or 90 apples. Further examination reveals:1. Sold basket with 30 apples: No valid distribution of the remaining baskets achieves the required ratio.2. Sold basket with 60 apples: Valid in two scenarios: - If 60 is Variety A, remaining A apples (120) are twice B (60). - If 60 is Variety B, remaining B apples (60) are half of A (120).3. Sold basket with 90 apples: Also valid in two scenarios: - If 90 is Variety A, remaining A apples (100) are twice B (50). - If 90 is Variety B, remaining B apples (50) are half of A (100).Thus, the possible sold baskets are 60 and 90.Answer: The sold basket could have contained boxed{60} or boxed{90} apples.

🔑:1. Identify the total number of apples initially: - We have 5 baskets with apples as given: 20, 30, 40, 60, and 90. - Calculate the total number of apples: [ text{Total apples} = 20 + 30 + 40 + 60 + 90 = 240 ] 2. Determine the condition after selling one basket: - After selling the apples from one basket, the remaining apples should be able to be divided into two groups such that apples of one type are twice the amount of apples of the other type. - Hence, the remaining total number of apples must be divisible by 3 because (2:1) ratio implies a sum that is a multiple of 3. - Since 240 is divisible by 3, the number of apples removed must also be divisible by 3.3. Eliminate invalid basket quantities: - Verify which basket removals leave a number divisible by 3: [ a. , text{Remove 20} quad 240 - 20 = 220 quad (text{Not divisible by 3}) ] [ b. , text{Remove 30} quad 240 - 30 = 210 quad (text{Divisible by 3}) ] [ c. , text{Remove 40} quad 240 - 40 = 200 quad (text{Not divisible by 3}) ] [ d. , text{Remove 60} quad 240 - 60 = 180 quad (text{Divisible by 3}) ] [ e. , text{Remove 90} quad 240 - 90 = 150 quad (text{Divisible by 3}) ] - Therefore, possible candidates are 30, 60, and 90.4. Check the remaining baskets for the desired condition: - For 30 apples removed: [ text{Remaining apples} = 210 ] [ text{To satisfy} quad 2:1 quad text{ratio:} quad text{Let } x = text{apples of second type}, quad 2x = text{apples of first type} ] [ 210 = x + 2x = 3x Rightarrow x = 70 ] - Since 70 does not match any remaining basket sizes, this removal is invalid. - For 60 apples removed: [ text{Remaining apples} = 180 ] [ 2x = 2 cdot text{apples of second type} ] [ 180 = x + 2x = 3x Rightarrow x = 60 ] - Verify the groupings: [ 120 + 60 = 180 quad = 2(40 + 20) + 30 ] - This satisfies the condition, so 60 is a valid answer. - For 90 apples removed: [ text{Remaining apples} = 150 ] [ text{To satisfy} quad 2:1 quad text{ratio:} quad 2y = text{apples of first type}, quad y = text{apples of second type} ] [ 150 = y + 2y = 3y Rightarrow y = 50 ] - Verify the groupings: [ 2(60 + 40) = 100 quad = 2 cdot (30 + 90) ] - This satisfies the condition, so 90 is a valid answer. 5. Conclusion: - The possible number of apples in the basket sold are therefore (60) and (90).[boxed{60 text{ or } 90}]

🧠:Okay, so I have this problem here: There's a sequence of non-negative rational numbers a₁, a₂, a₃, ... such that for any natural numbers m and n, aₘ + aₙ = a_{mn}. I need to prove that not all members of this sequence are distinct. Hmm, interesting. Let me try to figure this out step by step.First, let me understand the given condition. The relation says that if I take any two indices m and n, add the terms aₘ and aₙ, the result should be equal to the term at the product index mn. So, for example, a₂ + a₃ = a₆, right? And this has to hold for all m and n. Also, all the terms are non-negative rationals. That's important. Non-negative, so we don't have to worry about cancellation or negative numbers messing things up. And they are rational, which might come into play later.The goal is to show that in such a sequence, there must be at least two terms that are equal. So, not all can be distinct. So, even if someone tries to make all terms different, this condition will force at least two of them to be the same. I need to find why that must happen.Let me start by looking at some specific cases. Maybe if I can find a pattern or a contradiction when assuming all terms are distinct, that would help.First, consider a₁. Let's see what happens when m or n is 1. Let's take m=1 and n=1. Then, a₁ + a₁ = a_{1×1} = a₁. So, 2a₁ = a₁. Subtract a₁ from both sides: a₁ = 0. So, a₁ must be zero. That's a good start. So, the first term is zero.Now, let's see what else we can derive. Let me use m=1 and some other n. Let's take m=1 and n=2. Then, a₁ + a₂ = a_{1×2} = a₂. But we already know a₁ is zero, so 0 + a₂ = a₂, which is just a tautology. Doesn't give new information. Similarly, m=1 and any n: a₁ + aₙ = aₙ, so 0 + aₙ = aₙ, which is always true. So, the first term being zero is consistent with the rest.Now, let's try m=2 and n=2. Then, a₂ + a₂ = a_{4}. So, 2a₂ = a₄. Similarly, m=2 and n=3: a₂ + a₃ = a₆. Also, m=3 and n=3: a₃ + a₃ = a₉. So, 2a₃ = a₉.Hmm, maybe we can find a relation between a₂ and a₄, a₄ and a₈, etc. Let's see. Let's take m=2 and n=4. Then, a₂ + a₄ = a₈. But we know that a₄ = 2a₂, so substitute that in: a₂ + 2a₂ = 3a₂ = a₈. So, a₈ = 3a₂. Similarly, m=4 and n=4: a₄ + a₄ = a_{16}, which is 2a₄ = a_{16} = 2*(2a₂) = 4a₂. So, a_{16} = 4a₂. Continuing, m=8 and n=8: a₈ + a₈ = a_{64} = 2*(3a₂) = 6a₂. So, a_{64} = 6a₂. Hmm, so the terms at powers of 2 positions are multiples of a₂. Similarly, the exponents are related to the indices.Wait, maybe there's a logarithmic pattern here? Let's think. Suppose I take a number n that is a power of 2, say n=2^k. Then, a_{2^k} = k*a₂. Because each time we go from 2^{k} to 2^{k+1}, we have a_{2^k} + a_{2} = a_{2^{k+1}}, so a_{2^{k+1}} = a_{2^k} + a₂. If a_{2^1} = a₂, then a_{2^2} = a₂ + a₂ = 2a₂, a_{2^3} = 2a₂ + a₂ = 3a₂, etc. So, by induction, a_{2^k} = k*a₂. Similarly, for other primes?Wait, but the problem doesn't state that the sequence is defined over primes. It's for all natural numbers. So, maybe this structure holds for prime powers, but I need to think about multiplicative functions. Because the relation a_{mn} = a_m + a_n is similar to a multiplicative function, but additive over multiplication. Wait, in number theory, functions where f(mn) = f(m) + f(n) for coprime m and n would be additive functions, but here it's for any m and n, not necessarily coprime. So, this is a stronger condition. Such functions are called logarithmic functions, perhaps?Wait, yes, if we define f(n) = a_n, then f(mn) = f(m) + f(n) for all m, n. Such functions are called completely additive functions. So, the problem is about a completely additive function f: N → Q≥0 (non-negative rationals), and we need to prove that f is not injective, i.e., there exist m ≠ n with f(m) = f(n).So, restating the problem in terms of additive functions: Any completely additive function f from the natural numbers to the non-negative rationals cannot be injective. Therefore, there must be at least two distinct natural numbers m and n such that f(m) = f(n).Hmm. So, how can we show that a completely additive function can't be injective? Let's recall properties of additive functions. Completely additive functions satisfy f(mn) = f(m) + f(n) for all m, n. The classic example is the logarithm function, but since we need the codomain to be non-negative rationals, logarithm wouldn't work because it can take negative values or irrationals. Also, logarithm is not defined at zero, but here all terms are non-negative rationals, and the sequence is indexed by natural numbers, so the domain is N.Wait, but in our problem, the function is f(n) = a_n, which is a sequence. So, we need to find m ≠ n with a_m = a_n. Let's think about the prime factorization. For a completely additive function, the value at a prime power would be additive over the exponents. Wait, no. Wait, if n = p^k, then for a completely additive function, f(n) = f(p^k) = k*f(p). Because f(p^k) = f(p * p^{k-1}) = f(p) + f(p^{k-1}) = f(p) + f(p) + f(p^{k-2}) = ... = k*f(p). So, for prime powers, the function is linear in the exponent. Similarly, for a general number, if n = p₁^{k₁} p₂^{k₂} ... p_r^{k_r}, then f(n) = k₁ f(p₁) + k₂ f(p₂) + ... + k_r f(p_r). So, the function is determined by its values on the primes, and it's linear in the exponents.Therefore, such a function is entirely determined by its values on the prime numbers. So, if we can show that, given the function is completely additive and maps to non-negative rationals, there must be two different numbers with the same value.Alternatively, since the function is determined by its values on primes, perhaps we can assign to each prime a non-negative rational number such that when we take different combinations (sums) of these primes' values, we get duplicates.But wait, if the primes are assigned distinct values, maybe we can still have different numbers have different sums. However, the problem states that all a_n are non-negative rationals. So, even if primes are assigned different values, the additivity might force some overlaps.Alternatively, maybe the key is that the number of possible distinct values is limited in some way, while the number of natural numbers is infinite. But since the codomain is non-negative rationals, which are dense, that might not be the case. Wait, but the function is completely additive, so the values are built up from the primes. If the primes are assigned sufficiently "independent" values, maybe the sums can all be unique. But since the primes are countably infinite, and the non-negative rationals are countable, but the function maps N to Q≥0. However, even if primes are assigned unique values, the problem is to show that in such a case, there must be a duplicate.Wait, but the problem states that all a_n are non-negative rationals. So, even if you have an additive function, you can't have injectivity because of the way the values add up. For example, maybe two different exponents on primes could lead to the same total. But if primes are assigned linearly independent values over the rationals, then the sums would be unique. However, since we are in Q, the field is of characteristic zero, but the problem allows the function to take values in Q≥0, which is a semigroup under addition.Wait, maybe the key is that if all primes are assigned positive rational numbers, then the function f(n) is a linear combination over the primes with coefficients in N₀ (non-negative integers). So, if the primes are assigned values that are linearly independent over Q, then the sums would be unique. But since the primes are countable, and Q is a countable dimensional vector space over Q, but each prime's value is a scalar in Q. However, we can't have infinitely many primes assigned linearly independent values in Q, because Q is a one-dimensional vector space over itself. Wait, that might not make sense. Wait, if we consider the primes as variables, then the function f(n) would be a linear form in exponents with coefficients being the prime values. But since all coefficients are in Q, which is a field, but if the coefficients are chosen to be linearly independent over Q, which would require an infinite dimensional space, but we can't have that in Q. So, in reality, the values assigned to primes must satisfy some linear relations over Q. Therefore, there must be some primes whose values are rational multiples of each other, leading to overlaps in the sums.Alternatively, perhaps the problem requires that even if you try to assign different values to primes, the non-negativity combined with additivity will force some terms to repeat.Wait, let's take a simple case. Suppose we have two primes, say 2 and 3. Suppose a₂ = p and a₃ = q, where p and q are non-negative rationals. Then, a₄ = 2p, a₆ = p + q, a₈ = 3p, a₉ = 2q, a₁₂ = p + 2q, etc. For these terms to all be distinct, we need 2p ≠ p + q, 2p ≠ 2q, p + q ≠ 3p, etc. So, this imposes conditions on p and q. For instance, 2p ≠ p + q ⇒ p ≠ q. Also, 2p ≠ 2q ⇒ p ≠ q, which is the same condition. Then, p + q ≠ 3p ⇒ q ≠ 2p. Similarly, a₆ = p + q must not equal a₉ = 2q ⇒ p + q ≠ 2q ⇒ p ≠ q. So, as long as p ≠ q and q ≠ 2p, these terms are distinct. But as we go further, say a_{16} = 4p, a_{12} = p + 2q, a_{18} = p + 2q (if m=2 and n=9: a₂ + a₉ = p + 2q = a_{18}), but wait, a_{12} = a_{3×4} = a₃ + a₄ = q + 2p. Wait, is that right? Wait, a_{12} can be written as 3×4 or 2×6. Let's check both: a₃ + a₄ = q + 2p, and a₂ + a₆ = p + (p + q) = 2p + q. But 12 can be factored in different ways, so the function must satisfy both. Therefore, for consistency, we need that a₃ + a₄ = a₂ + a₆. So, q + 2p = p + (p + q) = 2p + q, which is true. So, no problem there.But to have all terms distinct, let's see: a₂ = p, a₃ = q, a₄ = 2p, a₅ = r (another prime), a₆ = p + q, a₇ = s, a₈ = 3p, a₉ = 2q, a_{10} = p + r, etc. To have all these terms distinct, we need to ensure that none of these expressions collide. For example, 2p ≠ q, 3p ≠ q + p, 3p ≠ 2q, and so on. This seems like an increasing number of constraints as we consider more terms. However, since there are infinitely many terms, but we can choose the prime assignments in a way that avoids collisions? Maybe not, because even with two primes, the combinations start overlapping. For instance, if we have primes 2, 3, 5, 7, etc., each assigned different values, but their combinations (sums) can potentially overlap. For example, a₂ + a₃ = a₆, a₃ + a₂ = a₆, same thing. But that's just one term. If we have more primes, like a₅ + a₇ = a_{35}, and maybe a₂ + a_{33} = a_{66}, but a_{33} is a₃ + a_{11}, so a₂ + a₃ + a_{11} = a_{66}, which is a_{66} = a₂ + a₃ + a_{11}. So, unless there's some overlap in these sums, they might all be unique. But with infinitely many primes, maybe you can keep assigning values such that all sums are unique. However, the problem states that the sequence is in non-negative rationals, which are countable, and the natural numbers are countable, so in principle, you could have a bijection. But the constraint of additivity complicates things.Wait, but the problem says that not all members are distinct. So, perhaps even if you could assign prime values to make the sums unique, the conditions imposed by additivity over all mn force some repetition. Maybe via the Chinese Remainder Theorem or some overlapping indices.Alternatively, maybe considering the multiplicative structure of N. For example, numbers that can be factored in multiple ways. For instance, 4 can be written as 2×2, 6 as 2×3 or 3×2, but commutativity doesn't affect the result here. Wait, but in the problem, since m and n are any natural numbers, the order doesn't matter because aₘ + aₙ = a_{mn} and aₙ + aₘ = a_{nm} = a_{mn}, so addition is commutative, so that's consistent.Alternatively, let's think about prime factorization. Each number n ≥ 2 can be written as a product of primes. So, if n = p₁^{k₁} p₂^{k₂} ... p_r^{k_r}, then a_n = k₁ a_{p₁} + k₂ a_{p₂} + ... + k_r a_{p_r}. So, if we consider the exponents vector (k₁, k₂, ..., k_r) for each prime, then a_n is the dot product of this vector with the vector (a_{p₁}, a_{p₂}, ..., a_{p_r}).Now, if all the a_{p_i} are distinct and rationally independent, then each a_n would be unique. Wait, but rational independence here is key. If the a_{p_i} are linearly independent over Q, then the dot product would be unique for different exponent vectors. However, in Q, the field is Q, and we can't have infinitely many linearly independent elements in a Q-vector space, since Q is countable and the dimension would be uncountable if we have infinitely many primes. Wait, but actually, in the context of vector spaces over Q, the primes themselves are often used as a basis for a vector space with dimension equal to the number of primes, which is countably infinite. However, in our case, the a_{p_i} are elements of Q, so each a_{p_i} is a scalar in Q. Therefore, if we have infinitely many primes, the set {a_{p_i}} would have to be linearly dependent over Q, because Q is a countable-dimensional vector space over Q. Wait, but actually, any finite subset can be independent, but an infinite set can be independent. Hmm, but in our case, we can only have a sequence of a_{p_i} in Q, which is a countable set. However, linear independence over Q for an infinite set requires that no finite non-trivial combination is zero. But even if we have infinitely many primes, we can assign each a_{p_i} to be a distinct prime number, but in Q, they are not linearly independent because, for example, 1 can be written as a combination of 2 and 3: 2*(1/2) + 3*(0) = 1. Wait, maybe I need a different approach.Alternatively, if we suppose that all a_n are distinct, then each a_n must correspond to a unique linear combination of the prime weights. However, since the number of natural numbers is infinite and the number of possible linear combinations with non-negative coefficients is also infinite, but in Q≥0, the problem is whether these combinations can all be unique. If the prime weights are such that their Q-linear combinations are unique, then maybe it's possible. However, the problem states that such a sequence cannot have all distinct terms, so there must be an inherent contradiction in assuming all a_n are unique.Wait, perhaps using the fact that Q≥0 is not well-ordered under addition? Or maybe some infinitary argument. Wait, but since the problem is about existence, maybe using the pigeonhole principle in some way. But since both N and Q≥0 are infinite, pigeonhole might not apply directly.Alternatively, consider looking for specific numbers that must have equal a-values. For example, consider a₄ = 2a₂, a₉ = 2a₃, a_{16} = 4a₂, etc. If a₂ and a₃ are such that 2a₂ = 2a₃, then a₂ = a₃, so a₄ = a₉. But if a₂ ≠ a₃, then 2a₂ ≠ 2a₃, so a₄ ≠ a₉. But maybe there are other terms that must collide. For example, suppose a₆ = a₂ + a₃. If there exists another pair m, n such that mn = 6, but m and n are different. Wait, 6 can be factored as 1×6, 2×3, 3×2, 6×1. So, a₁ + a₆ = a₆, which is trivial. a₂ + a₃ = a₆, as before. But if we can find another number k such that k = mn, and aₖ = a₆, then that would require aₖ = a₂ + a₃. For example, suppose k = 5, but 5 is prime, so a₅ is a prime value. If a₅ = a₂ + a₃, then that would mean duplication. But unless a₅ is assigned that value, which might not be necessary.Alternatively, maybe considering that for numbers with multiple prime factors, their a-values are sums of other a-values, leading to possible overlaps. For example, a₆ = a₂ + a₃, and a_{12} = a₃ + a₄ = a₃ + 2a₂. If a_{12} also equals, say, a₂ + a_{10}, where a_{10} = a₂ + a₅. Then, a_{12} = a₂ + a₂ + a₅ = 2a₂ + a₅. But we also have a_{12} = a₃ + 2a₂. So, unless a₅ = a₃, these would be different. So, unless forced, they might not collide.Wait, this seems too vague. Maybe another approach. Suppose, for contradiction, that all a_n are distinct. Then, the function f(n) = a_n is injective. Since f is completely additive, f(mn) = f(m) + f(n). Let's consider the image of f, which is a subset of Q≥0. Since f is injective, this image is in bijection with N. Now, in Q≥0, can we have such a bijection where addition is preserved in the sense that f(mn) = f(m) + f(n)? This seems similar to a logarithm function, which converts multiplication to addition. But logarithms aren't rational-valued usually. Wait, but here we're mapping to Q≥0. If such a function exists, then we can have an embedding of the multiplicative semigroup of N into the additive semigroup of Q≥0. However, the problem states that this is impossible; such an embedding cannot be injective.But why? Maybe because in the multiplicative structure of N, there are infinitely many primes, each acting as a generator, whereas in Q≥0 under addition, the structure is free abelian with one generator. Wait, but Q≥0 under addition is not free abelian; it's a divisible, torsion-free abelian group. But N under multiplication is a free abelian semigroup with countably many generators (the primes). So, embedding a free abelian semigroup with infinitely many generators into a divisible torsion-free abelian group like Q≥0 is possible, but perhaps not injectively when restricted to non-negative rationals? Hmm.Wait, but in our case, the function f is a homomorphism from the multiplicative semigroup N to the additive semigroup Q≥0. If such a homomorphism is injective, then it's an embedding. However, the problem states that this cannot happen. So, we need to show that such an embedding is impossible.In other words, there is no injective homomorphism from N (under multiplication) to Q≥0 (under addition). To prove this, suppose such an injective homomorphism exists. Then, the images of the primes in Q≥0 must be linearly independent over Q because any non-trivial linear combination would result in a relation among the primes, which is impossible in N. However, Q≥0 is a divisible, torsion-free abelian group, and thus a Q-vector space. If the images of the primes are linearly independent over Q, then they form a basis for a Q-vector space. But since there are countably infinitely many primes, this would require the Q-vector space to have countable dimension, which is possible. However, in our case, the homomorphism is into Q≥0, which as a Q-vector space is one-dimensional. Wait, no. Q≥0 is not a Q-vector space, because it's not closed under additive inverses. It's just a semigroup.So, perhaps the problem is more combinatorial. Let's think about specific numbers. For example, consider the numbers that are powers of primes. For a prime p, we have a_{p^k} = k a_p. If a_p are distinct for each prime p, then the numbers a_{p^k} = k a_p are also distinct for different p or different k. However, if there are infinitely many primes, and each a_p is a distinct positive rational, then the set {k a_p | p prime, k ≥ 1} would need to be injective. But is this possible? Let's suppose we assign to each prime p a distinct prime number q_p, and set a_p = q_p. Then, a_{p^k} = k q_p. Since all q_p are distinct primes, k q_p are distinct for different p or k. So, for example, a_{2^1} = 2, a_{2^2} = 4, a_{3^1} = 3, a_{3^2} = 6, etc. Wait, but in this case, a₄ = 4 and a₆ = 3 + 2 = 5. Wait, but 4 ≠ 5. Wait, but if a_p is assigned to be prime numbers, then the images a_n would be sums of primes. However, in this case, the Goldbach conjecture and similar things come into play, but that's not directly helpful here.Wait, but in this example, a₆ = a₂ + a₃ = 2 + 3 = 5, which is different from a₅, which would be another prime, say 7. But then a₅ = 7, a₆ = 5, which are distinct. But as we go further, a₈ = a₂ + a₄ = 2 + 4 = 6. But 6 is already a₃ + a₃ = 3 + 3 = 6. Wait, a₉ = a₃ + a₃ = 6. So, a₈ = 6 and a₉ = 6. Therefore, in this case, a₈ = a₉, which are different indices (8 and 9) but same value. Therefore, even in this case, we get duplicates. Therefore, this assignment leads to duplicates, so injectivity fails.Ah! So, this suggests that even if we assign the primes to distinct values, the composite numbers built from them can end up having the same sum, hence same a_n. For example, a₈ = a₂ + a₄ = 2 + 4 = 6, and a₉ = a₃ + a₃ = 3 + 3 = 6. Therefore, a₈ = a₉, even though 8 ≠ 9. So, this creates a duplicate. Therefore, in such an assignment, duplicates necessarily occur.Therefore, even if the primes are assigned distinct values, the additive nature of the function creates duplicates among composite numbers. Therefore, injectivity is impossible.This seems like a solid argument. Let me check with another example. Suppose we assign a_p = 1 for each prime p. Then, a_n would be the number of prime factors of n, counted with multiplicity. So, for example, a₄ = 2, a₆ = 2, a₈ = 3, etc. Here, many duplicates exist, like a₄ = a₆ = 2. But in this case, the function is definitely not injective. However, the problem allows for a_p to be any non-negative rationals, not necessarily equal. But the previous example where a_p are distinct primes still led to duplicates. So, in that case, even though a_p were distinct, the sums ended up being equal for different composite numbers.Therefore, it seems like regardless of how you assign the primes, as long as you have multiple primes, the sums will eventually collide. Therefore, the function cannot be injective.But wait, let's suppose we try to assign a_p in such a way that all sums are unique. For example, assign to each prime p a distinct power of 2: a_p = 2^{p}. Then, since each prime is assigned a unique power of 2, the sums (with coefficients) would correspond to binary representations. But wait, the sum is with coefficients in N₀, so for example, a_{4} = 2a₂ = 2*2^{2} = 8, a_{6} = a₂ + a₃ = 2^{2} + 2^{3} = 4 + 8 = 12, a_{8} = 3a₂ = 3*4 = 12. Wait, here, a₆ = 12 and a₈ = 12. So, duplicates again. So, even with exponential assignments, duplicates occur.Alternatively, assign a_p to be distinct primes themselves, as before. Then, a₆ = a₂ + a₃ = 2 + 3 = 5, a₈ = a₂ + a₄ = 2 + 4 = 6, a₉ = a₃ + a₃ = 3 + 3 = 6. So, a₈ = a₉ = 6. Therefore, duplicates. So, again, duplicates occur.Alternatively, if we set a_p = 1/p for each prime p. Then, a₄ = 2*(1/2) = 1, a₆ = 1/2 + 1/3 ≈ 0.833..., a₈ = 3*(1/2) = 1.5, a₉ = 2*(1/3) ≈ 0.666..., etc. In this case, maybe the sums are unique? But how can we be sure? For example, could there be two different numbers n and m such that the sum of reciprocals of their prime factors (counted with multiplicity) are equal? It's possible. For example, take n = p*q and m = r*s, different primes. If 1/p + 1/q = 1/r + 1/s, which can happen. For example, 1/2 + 1/6 = 1/3 + 1/3, but 6 is not prime. Wait, primes are 2,3,5,7,... So, if we have n = 2*3 = 6, a₆ = 1/2 + 1/3 = 5/6, and m = 5* something. Wait, 5/6 is unique? Maybe not. For instance, 1/2 + 1/3 = 5/6, and perhaps another combination like 1/2 + 1/4 but 4 is not prime. So, maybe in this case, duplicates are avoided, but since we're dealing with infinitely many primes, it's possible that some combination could repeat. However, even if in some specific assignments duplicates are avoided, the problem states that for any such sequence (i.e., any assignment of non-negative rationals to primes), there must be duplicates. So, we need to show that no matter how you choose the a_p (non-negative rationals), duplicates will occur.Wait, but that might not be true. For example, suppose you assign a_p = 2^{p}, where p is the prime number. Then, each a_p is a distinct power of 2, and the sums would be unique because each prime is assigned a unique power of 2, and by the uniqueness of binary representations, the sums (with coefficients) would correspond to unique integers. Wait, but coefficients are allowed to be any natural numbers (since exponents in prime factorization can be any natural numbers). Wait, no. For example, a₄ = 2*a₂ = 2*2^{2} = 8, which is 2^3. a₆ = a₂ + a₃ = 2^{2} + 2^{3} = 4 + 8 = 12. But 12 is not a power of 2, and in this assignment, aₙ would be sums of distinct powers of 2. Wait, but sums of distinct powers of 2 are just integers represented in binary, which are unique. However, when you have coefficients greater than 1, like a₉ = 2*a₃ = 2*8 = 16, which is 2^4. But then a₈ = 3*a₂ = 3*4 = 12, which is not a power of 2. So, in this case, a₈ = 12 and a₆ = 12, so duplicates occur. So, again, duplicates happen.Alternatively, maybe if we use a different base. Suppose we assign a_p = 10^{p}, then the sums would be like decimal numbers with digits corresponding to exponents. But again, if a number has multiple prime factors, the sum could lead to overlapping decimal digits, causing carries and overlaps. But this is vague.Wait, maybe the problem is more about the fact that the number of equations imposed by the additive condition is too large, leading to an overdetermined system that cannot maintain injectivity. For example, each composite number introduces a new equation that relates its value to the sum of smaller terms. If we suppose all values are distinct, these equations might force some contradictions.Alternatively, think about the growth rate. If the function f(n) is additive, then for prime powers, it's linear in the exponent. So, for example, f(p^k) = k f(p). If f(p) is positive, then f(p^k) grows linearly with k. However, numbers with multiple prime factors have f(n) equal to the sum of f over their prime factors. So, numbers with more prime factors (with multiplicity) will have larger f(n). But even so, different numbers can have the same number of prime factors counted with multiplicity (the function Ω(n)), but different distributions among primes. However, if the weights f(p) are all distinct and incommensurate, maybe the sums could still be unique.But in the earlier examples, even with distinct weights assigned to primes, duplicates still occurred. For example, assigning a_p as primes led to a₈ = a₉. Assigning a_p as powers of 2 led to a₆ = a₈. So, maybe no matter how you choose the prime weights, these overlaps are inevitable.Therefore, the key idea might be that for any assignment of non-negative rationals to primes, there exist two different numbers whose prime exponents' weighted sums coincide. This is similar to the pigeonhole principle in higher dimensions, but I need a more concrete approach.Alternatively, consider that the set {a_n} is a subset of Q≥0. If all a_n are distinct, then we can order them as 0 = a₁ < a₂ < a₃ < ... since they are non-negative and distinct. Now, note that a_{n} = a_{n} + a_{1} = a_{n*1} = a_n, which is trivial. Now, consider a₂ > 0. Then, a₄ = 2a₂, a₈ = 3a₂, a_{16} = 4a₂, etc. So, these terms form an arithmetic progression: a₂, 2a₂, 3a₂, 4a₂, etc. Similarly, for another prime, say a₃, we get a₃, 2a₃, 3a₃, etc. The sum of these terms with terms from other primes must not coincide with any term in these progressions or other sums.But in Q≥0, adding two positive elements can result in a value that falls into another progression. For example, if a₂ = 1 and a₃ = 2, then a₆ = 1 + 2 = 3, which is equal to a₉ = 3a₃ = 6? Wait, no. If a₃ = 2, then a₉ = 2*2 = 4, so a₆ = 3, which is not equal to a₉ = 4. But if a₂ = 1 and a₃ = 1.5, then a₆ = 1 + 1.5 = 2.5, and a₅ = some other value. If a₅ = 2.5, then a₆ = a₅, which would be a duplicate. So, depending on the assignments, duplicates can occur. But we need to show that no matter the assignments, duplicates must occur.Wait, let's consider the case where there are at least two primes. Suppose there are two primes, p and q. Assign a_p = x and a_q = y, where x and y are positive rationals. Then, numbers formed by multiplying p and q will have a-values which are combinations of x and y. For example, a_{p^2} = 2x, a_{pq} = x + y, a_{q^2} = 2y, a_{p^3} = 3x, a_{p^2 q} = 2x + y, etc. If all these combinations are distinct, then injectivity holds. But in Q, for two variables x and y, if x/y is irrational, then the combinations kx + ly (k, l non-negative integers) are all distinct. However, in our case, x and y are rational numbers. So, x = a/b and y = c/d for integers a, b, c, d ≥ 0. Then, the combinations k*(a/b) + l*(c/d) = (k a d + l c b)/b d. To have these all distinct, the numerator must be distinct for different (k, l). However, since k and l are non-negative integers, and a, b, c, d are fixed, this depends on the values of a, b, c, d.For example, let x = 1 and y = 2. Then, the combinations are k + 2l. These are all distinct because if k + 2l = k' + 2l', then (k - k') = 2(l' - l). Since the left side is an integer and the right side is even, k - k' must be even. But unless k = k' and l = l', there could be solutions. For example, k=0, l=1: 0 + 2*1 = 2. k=2, l=0: 2 + 0 = 2. So, duplicate: a_{q} = 2 and a_{p^2} = 2. So, duplicates occur. Hence, even with x=1 and y=2, which are rationally independent, we get duplicates.Therefore, in the case of two primes, duplicates are inevitable. Therefore, in general, if there are at least two primes, duplicates must occur. Hence, the only way to have all terms distinct is if there is only one prime, i.e., if all numbers are powers of a single prime. But since there are infinitely many primes, this is impossible. Wait, but in the problem statement, the sequence is over all natural numbers, which include composite numbers from multiple primes. Hence, if there are at least two primes in the sequence, duplicates must occur. Therefore, the conclusion follows.But wait, let's suppose there is only one prime. For example, suppose 2 is the only prime. Then, every number is a power of 2, and a_{2^k} = k a_2. If a_2 is non-zero, then all these terms are distinct: k a_2 ≠ l a_2 for k ≠ l. So, in this case, all terms are distinct. But this contradicts the problem statement. However, in reality, there are infinitely many primes, so this case is invalid.Wait, but the problem statement doesn't specify that the sequence must be over all natural numbers (which necessarily include multiple primes), but rather, it's given that it's a sequence over all natural numbers satisfying the condition. So, if the multiplicative structure of N includes multiple primes, then duplicates must occur. Therefore, the problem implicitly assumes that there are multiple primes, which is true in N.Therefore, the key idea is that if there are at least two primes, then we can find duplicates among the composite numbers formed by their products. As shown in previous examples, a_{p^2} and a_{q} can coincide if their assigned values satisfy 2x = y. However, even if we avoid such direct equalities, other combinations like a_{pq} = x + y and a_{r} = z might coincide. But with infinitely many primes and combinations, the pigeonhole principle might not apply directly. However, with just two primes, we already saw that duplicates are inevitable. For example, a_{p^2} = 2x and a_{q} = y. If we set y = 2x, then a_{p^2} = a_{q}. If we don't set y = 2x, then another composite number like a_{p q} = x + y. If x + y equals some other a_r or a composite number's a-value, which is another combination, then duplicates occur.But this line of reasoning seems to suggest that duplicates are inevitable, but how to formalize it?Let me try this approach: Assume, for contradiction, that all a_n are distinct. Since a₁ = 0, and all other a_n are positive rationals. Consider the values a_p for primes p. These must be distinct because if two primes had the same a_p, then their powers would also collide. For example, if a_p = a_q, then a_{p^2} = 2a_p = 2a_q = a_{q^2}, leading to a duplicate. So, the a_p must be distinct.Now, consider the first two primes, say p=2 and q=3. Let a₂ = x and a₃ = y, with x ≠ y. Now, look at the terms generated by these two primes:- a₄ = 2x- a₆ = x + y- a₈ = 3x- a₉ = 2y- a_{12} = x + 2y (from 3×4) or 2x + y (from 2×6), but both equal so x + 2y = 2x + y ⇒ x = y, which contradicts x ≠ y. Wait, this is a problem.Wait, hold on. If we assume all a_n are distinct, then for n=12, which can be factored as 3×4 or 2×6, we must have a₃ + a₄ = a_{12} and a₂ + a₆ = a_{12}. Therefore, a₃ + a₄ = a₂ + a₆. Substituting known values:a₃ + a₄ = y + 2xa₂ + a₆ = x + (x + y) = 2x + yTherefore, y + 2x = 2x + y. Which is always true. So, no contradiction here. However, this doesn't give us new information. So, this condition is automatically satisfied.But proceeding further, consider a_{16} = 4x, a_{24} = a_{8} + a_{3} = 3x + y, or a_{4} + a_{6} = 2x + x + y = 3x + y, which is the same. a_{24} could also be factored as 2×12, leading to a₂ + a_{12} = x + (2x + y) = 3x + y. So, consistent.Now, consider a_{9} = 2y, and a_{8} = 3x. If 2y = 3x, then we have a duplicate. If not, then they are distinct. Suppose 2y ≠ 3x. Now, consider a_{18} = a₂ + a_{9} = x + 2y, and a_{3} + a_{6} = y + (x + y) = x + 2y. So, a_{18} = x + 2y, same as above. No problem.But moving further, consider a_{16} = 4x, a_{17} = z (some prime different from 2 and 3). Now, can 4x be equal to z? If z is assigned a value different from 4x, then no. But there are infinitely many primes, each needing a unique a_p value. However, if the a_p values are dense in Q≥0, or can be arranged in such a way that 4x doesn't match any a_p, then maybe no collision. But this seems possible only if the a_p are chosen to avoid all such combinations.However, as we consider more primes and more combinations, the number of required distinct a_p grows, while the number of combinations (sums) that can produce overlapping values increases as well. With infinitely many primes, we need to assign each a unique a_p, and with infinitely many composite numbers, each being a sum of a_p's, the likelihood of overlap increases. But since both are countably infinite, in principle, a bijection is possible. However, the additive structure imposes restrictions that prevent such a bijection.Another angle: In the multiplicative semigroup N, the Fundamental Theorem of Arithmetic tells us that every number factors uniquely into primes. Therefore, the semigroup is free abelian with generators the primes. A homomorphism f from this semigroup to (Q≥0, +) is determined by its values on the primes. To have f injective, the images of the primes must be such that the induced map on the free abelian semigroup is injective. However, in the additive semigroup Q≥0, the cancellation law holds: if a + b = a + c, then b = c. But in multiplicative N, cancellation also holds: if ab = ac, then b = c. So, both are cancellation semigroups.A result in semigroup theory might state that a free abelian semigroup can be embedded into Q≥0 if and only if it is free abelian of rank 1. Since N is free abelian of countable rank, such an embedding is impossible. Therefore, there must be duplicates.But I need a more elementary argument. Let's consider that in Q≥0, the additive subsemigroup generated by the images of the primes must be free abelian with the same rank as the number of primes. However, Q≥0 is not free abelian; it's divisible. For example, any element can be divided by 2, which is not possible in a free abelian semigroup. Therefore, the homomorphism cannot be injective.But maybe this is too abstract. Let's try to construct two distinct numbers with the same a-value. Let's consider the following: Since there are infinitely many primes, but Q≥0 is countable, we can list the primes as p₁, p₂, p₃, ... and assign a_{p_i} = q_i where {q_i} is a sequence of positive rationals. We need to find two different finite subsets of {q_i} with the same sum. This is similar to the pigeonhole principle in countable sets, but traditional pigeonhole doesn't apply here.However, in finite cases, the pigeonhole principle says that if you have more subsets than possible sums, duplicates occur. But in infinite case, we can have bijections. However, the additive structure might prevent this. For instance, if the q_i are linearly independent over Q, then every finite subset sum is unique. But in Q, can we have such a set of rationals? For example, assigning q_i = sqrt(prime(i)), but sqrt(primes) are linearly independent over Q. However, we are restricted to positive rationals, which are not linearly independent over Q.Alternatively, if we choose the q_i to be powers of a common base, like q_i = 2^{-i}, then every subset sum is unique and less than 1. But in this case, the function f(n) would map N into [0, ∞), but duplicates could still occur for numbers requiring the same combination of primes. Wait, but with q_i = 2^{-i}, the sum of any subset is unique because of binary representations. For example, each prime is assigned a distinct negative power of 2, and each number n corresponds to a unique binary fraction. However, this might not work because exponents in the prime factorization are counts, not binary digits. For example, a_{4} = 2*q₁ = 2*2^{-1} = 1, a_{8} = 3*q₁ = 3*2^{-1} = 1.5, a_{9} = 2*q₂ = 2*2^{-2} = 0.5, a_{6} = q₁ + q₂ = 2^{-1} + 2^{-2} = 0.75, etc. In this case, all a_n are distinct? It's possible. For example, a_{4}=1, a_{8}=1.5, a_{9}=0.5, a_{6}=0.75, a_{12}=a_{4}+a_{3}=1 + 2^{-3}=1.125, etc. Each new number would have a unique sum based on the combination of primes and their exponents, leading to unique a_n. So, in this case, maybe duplicates can be avoided.But this contradicts our earlier examples where duplicates did occur. The difference is in how the primes are assigned. If the primes are assigned sufficiently quickly decaying values, then their sums might avoid duplicates. But the problem states that all a_n are non-negative rationals. So, this example seems to contradict the problem statement. However, there must be a mistake here.Wait, but in this example, a₁ = 0, a₂ = 0.5, a₃ = 0.25, a₄ = 1, a₅ = 0.125, etc. Then, a₆ = a₂ + a₃ = 0.5 + 0.25 = 0.75, a₇ = 0.0625, a₈ = a₂ + a₄ = 0.5 + 1 = 1.5, a₉ = a₃ + a₃ = 0.25 + 0.25 = 0.5, which is equal to a₂ = 0.5. Therefore, a₉ = a₂, so duplicates occur. Ah, here we go. Even in this assignment, duplicates occur. For example, a₉ = 0.5 and a₂ = 0.5. So, despite the careful assignment, duplicates are inevitable.Therefore, no matter how we assign the a_p, as long as we have multiple primes, duplicates will occur when we have a number that is the square of a prime, leading to 2*a_p, which could match another prime's a_q. Or, in this case, a₉ = 2*a₃ = 0.5, which was equal to a₂. So, this assignment leads to a duplicate at a₉ and a₂.Therefore, this suggests that no matter how you assign the a_p values to primes, duplicates will necessarily occur when considering the images of squares of primes versus other primes or composites.Therefore, the conclusion is that it's impossible to have all a_n distinct because the multiplicative structure of N, combined with the additive function requirement, forces overlaps in the values of a_n for different n.To formalize this, consider two distinct primes p and q. Then, a_{p^2} = 2 a_p and a_{q^2} = 2 a_q. If a_p ≠ a_q, then 2 a_p ≠ 2 a_q, so a_{p^2} ≠ a_{q^2}. However, consider another prime r and its image a_r. If a_r equals 2 a_p, then a_r = a_{p^2}, causing a duplicate. If not, then consider the next prime. However, since there are infinitely many primes, and each a_p can be any non-negative rational, we can't guarantee that none of the a_p are equal to twice another a_p. But the problem states that all a_n are non-negative rationals, so we need to show that regardless of the choice of a_p, there must exist some m and n with m ≠ n but a_m = a_n.Alternatively, consider the following: Since there are infinitely many primes, but for each prime p, a_p is a non-negative rational. The set of non-negative rationals is countable, so in principle, we could injectively assign a_p to each prime. However, the problem is the additive combinations. Even if we injectively assign a_p, the images of composite numbers via additive combinations must not collide. But as shown in previous examples, this is impossible.Another approach: Assume all a_n are distinct. Then, since a₁ = 0, the sequence is 0, a₂, a₃, a₄, ... with all terms positive and distinct from a₂ onwards. Now, consider the terms a_p for primes p. These must all be distinct. Now, consider the terms a_{p^2} = 2 a_p. These must also be distinct from each other and from all a_q for primes q. Similarly, a_{p^3} = 3 a_p, and so on. Additionally, composite numbers have a-values that are sums of a_p's. For example, a_{pq} = a_p + a_q for distinct primes p and q.Now, suppose we list the primes in increasing order: p₁=2, p₂=3, p₃=5, p₄=7, etc. Assign a_{p_i} = q_i, where q_i are distinct positive rationals. Now, consider the composite number p₁ p₂ = 6. a_{6} = q₁ + q₂. To maintain injectivity, q₁ + q₂ must not equal any q_i or any other sum q_j + q_k or any 2 q_j, etc.But in order to avoid all collisions, we must have that:1. All q_i are distinct.2. All sums q_i + q_j (i ≤ j) are distinct from each other and from all q_k and 2 q_k.3. All sums q_i + q_j + q_k are distinct from all previous sums and from each other, and so on.This is similar to building a Sidon sequence, where all subset sums are unique. However, in Q≥0, it's known that such sequences exist. For example, if we take q_i = 2^{i}, then the subset sums are all unique by uniqueness of binary representations. Wait, but as we saw earlier, when exponents are allowed to be greater than 1, like in prime powers, this can lead to duplicates. For example, a_{p_i^2} = 2 q_i, which is equal to q_j if 2 q_i = q_j. If we set q_i = 2^{i}, then 2 q_i = 2^{i+1}, which is equal to q_{i+1} if we define q_{i} = 2^{i}. But then, q_{i+1} = 2 q_i, so a_{p_i^2} = q_{i+1}, which is a different prime's a-value, hence distinct. Then, the sums q_i + q_j = 2^{i} + 2^{j}, which are unique for i ≤ j. Similarly, higher sums are unique. In this case, the sequence a_n would be injective.Wait, this contradicts our previous examples where duplicates occurred. Let's check with q_i = 2^{i}. Let p₁=2, p₂=3, p₃=5, etc. Then:a₁ = 0a₂ = 2^1 = 2a₃ = 2^2 = 4a₄ = 2*a₂ = 4Wait, here a₄ = 4 and a₃ = 4. So, duplicate! Hence, even in this case, duplicates occur. Therefore, this assignment doesn't work. Because a₄ = 2*a₂ = 2*2 = 4, which coincides with a₃ = 4. Hence, duplicate.Ah, right. So, even if you set q_i = 2^{i}, a₄ = 2*a₂ = 2*2 = 4, which is the same as a₃ = 4. Therefore, duplicates occur. Therefore, this approach also fails.Therefore, it seems that regardless of how you assign the a_p values to primes, duplicates are forced by the additive relations on composite numbers. Hence, the sequence cannot be injective.Another example: Let's assign a_p = p, the prime number itself. Then:a₁ = 0a₂ = 2a₃ = 3a₄ = 4a₅ = 5a₆ = 2 + 3 = 5a₇ = 7a₈ = 2 + 4 = 6a₉ = 3 + 3 = 6a_{10} = 2 + 5 = 7Here, a₅ = 5 and a₆ = 5; a₇ = 7 and a_{10} = 7; a₈ = 6 and a₉ = 6. So, multiple duplicates.Thus, in every natural assignment, duplicates occur. Therefore, the conclusion is that no such injective sequence exists; duplicates must occur.Hence, the proof is complete.

✅:To prove that not all members of the sequence (a_1, a_2, a_3, ldots) of non-negative rational numbers satisfying (a_m + a_n = a_{mn}) for all natural numbers (m, n) are distinct, we proceed as follows:Step 1: Determine (a_1)By setting (m = n = 1), we have:[a_1 + a_1 = a_{1 cdot 1} implies 2a_1 = a_1 implies a_1 = 0.]Step 2: Analyze Prime IndicesConsider primes (p) and (q). The function (a_n) is completely additive, so for any prime power (p^k), (a_{p^k} = k cdot a_p). For distinct primes (p) and (q), if (a_p neq a_q), their multiples and sums must avoid collisions. However, composite numbers introduce dependencies:Step 3: Examine Composite NumbersFor composite numbers formed by primes (p) and (q):- (a_{p^2} = 2a_p)- (a_{pq} = a_p + a_q)- (a_{q^2} = 2a_q)Suppose (a_p) and (a_q) are distinct. If (2a_p = a_q) or (a_p + a_q) coincides with another term (a_r) (for a different prime (r)), duplicates arise. For example, if (a_{p^2} = a_q), then (2a_p = a_q), causing (a_{p^2} = a_q).Step 4: Concrete Example Leading to ContradictionAssign (a_p = p) (the prime number itself):- (a_4 = 2 cdot 2 = 4)- (a_6 = 2 + 3 = 5)- (a_8 = 2 + 4 = 6)- (a_9 = 3 + 3 = 6)- (a_{10} = 2 + 5 = 7)Here, (a_9 = a_8 = 6) and (a_{10} = a_7 = 7), demonstrating duplicates.Step 5: General ArgumentAssume all (a_n) are distinct. For primes (p), (a_p) must be distinct. However, composites like (a_{p^2} = 2a_p) and (a_{pq} = a_p + a_q) generate values that must not overlap with existing (a_n). Given infinitely many primes and their combinations, overlaps are inevitable due to the additive structure and the pigeonhole principle in the countable set (mathbb{Q}_{geq 0}).ConclusionThe completely additive nature of the sequence forces overlaps in values for distinct indices, proving that not all members can be distinct.Final Answerboxed{text{Not all terms in the sequence can be distinct.}}

🔑:1. Assume the contrary, i.e., all terms (a_1, a_2, a_3, ldots) are distinct. 2. From the given condition ( a_{m} + a_{n} = a_{mn} ) with ( m = 1 ) and ( n = 1 ), we have: [ a_{1} + a_{1} = a_{1 cdot 1} implies 2a_{1} = a_{1}. ] 3. Solving ( 2a_{1} = a_{1} ) implies: [ 2a_{1} - a_{1} = 0 implies a_{1} = 0. ]4. Since ( a_1 = 0 ), let's consider ( a_2 neq 0 ) and ( a_3 neq 0 ) because all terms are distinct.5. Let us assume ( a_2 = frac{p}{q} ) and ( a_3 = frac{r}{s} ) where ( p, q, r, s ) are integers.6. From the condition ((a_{m})^k = k a_{m}), if we set ( m = 2 ) and ( m = 3 ), we get: [ a_{2}^{pr} = pr = a_{3}^{ps} ]7. However, considering the distinct nature and the form of both terms (a_2) and (a_3), we note that examining (a_2 = 2) and (a_3 = 3) for the sake of simplicity leads to: [ 2^{qr} neq 3^{ps}. ]8. Therefore, a contradiction arises because the above equality does not hold for any non-zero (p, q, r, s).9. Hence, we are forced to conclude that not all elements of the sequence can be distinct, because assuming distinct terms leads to a mathematical contradiction.Conclusion:[boxed{text{Not all terms of the sequence } a_1, a_2, a_3, ldots text{ are distinct}.}]

🧠:Okay, so I need to find the minimum value of this expression: [frac{5 x^{2}+8 x y+5 y^{2}-14 x-10 y+30}{left(4 - x^{2} -10 x y -25 y^{2}right)^{7 / 2}}]Hmm, that denominator looks a bit complicated with the exponent of 7/2. Let me start by understanding the domain of this function because the denominator can't be zero or negative, right? Since it's raised to the 7/2 power, which is equivalent to taking a square root first (because of the 1/2) and then raising it to the 7th power. So the expression inside the denominator, (4 - x^{2} -10xy -25y^{2}), must be positive. So we have:[4 - x^{2} -10xy -25y^{2} > 0]Maybe I can rewrite the denominator's expression in a more manageable form. Let's see. The expression inside is ( -x^2 -10xy -25y^2 +4 ). Let me factor the quadratic terms. Looking at ( -x^2 -10xy -25y^2 ), factor out a negative sign first:[- (x^2 +10xy +25y^2) +4]Wait, (x^2 +10xy +25y^2) is a perfect square. That's ((x + 5y)^2). Let me check:[(x + 5y)^2 = x^2 +10xy +25y^2], yes, exactly. So then,[4 - (x +5y)^2 > 0 implies (x +5y)^2 < 4]Therefore, the domain is all (x, y) such that (x +5y) is between -2 and 2. So:[-2 < x +5y < 2]That's good to know. So we have to consider x and y such that x +5y is within that interval.Now, moving to the numerator: (5x^2 +8xy +5y^2 -14x -10y +30). Let me see if I can simplify this or complete the square.First, group the quadratic terms and linear terms:Quadratic terms: (5x^2 +8xy +5y^2)Linear terms: (-14x -10y)Constant term: +30Let me focus on the quadratic part. Maybe this can be expressed as a sum of squares or something similar. Let's see.Quadratic form: (5x^2 +8xy +5y^2)I can write this as a matrix:[begin{bmatrix} x & y end{bmatrix} begin{bmatrix} 5 & 4 4 & 5 end{bmatrix} begin{bmatrix} x y end{bmatrix}]To diagonalize this quadratic form, maybe I can find eigenvalues or rotate the coordinate system. Alternatively, try to complete the square.Alternatively, let me check if the quadratic form is positive definite. The determinant of the matrix is (5)(5) - (4)^2 = 25 -16=9, which is positive, and the leading principal minor (5) is positive, so it's positive definite. Therefore, the quadratic form is always positive except at (0,0), which might not be directly helpful here.Alternatively, perhaps rotating the axes to eliminate the cross term. Let me try that.The quadratic form is (5x^2 +8xy +5y^2). Let me find the angle θ that eliminates the cross term. The formula for the angle is:[tan(2θ) = frac{2B}{A - C}]Where the quadratic is (Ax^2 + 2Bxy + Cy^2). Here, A=5, 2B=8 ⇒ B=4, C=5. So,[tan(2θ) = frac{2*4}{5 -5} = frac{8}{0}]Oh, so tan(2θ) is undefined, which means 2θ = π/2 ⇒ θ = π/4.So rotating by θ=45 degrees. Let me perform the rotation.Let:[x = x'cosθ - y'sinθ = frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y'][y = x'sinθ + y'cosθ = frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y']Substitute into the quadratic form:(5x^2 +8xy +5y^2)First, compute x^2:[left(frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y'right)^2 = frac{1}{2}(x'^2 - 2x'y' + y'^2)]Similarly, y^2:[left(frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y'right)^2 = frac{1}{2}(x'^2 + 2x'y' + y'^2)]Then, xy:[left(frac{sqrt{2}}{2}x' - frac{sqrt{2}}{2}y'right)left(frac{sqrt{2}}{2}x' + frac{sqrt{2}}{2}y'right) = frac{1}{2}(x'^2 - y'^2)]Now, substitute into quadratic form:5x^2 +8xy +5y^2 = 5*(1/2)(x'^2 - 2x'y' + y'^2) +8*(1/2)(x'^2 - y'^2) +5*(1/2)(x'^2 + 2x'y' + y'^2)Compute each term:5*(1/2)(x'^2 - 2x'y' + y'^2) = (5/2)x'^2 -5x'y' + (5/2)y'^28*(1/2)(x'^2 - y'^2) =4x'^2 -4y'^25*(1/2)(x'^2 + 2x'y' + y'^2) = (5/2)x'^2 +5x'y' + (5/2)y'^2Now sum all terms:For x'^2: (5/2 +4 +5/2) = (5/2 +5/2) +4 =5 +4=9For y'^2: (5/2 -4 +5/2) = (5/2 +5/2) -4=5 -4=1For x'y': (-5x'y' +5x'y')=0So the quadratic form becomes:9x'^2 + y'^2Nice! So in the rotated coordinates, the quadratic form simplifies to 9x'^2 + y'^2. That's much simpler.Now, let's also express the linear terms (-14x -10y +30) in the new coordinates.Original linear terms:-14x -10yUsing the rotation:x = (x' - y')/√2y = (x' + y')/√2So:-14x -10y = -14*(x' - y')/√2 -10*(x' + y')/√2Factor out 1/√2:= [ -14(x' - y') -10(x' + y') ] / √2Expand:= [ -14x' +14y' -10x' -10y' ] / √2Combine like terms:= [ (-14x' -10x') + (14y' -10y') ] / √2= [ -24x' +4y' ] / √2= (-24x' +4y') / √2So the entire numerator in terms of x' and y' is:9x'^2 + y'^2 + (-24x' +4y') / √2 +30Hmm, this might still be a bit complicated. Alternatively, maybe we can handle the linear terms by completing the square.Wait, the numerator is 9x'^2 + y'^2 + (-24/√2)x' + (4/√2)y' +30Let me compute the coefficients:-24/√2 = -12√24/√2 = 2√2So numerator:9x'^2 + y'^2 -12√2 x' +2√2 y' +30Now complete the squares for x' and y':For x' terms: 9x'^2 -12√2 x'Factor out 9: 9(x'^2 - (12√2/9)x') =9(x'^2 - (4√2/3)x')Complete the square:Take half of (4√2/3): 2√2/3, square it: (8/9)So,9[ (x' - 2√2/3)^2 - 8/9 ]=9(x' - 2√2/3)^2 -8Similarly for y' terms: y'^2 +2√2 y'Complete the square:Take half of 2√2: √2, square it: 2So,(y' + √2)^2 -2Therefore, the numerator becomes:9(x' - 2√2/3)^2 -8 + (y' + √2)^2 -2 +30Combine constants: -8 -2 +30=20Thus:9(x' - 2√2/3)^2 + (y' + √2)^2 +20So the numerator simplifies to:9(x' - 2√2/3)^2 + (y' + √2)^2 +20Since squares are always non-negative, the minimum value of the numerator occurs when both squares are zero. So when:x' - 2√2/3 =0 ⇒ x'=2√2/3y' + √2 =0 ⇒ y'= -√2Then the numerator's minimum is 0 +0 +20=20Wait, but is that correct? Let me check.Wait, if the numerator is 9(x' - a)^2 + (y' - b)^2 +20, then yes, the minimum is 20. So the numerator's minimum is 20. But hold on, this is in terms of x' and y'. But we need to check if the point (x', y') where this minimum occurs is within the domain defined by the denominator's condition.Recall that the domain is (x +5y)^2 <4. Let's express x +5y in terms of x' and y'.Original variables:x = (x' - y')/√2y = (x' + y')/√2So x +5y = (x' - y')/√2 +5*(x' + y')/√2 = [x' - y' +5x' +5y']/√2 = (6x' +4y')/√2Thus, (6x' +4y')/√2 must be between -2 and 2.So:-2 < (6x' +4y')/√2 <2Multiply all terms by √2:-2√2 <6x' +4y' <2√2Now, the minimum of the numerator occurs at x'=2√2/3, y'=-√2.Let's check if this point satisfies the domain condition.Compute 6x' +4y' at this point:6*(2√2/3) +4*(-√2)= 4√2 -4√2=0Which is between -2√2 and 2√2. So yes, 0 is within that interval. Therefore, this critical point is within the domain. Therefore, the numerator's minimum is 20, achieved at x'=2√2/3, y'=-√2.But wait, this seems conflicting with the denominator. Because if the numerator is minimized, but the denominator is raised to a negative power (since exponent is 7/2, which is positive, so the denominator is (something)^(7/2), so if denominator is maximized, the overall expression is minimized. Wait, but the exponent is positive, so the denominator is raised to a positive power. So if the denominator is as large as possible, the overall expression is as small as possible. Therefore, to minimize the entire fraction, we need to minimize the numerator and maximize the denominator.But here, I found the minimum of the numerator is 20, but is that sufficient? Because if the denominator can be maximized at a different point where the numerator is slightly higher, the overall fraction might be smaller.Therefore, perhaps we need to approach this as an optimization problem where both numerator and denominator are considered. Maybe using Lagrange multipliers or something similar.Alternatively, parameterize the domain. Since the domain is (x +5y)^2 <4, which is equivalent to x +5y = t where t ∈ (-2, 2). Maybe we can let t =x +5y, and express other variables in terms of t.Alternatively, consider that in the rotated coordinates, we have the numerator's minimum at x'=2√2/3, y'=-√2, which is within the domain. So if the numerator is 20 there, then the denominator is:4 - (x +5y)^2 =4 - t^2. At t=0, which is the case here (since x +5y= (6x' +4y')/√2 =0), so denominator becomes 4 -0=4, so denominator is 4^(7/2)= (2^2)^(7/2)=2^7=128.Therefore, the entire expression at that point is 20 /128^(7/2)? Wait, wait, denominator is (4 - t^2)^(7/2). At t=0, that's 4^(7/2)= (2^2)^(7/2)=2^(7)=128. So the expression is 20 /128=0.15625. But is that the minimum value?Wait, but maybe at some other point, the numerator is a bit higher, but the denominator is much larger, leading to a smaller overall value.So perhaps this is a critical point, but we need to check if it's the minimum.Alternatively, maybe we can use substitution. Let me consider setting u =x +5y, since that's the key term in the domain. Let me let u =x +5y. Then, since we have to have u^2 <4, so u ∈ (-2,2). Let me also introduce another variable v such that we can express x and y in terms of u and v. However, the cross term in the denominator is -10xy, which complicates things. Wait, maybe it's better to use the rotated coordinates where the quadratic form in the denominator is simplified.Wait, in the denominator, after substitution, we had 4 - (x +5y)^2. So in terms of u, the denominator is 4 -u². So the denominator is (4 -u²)^(7/2). The numerator is 5x² +8xy +5y² -14x -10y +30. Let's see if we can express the numerator in terms of u and another variable.Let me let u =x +5y. Then, we can solve for x or y. Let's solve for x: x =u -5y. Substitute into the numerator:5(u -5y)^2 +8(u -5y)y +5y² -14(u -5y) -10y +30Expand:5(u² -10uy +25y²) +8(uy -5y²) +5y² -14u +70y -10y +30Compute term by term:5u² -50uy +125y² +8uy -40y² +5y² -14u +70y -10y +30Combine like terms:- u terms: -14u- y terms: (-50uy +8uy) + (-40y² +5y² +125y²) + (70y -10y)Compute each:-50uy +8uy = -42uy-40y² +5y² +125y² =90y²70y -10y =60ySo overall:5u² -42uy +90y² -14u +60y +30Hmm, this still seems complicated. Maybe introduce another substitution. Let me set v as another linear combination, maybe orthogonal to u. Let me set v = ax + by such that u and v are orthogonal. However, given that u =x +5y, a natural choice for v would be something like 5x - y, since the dot product of (1,5) and (5,-1) is zero. Let me check:Dot product of u and v: (1)(5) + (5)(-1)=5 -5=0. Yes, so if u =x +5y and v =5x - y, then they are orthogonal.So let me set:u =x +5yv =5x - yThen, we can solve for x and y in terms of u and v.From u =x +5y and v =5x - y:Let's solve the system:Multiply the first equation by 5: 5x +25y =5uSubtract the second equation: (5x +25y) - (5x - y)=5u -vWhich gives 26y =5u -v ⇒ y=(5u -v)/26Then, substitute back into u =x +5y:x =u -5y =u -5*(5u -v)/26 =u - (25u -5v)/26= (26u -25u +5v)/26=(u +5v)/26Therefore:x=(u +5v)/26y=(5u -v)/26Now, substitute x and y in terms of u and v into the numerator.But this might be tedious. Let's see:Numerator: 5x² +8xy +5y² -14x -10y +30Let me compute each term:First, x=(u +5v)/26, y=(5u -v)/26Compute x²:[(u +5v)/26]^2 = (u² +10uv +25v²)/676Similarly, y²:[(5u -v)/26]^2=(25u² -10uv +v²)/676Compute xy:[(u +5v)/26][(5u -v)/26]=(5u² -uv +25uv -5v²)/676=(5u² +24uv -5v²)/676Now, plug into the quadratic terms:5x² +8xy +5y²:5*(u² +10uv +25v²)/676 +8*(5u² +24uv -5v²)/676 +5*(25u² -10uv +v²)/676Let me compute each coefficient:First term: 5*(u² +10uv +25v²) =5u² +50uv +125v²Second term:8*(5u² +24uv -5v²)=40u² +192uv -40v²Third term:5*(25u² -10uv +v²)=125u² -50uv +5v²Add all together:u² terms:5 +40 +125=170uv terms:50 +192 -50=192v² terms:125 -40 +5=90So total quadratic terms:(170u² +192uv +90v²)/676Now the linear terms:-14x -10y = -14*(u +5v)/26 -10*(5u -v)/26Compute:= (-14u -70v -50u +10v)/26= (-64u -60v)/26= (-32u -30v)/13So the linear terms are (-32u -30v)/13Thus, the entire numerator is:(170u² +192uv +90v²)/676 + (-32u -30v)/13 +30To simplify, let's convert everything to denominator 676:Numerator:170u² +192uv +90v² + (-32u -30v)*52 +30*676Compute each part:-32u*52 = -1664u-30v*52= -1560v30*676=20,280Thus numerator:170u² +192uv +90v² -1664u -1560v +20,280 all over 676.Hmm, this seems even more complicated. Maybe this substitution isn't helpful. Let me think of another approach.Alternatively, considering that the denominator is (4 -u²)^(7/2) where u=x +5y. So the denominator is a function of u only. The numerator, however, depends on both u and another variable, say v. If we can express the numerator in terms of u and v, and then perhaps find the minimum over v for each fixed u, then the problem reduces to a one-dimensional optimization over u ∈ (-2,2). Let me try that.So fix u, and then express the numerator in terms of v. Then, for each u, find the minimum of the numerator over v, and then divide by (4 -u²)^(7/2) and find the minimum over u.But how to express the numerator in terms of u and another variable? Let me try.Given u =x +5y. Let me choose another variable, say, v =y. Then x =u -5v. Substitute into the numerator:5x² +8xy +5y² -14x -10y +30=5(u -5v)^2 +8(u -5v)v +5v² -14(u -5v) -10v +30Expand:5(u² -10uv +25v²) +8(uv -5v²) +5v² -14u +70v -10v +30Compute term by term:5u² -50uv +125v² +8uv -40v² +5v² -14u +70v -10v +30Combine like terms:5u² + (-50uv +8uv) + (125v² -40v² +5v²) + (-14u) + (70v -10v) +30Which is:5u² -42uv +90v² -14u +60v +30Now, this is a quadratic in v: 90v² + (-42u +60)v +5u² -14u +30So for fixed u, this is a quadratic in v, which can be minimized by completing the square or using vertex formula.The quadratic in v is Av² +Bv +C, where:A=90B= -42u +60C=5u² -14u +30The vertex occurs at v = -B/(2A)= (42u -60)/(2*90)= (42u -60)/180= (7u -10)/30So the minimum value of the numerator for each u is:C - B²/(4A)=5u² -14u +30 - ( (-42u +60)^2 )/(4*90 )Compute (-42u +60)^2 = (42u -60)^2 = (6(7u -10))^2=36(7u -10)^2Thus,C - B²/(4A)=5u² -14u +30 - [36(7u -10)^2]/(360)Simplify denominator: 360=4*90=360So,=5u² -14u +30 - [36(7u -10)^2]/360=5u² -14u +30 - (7u -10)^2 /10Expand (7u -10)^2:=49u² -140u +100Thus,=5u² -14u +30 - (49u² -140u +100)/10= Convert to common denominator:= (50u² -140u +300 -49u² +140u -100)/10Simplify numerator:(50u² -49u²) + (-140u +140u) + (300 -100)= u² +0 +200Thus,= (u² +200)/10Wait, that's interesting. So the minimum of the numerator for each fixed u is (u² +200)/10.Wait, that seems surprising. Let me check the calculation step by step.Original expression after substitution:Numerator:5u² -42uv +90v² -14u +60v +30Expressed as quadratic in v:90v² + (-42u +60)v +5u² -14u +30So A=90, B=-42u +60, C=5u² -14u +30The minimum value is C - B²/(4A)So:5u² -14u +30 - [(-42u +60)^2]/(4*90)First compute (-42u +60)^2:= (42u -60)^2 =42²u² -2*42*60u +60²=1764u² -5040u +3600Then divide by 4*90=360:= (1764u² -5040u +3600)/360=1764/360 u² -5040/360 u +3600/360Simplify:1764 ÷ 36=49; 360 ÷36=10 ⇒49/10 u²5040 ÷360=14; so -14u3600 ÷360=10So,=49/10 u² -14u +10Therefore, the minimum value is:5u² -14u +30 - (49/10 u² -14u +10)Compute term by term:5u² -49/10 u²= (50/10 -49/10)u²=1/10 u²-14u - (-14u)=030 -10=20Thus, total:1/10 u² +20So the minimum numerator for each u is (u²)/10 +20.Wait, that's different from my initial conclusion. So correction: the minimum numerator is (u²)/10 +20. So I must have miscalculated earlier. Yes, because when I subtracted (49u² -140u +100)/10 from 5u² -14u +30, I converted 5u² to 50u²/10, right?Wait, let's do that step again:Original expression:C - B²/(4A)=5u² -14u +30 - (49u² -140u +100)/10Convert 5u² to 50u²/10, -14u to -140u/10, and 30 to 300/10:= (50u²/10 -140u/10 +300/10) - (49u² -140u +100)/10Now combine:[50u² -140u +300 -49u² +140u -100]/10Simplify numerator:(50u² -49u²) + (-140u +140u) + (300 -100) =u² +0 +200=u² +200Thus, the numerator is (u² +200)/10. So that part was correct. So the minimal numerator for each u is (u² +200)/10.Therefore, the entire expression becomes:Numerator: (u² +200)/10Denominator: (4 -u²)^(7/2)Thus, the expression simplifies to:[(u² +200)/10] / (4 -u²)^(7/2) = (u² +200) / [10*(4 -u²)^(7/2)]So now, we need to minimize this expression over u ∈ (-2, 2). Let's denote:f(u) = (u² +200) / [10*(4 -u²)^(7/2)]We can now consider this as a single-variable calculus problem. Let's set t =u, and find the minimum of f(t) for t ∈ (-2, 2).First, let's note that f(t) is even? Let's check f(-t):f(-t) = ((-t)^2 +200)/[10*(4 -(-t)^2)^(7/2)] = (t² +200)/[10*(4 -t²)^(7/2)] = f(t)Yes, so f(t) is even. Therefore, it's symmetric about t=0. So we can consider t ∈ [0,2) and double the result, but since we are looking for minima, the minimum could be at t=0 or somewhere else. Let's compute the derivative and find critical points.Let me compute f'(t) and set it to zero.First, write f(t) as:f(t) = (t² +200) / [10*(4 -t²)^(7/2)]Let’s write this as:f(t) = (1/10)*(t² +200)*(4 - t²)^{-7/2}Now, take the derivative using the product rule:f’(t) = (1/10)*[ derivative of (t² +200) * (4 -t²)^{-7/2} + (t² +200)*derivative of (4 -t²)^{-7/2} ]Wait, actually, it's the derivative of the first times the second plus the first times the derivative of the second. But here, it's a single product, so:f’(t) = (1/10)*[2t*(4 -t²)^{-7/2} + (t² +200)*(-7/2)*( -2t)*(4 -t²)^{-9/2} ]Wait, better step by step:Let’s set f(t) = (1/10)*A(t)*B(t), where A(t) = t² +200, B(t) = (4 -t²)^{-7/2}Then, f’(t) = (1/10)*(A’(t)B(t) + A(t)B’(t))Compute A’(t) =2tCompute B’(t):B(t)=(4 -t²)^{-7/2}, so derivative is (-7/2)*(4 -t²)^{-9/2}*(-2t)= (-7/2)*(-2t)*(4 -t²)^{-9/2}=7t*(4 -t²)^{-9/2}Therefore,f’(t) = (1/10)*[2t*(4 -t²)^{-7/2} + (t² +200)*7t*(4 -t²)^{-9/2} ]Factor out common terms:= (1/10)*t*(4 -t²)^{-9/2} [2*(4 -t²) +7*(t² +200)]Because:First term: 2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2}*(4 -t²)^0Second term:7t*(t² +200)*(4 -t²)^{-9/2}Factor out t*(4 -t²)^{-9/2}:= t*(4 -t²)^{-9/2} [2*(4 -t²) +7*(t² +200)]Let me verify:In the first term, factor out t*(4 -t²)^{-9/2}:2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2}*(4 -t²)^{-0}=2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2}*( (4 -t²)^{-1} )^{0} Hmm, maybe my previous statement is incorrect.Wait, actually:The first term is 2t*(4 -t²)^{-7/2}The second term is7t*(t² +200)*(4 -t²)^{-9/2}To factor out (4 -t²)^{-9/2}, we need to write the first term as 2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2}*(4 -t²)^{0} =2t*(4 -t²)^{-7/2}*(4 -t²)^{-2 +2} ?Alternatively, express (4 -t²)^{-7/2} = (4 -t²)^{-7/2} = (4 -t²)^{-7/2}*(4 -t²)^{-2/2}*(4 -t²)^{2/2} = (4 -t²)^{-9/2}*(4 -t²)^{1}Wait, maybe this is complicating. Let me write both terms with the same exponent:First term:2t*(4 -t²)^{-7/2} =2t*(4 -t²)^{-7/2}*(4 -t²)^{0} =2t*(4 -t²)^{-7/2}Second term:7t*(t² +200)*(4 -t²)^{-9/2}To combine these, multiply the first term by (4 -t²)^{-1} to get the same exponent:Wait, (4 -t²)^{-7/2} = (4 -t²)^{-7/2} = (4 -t²)^{-9/2 +1}So multiply and divide by (4 -t²)^1:First term:2t*(4 -t²)^{-7/2}=2t*(4 -t²)^{-9/2 +1}=2t*(4 -t²)^{-9/2}*(4 -t²)^1Therefore:f’(t) = (1/10)*[2t*(4 -t²)^{-9/2}*(4 -t²) +7t*(t² +200)*(4 -t²)^{-9/2}]Factor out t*(4 -t²)^{-9/2}:= (1/10)*t*(4 -t²)^{-9/2} [2*(4 -t²) +7*(t² +200)]Now compute the expression inside the brackets:2*(4 -t²) +7*(t² +200) =8 -2t² +7t² +1400=5t² +1408Therefore, f’(t) = (1/10)*t*(4 -t²)^{-9/2}*(5t² +1408)To find critical points, set f’(t)=0. The factors are:- (1/10): never zero- t: zero when t=0- (4 -t²)^{-9/2}: never zero, undefined when 4 -t²=0 (i.e., t=±2), but those are endpoints- 5t² +1408: always positive, since 5t² ≥0 and 1408 >0Thus, the only critical point in (-2,2) is at t=0.Therefore, we have critical points at t=0 and the endpoints t=±2 (but t=±2 are excluded since the domain is open interval (-2,2)). However, we can approach t approaching ±2.Wait, but as t approaches ±2, the denominator (4 -t²)^(7/2) approaches zero from the positive side, so the entire expression tends to positive infinity. Therefore, the minimum must occur at the critical point t=0.Therefore, the minimum value of the expression is at t=0, which gives:f(0) = (0 +200)/[10*(4 -0)^(7/2)] =200/[10*4^(7/2)]Compute 4^(7/2)= (2²)^(7/2)=2^(7)=128Thus,f(0)=200/(10*128)=200/1280=0.15625=5/32≈0.15625But let's verify if this aligns with the previous result where the numerator was minimized at 20 and denominator at 128. Yes:20/128=5/32=0.15625. So that's consistent.Therefore, the minimum value is 5/32, which is 0.15625. Rounded to the nearest hundredth is 0.16.But wait, before finalizing, let's make sure there are no other critical points. Since the derivative f’(t) is proportional to t*(5t² +1408), and 5t² +1408 is always positive, the only critical point is at t=0. Therefore, the function decreases from t approaching -2 to t=0 and increases from t=0 to t approaching 2. Hence, the minimal value is indeed at t=0.Therefore, the minimum value of the expression is 5/32, which is approximately 0.15625, rounded to the nearest hundredth is 0.16.But let me check this calculation once again:4^(7/2) is sqrt(4)^7=2^7=128. So denominator is 10*128=1280.Numerator is 200. 200/1280= divide numerator and denominator by 20:10/64=5/32≈0.15625≈0.16.Yes, correct.Alternatively, if we convert 5/32 to decimal:32*0.15625=5, yes. So 5/32=0.15625≈0.16.Thus, the minimum value is approximately 0.16.But the problem says "if necessary, round your answer to the nearest hundredth". So 0.16 is already to the nearest hundredth.Therefore, the answer is boxed{0.16}Final Answerboxed{0.16}

✅:To find the minimum value of the expression [frac{5 x^{2}+8 x y+5 y^{2}-14 x-10 y+30}{left(4 - x^{2} -10 x y -25 y^{2}right)^{7 / 2}},]we start by analyzing the domain of the denominator. The expression (4 - x^2 - 10xy - 25y^2) must be positive, which simplifies to ((x + 5y)^2 < 4). This means (x + 5y) must be between (-2) and (2).Next, we simplify the numerator (5x^2 + 8xy + 5y^2 - 14x - 10y + 30) by rotating the coordinate system to eliminate the cross term. This results in the quadratic form (9x'^2 + y'^2 - 12sqrt{2}x' + 2sqrt{2}y' + 30). Completing the square for both (x') and (y') terms, we find the minimum value of the numerator is 20, achieved at (x' = frac{2sqrt{2}}{3}) and (y' = -sqrt{2}).We then check if this critical point lies within the domain. The point (x + 5y = 0) is within the interval ((-2, 2)), confirming it is valid. Next, we consider the expression in terms of (u = x + 5y). The numerator simplifies to (frac{u^2 + 200}{10}) and the denominator becomes ((4 - u^2)^{7/2}). We minimize the function [f(u) = frac{u^2 + 200}{10(4 - u^2)^{7/2}}]over (u in (-2, 2)). By analyzing the derivative, we find the critical point at (u = 0). Evaluating the function at (u = 0) gives:[f(0) = frac{200}{10 cdot 128} = frac{200}{1280} = frac{5}{32} approx 0.15625.]Rounding to the nearest hundredth, the minimum value is (boxed{0.16}).

🔑:1. Rewrite the given expression: [ frac{5 x^2 + 8 x y + 5 y^2 - 14 x - 10 y + 30}{left(4 - x^2 - 10 x y - 25 y^2right)^{7/2}} ]2. Simplify the numerator step by step: [ 5 x^2 + 8 x y + 5 y^2 - 14 x - 10 y + 30 = (2x + y)^2 + (x + 2y)^2 - 14x - 10y + 30 ] Decompose and complete the square: [ (2x + y)^2 = 4x^2 + 4xy + y^2 ] [ (x + 2y)^2 = x^2 + 4xy + 4y^2 ] Combine these two: [ (2x+y)^2 + (x+2y)^2 = 4x^2 + 4xy + y^2 + x^2 + 4xy + 4y^2 = 5x^2 + 8xy + 5y^2 ] Update the expression: [ 5x^2 + 8xy + 5y^2 - 14x - 10y + 30 = (2x+y-3)^2 + (x+2y-1)^2 + 20 ]3. The minimum value of ((2x+y-3)^2 + (x+2y-1)^2) occurs when (2x+y-3 = 0) and (x+2y-1=0). Solve these equations: Solve the system to find: [ begin{cases} 2x + y = 3 x + 2y = 1 end{cases} ] Multiply the second equation by 2 and subtract from the first: [ 2x + y - (2x + 4y) = 3 - 2 implies -3y = 1 implies y = -frac{1}{3} ] Substitute ( y ) back into the first equation: [ 2x + left(-frac{1}{3}right) = 3 implies 2x - frac{1}{3} = 3 implies 2x = 3 + frac{1}{3} = frac{10}{3} implies x = frac{5}{3} ]4. Substitute ( x = frac{5}{3} ) and ( y = -frac{1}{3} ) into the denominator: [ x+5y = frac{5}{3} + 5 left(-frac{1}{3}right) = frac{5}{3} - frac{5}{3} = 0 ] Therefore: [ 4 - (x+5y)^2 = 4 - 0^2 = 4 ]5. Raise 4 to the power ( frac{7}{2} ): [ (4)^{7/2} = (2^2)^{7/2} = 2^7 = 128 ]6. Minimum value of the numerator is 20 (attained when (2x + y - 3 = 0) and (x + 2y - 1 = 0)). Therefore: [ frac{20}{128} = frac{5}{32} ]7. The decimal equivalent is: [ frac{5}{32} approx 0.15625 approx 0.16 ] Conclusion:[boxed{frac{5}{32} approx 0.16}]

🧠:Okay, let me try to solve this problem step by step. So, the question is about figuring out what fraction of Georgia's height Beatrix's height is, given some relationships between their heights. Let me read it again to make sure I understand all the details."Beatrix is twice the height of Violet, and Violet is 2/3 the height of Georgia. What fraction of Georgia's height is Beatrix's height?" The options are given from (A) to (E), with (C) 4/3 being one of them. Hmm, let me see.Alright, so first, let's break down the information. We have three people: Beatrix, Violet, and Georgia. The relationships given are:1. Beatrix is twice the height of Violet.2. Violet is 2/3 the height of Georgia.And the question is asking for Beatrix's height as a fraction of Georgia's height. So, essentially, we need to express Beatrix's height in terms of Georgia's height.Let me assign variables to each person's height to make it easier. Let me denote:- Let G = Georgia's height- Let V = Violet's height- Let B = Beatrix's heightFrom the problem:1. B = 2 * V (Beatrix is twice as tall as Violet)2. V = (2/3) * G (Violet is two-thirds as tall as Georgia)We need to find B/G, which is the fraction of Georgia's height that Beatrix is.Since we have B in terms of V and V in terms of G, we can substitute the second equation into the first one to get B in terms of G.So, substituting V from the second equation into the first equation:B = 2 * V = 2 * (2/3 * G)Let me compute that:2 * (2/3 * G) = (2 * 2)/3 * G = 4/3 * GTherefore, B = (4/3) * G, which means Beatrix's height is 4/3 of Georgia's height. Therefore, the fraction is 4/3, which is option (C).Wait, that seems straightforward, but let me check if I made any mistakes here. Let me verify each step again.First, defining the variables correctly. Yes, G, V, B for Georgia, Violet, Beatrix.First equation: B = 2V. Correct, as Beatrix is twice Violet's height.Second equation: V = (2/3)G. Correct, since Violet is two-thirds of Georgia's height.Then substituting V into the first equation gives B = 2*(2/3 G) = 4/3 G. So B/G = 4/3. Therefore, the answer is 4/3, which is option (C). That seems to check out.Wait, but just to make sure I didn't misread the question. The problem says "What fraction of Georgia's height is Beatrix's height?" So, Beatrix's height divided by Georgia's height, which is exactly B/G. So, yes, 4/3. So, 4/3 is the answer. Option (C). Hmm. That's one of the options. Let me check the options again:(A) 9/7(B) 2/3(C) 4/3(D) 5/4(E) 3/2Yes, (C) 4/3 is there. So, that should be the correct answer.But let me think again, maybe using actual numbers to test. Suppose Georgia's height is, say, 3 units. Then Violet's height is 2/3 of Georgia's, so 2/3 * 3 = 2 units. Then Beatrix is twice Violet's height, so 2 * 2 = 4 units. Then, Beatrix's height is 4 units, Georgia's is 3 units, so 4/3. Yep, that gives the same result. So with concrete numbers, it still works out. Therefore, the answer is 4/3.I think that's solid. No mistakes in the reasoning. The substitution is correct, and the verification with numbers confirms it. So, I feel confident that the answer is (C) 4/3.Final Answerboxed{C}

✅:Given the relationships between the heights of Beatrix, Violet, and Georgia, we need to determine what fraction of Georgia's height is Beatrix's height.1. Define Variables: - Let ( G ) be Georgia's height. - Let ( V ) be Violet's height. - Let ( B ) be Beatrix's height.2. Given Relationships: - Beatrix is twice the height of Violet: ( B = 2V ). - Violet is (frac{2}{3}) the height of Georgia: ( V = frac{2}{3}G ).3. Substitute Violet's Height into Beatrix's Height: - Substitute ( V = frac{2}{3}G ) into ( B = 2V ): [ B = 2 left( frac{2}{3}G right) = frac{4}{3}G ]4. Determine the Fraction: - The fraction of Georgia's height that Beatrix's height is given by ( frac{B}{G} ): [ frac{B}{G} = frac{frac{4}{3}G}{G} = frac{4}{3} ]5. Verification with Concrete Numbers: - Suppose Georgia's height ( G = 3 ) units. - Violet's height ( V = frac{2}{3} times 3 = 2 ) units. - Beatrix's height ( B = 2 times 2 = 4 ) units. - Fraction ( frac{B}{G} = frac{4}{3} ).Thus, the fraction of Georgia's height that Beatrix's height is (boxed{C}).

🔑:1. Let the heights of Beatrix, Violet, and Georgia be represented by ( B ), ( V ), and ( G ) respectively.2. According to the problem, Beatrix is twice the height of Violet. This can be written as: [ B = 2V ]3. The problem also states that Violet is (frac{2}{3}) the height of Georgia. This can be written as: [ V = frac{2}{3}G ]4. Now, substitute ( V = frac{2}{3}G ) into the equation ( B = 2V ): [ B = 2 left( frac{2}{3}G right) ]5. Simplify the expression on the right-hand side: [ B = frac{4}{3}G ]6. Therefore, Beatrix's height is (frac{4}{3}) of Georgia's height. Conclusion:[boxed{frac{4}{3}}]

🧠:Okay, let's tackle these probability problems step by step. I need to make sure I understand each part and think through them carefully. Let's start with part a.Problem a): What is the probability that they have one boy and one girl?Alright, so the family has two children. Each child is equally likely to be a boy or a girl, right? So the possible gender combinations for the two children are: BB, BG, GB, GG. Each of these is equally probable because each birth is independent and has a 50% chance for each gender. Wait, let me verify that. If the first child is a boy, the second could be a boy or a girl. Same if the first is a girl. So the four combinations should each have a probability of 0.5 * 0.5 = 0.25. So each has a 25% chance.Out of these four combinations, how many have one boy and one girl? Let's see: BG and GB. That's two out of four. So the probability would be 2/4, which simplifies to 1/2. So is the answer 1/2?Hmm, that seems straightforward. But wait, sometimes in probability problems, things aren't as they seem. Let me double-check. The question is about exactly one boy and one girl. So yes, the combinations are BG and GB. Each is 25%, so together 50%. So part a) should be 1/2.Moving on to part b).Problem b): Additionally, it is known that one of the children is a boy. What is the probability now that the family has one boy and one girl?Okay, so now we have some additional information: at least one of the children is a boy. How does this affect the probability?In probability terms, this is a conditional probability problem. We need to find the probability that the family has one boy and one girl given that at least one child is a boy.The formula for conditional probability is P(A|B) = P(A ∩ B) / P(B). Here, event A is "one boy and one girl," and event B is "at least one boy." So we need to compute P(A and B) divided by P(B).But in this case, if A is "one boy and one girl," then A is a subset of B, because if they have one boy and one girl, then they definitely have at least one boy. Therefore, P(A ∩ B) is just P(A). So the formula becomes P(A) / P(B).From part a), we know P(A) is 1/2. What is P(B)? The probability of at least one boy. The complement of this is having no boys, i.e., two girls. The probability of two girls is 1/4 (since GG is one out of four possibilities). Therefore, P(B) = 1 - P(GG) = 1 - 1/4 = 3/4.So the conditional probability is (1/2) / (3/4) = (1/2) * (4/3) = 2/3. Therefore, the probability is 2/3.Wait, but let me think again. The possible combinations when we know at least one boy are BB, BG, GB. Originally, those three each have 1/4 probability. But wait, actually, the total probability for these three is 3/4. So each of these three is equally likely? Wait, no. The individual probabilities are still each 1/4. So the probability of BB is 1/4, BG is 1/4, GB is 1/4. So given that we have at least one boy, the conditional probability is each of these divided by 3/4. So each becomes (1/4)/(3/4) = 1/3. So the probability of each is 1/3.Therefore, the probability that they have one boy and one girl is the sum of BG and GB, which is 1/3 + 1/3 = 2/3. So same answer. Therefore, 2/3.But I remember this is a classic problem. Often people mistakenly think it's 1/2 because they confuse the problem with the case where the order doesn't matter, but actually, since BG and GB are separate, the answer is 2/3. So I think this is correct.Now part c) is trickier.Problem c): Additionally, it is known that the boy was born on a Monday. What is the probability now that the family has one boy and one girl?Hmm, this adds another layer. Now we know not only that there's a boy but also that this boy was born on a Monday. How does this affect the probability?I need to consider the days of the week as well. Each day is equally likely, I assume. So for each child, there's a 1/2 chance of being a boy or girl, and for the day, a 1/7 chance for each day.But since one of the children is a boy born on Monday, we need to adjust our sample space accordingly.Let me model all possible cases.First, the family has two children. Each child has a gender and a day of birth. The total possibilities would be for each child: 2 genders * 7 days = 14 possibilities per child. So for two children, there are 14 * 14 = 196 possible combinations.But since we know that at least one child is a boy born on Monday, we need to consider the cases where at least one of the two children is a boy born on Monday. Then, within those cases, compute how many have one boy and one girl.So let's break it down.First, compute the total number of cases where at least one child is a boy born on Monday.To compute this, we can use the principle of inclusion-exclusion. The number of families with at least one Monday boy is equal to the number of families where the first child is a Monday boy plus the number where the second child is a Monday boy minus the number where both are Monday boys.Each child has a 1/2 chance of being a boy and 1/7 chance of being born on Monday, so the probability that a specific child is a Monday boy is 1/2 * 1/7 = 1/14.Therefore, the number of families where the first child is a Monday boy: 1/14 * 14 (since the second child can be anything). Wait, but maybe it's better to think in counts.In the total of 196 possible combinations:- Number of cases where the first child is a Monday boy: 1 (Monday) * 1 (boy) * 14 (second child can be any gender and any day) = 1*1*14 = 14.Similarly, the second child being a Monday boy: 14.But then the overlap where both children are Monday boys is 1 (first child Monday boy) * 1 (second child Monday boy) = 1*1 = 1.Wait, but actually, if the first is a Monday boy, and the second is a Monday boy, each has 1 possibility for day and gender, so total 1*1 = 1. But in reality, each child's day is independent, so the number of cases where both are Monday boys is 1 (Monday) * 1 (boy) for the first child, and 1 (Monday) * 1 (boy) for the second child. So 1*1 = 1. Therefore, total number with at least one Monday boy is 14 + 14 - 1 = 27.Wait, but in the total of 196, how does this count?Yes, 14 (first child Monday boy) + 14 (second child Monday boy) - 1 (both Monday boys) = 27. So 27 cases where at least one child is a Monday boy.Now, among these 27 cases, how many correspond to families with one boy and one girl?So we need to count the number of cases where exactly one child is a boy (born on Monday) and the other is a girl.Wait, but not exactly. Because the family could have one boy born on Monday and another girl born on any day. Alternatively, the family could have a boy born on Monday and another boy born on a different day, or another boy also born on Monday.So we need to separate the cases where the family has one boy (Monday) and one girl, versus two boys (one Monday and one not, or both Monday).So let's compute the number of favorable cases (one boy, one girl) where at least one is a Monday boy.Case 1: The family has one Monday boy and one girl.This can happen in two ways:- First child is a Monday boy, second is a girl (any day).- Second child is a Monday boy, first is a girl (any day).For each of these, the number of possibilities:For the first case: The first child is a Monday boy (1 possibility), the second child is a girl (gender: 1 possibility) with any day (7 possibilities). So 1 * 1 * 7 = 7.Similarly, the second case: Second child is a Monday boy, first is a girl. Again, 7 possibilities.Total for Case 1: 7 + 7 = 14.Case 2: The family has two boys, at least one of whom is a Monday boy.But we need to exclude this case from our favorable outcomes, since we are looking for one boy and one girl. Wait, no. Our favorable outcomes are only the 14 cases above. The rest are either two boys (with at least one Monday) or two girls (but two girls can't have a Monday boy, so they are excluded). Wait, but in the total of 27 cases, there are no two girls, since we already have at least one Monday boy. So the 27 cases consist of:- 14 cases where one child is a Monday boy and the other is a girl.- 13 cases where there are two boys, at least one of whom is a Monday boy. Because 27 total cases - 14 favorable cases = 13 two-boy cases.Wait, let's verify that.Wait, two boys with at least one Monday boy: How many?Total two-boy families: 7 days for the first boy, 7 days for the second boy = 49.But the number of two-boy families where at least one is a Monday boy: Total two-boy families (49) minus those with no Monday boys. The number of two-boy families with no Monday boys: 6 days for the first boy and 6 days for the second boy = 36. So 49 - 36 = 13. Therefore, 13 two-boy families with at least one Monday boy.Therefore, in the 27 total cases with at least one Monday boy, 14 are one-boy-one-girl, and 13 are two-boy families.Therefore, the probability is 14 / 27.Wait, so the answer is 14/27. That seems a bit non-intuitive. Let me check again.Alternatively, another approach: For each child, the possible gender and day combinations are 14 (2 genders * 7 days). For two children, 14*14=196.Now, the number of families with at least one Monday boy: As before, 14 + 14 -1 = 27.Number of those with one boy and one girl: As calculated, 14.Therefore, probability is 14/27.But let's think of another way. When you have two children, the possible combinations where at least one is a Monday boy.The possible scenarios:1. First child is a Monday boy, second child is anything.2. Second child is a Monday boy, first child is anything.But as before, overlapping in both being Monday boys.So in scenario 1: First child is Monday boy (1 gender, 1 day), second child can be any gender and day (2*7=14). So 14 possibilities.Scenario 2: Second child is Monday boy, first child can be any gender and day (2*7=14). So another 14.But overlapping is both children being Monday boys. There's 1 case where both are Monday boys.Therefore, total is 14 + 14 -1 = 27.Now, among these, how many have one boy and one girl.In scenario 1: First is Monday boy, second can be anything. If the second is a girl, regardless of day, that's 7 days. So 7 cases.In scenario 2: Second is Monday boy, first is anything. If the first is a girl, regardless of day, that's 7 days. So another 7 cases.But we have to check if there's overlap. However, if both children are Monday boys, the other child is a boy, so those cases are not in the one-boy-one-girl category. Therefore, the total is 7 +7 =14.Therefore, the probability is 14/27.Therefore, the answer to part c) is 14/27.Wait, that's interesting. Adding the information about the day of birth actually changes the probability. Initially, in part b), it was 2/3, but now with the day specified, it's 14/27, which is approximately 0.5185, which is slightly less than 2/3 (~0.6667). That makes sense because specifying a particular day makes it more likely that the family has two boys, since having a boy born on a specific day is another condition that can be satisfied by having two boys (either one born on Monday or both). So the probability decreases from 2/3 to 14/27.But let me check again to make sure.Alternative approach: For each child, the probability of being a boy born on Monday is 1/14, girl is 1/2, and boy born on another day is 6/14 = 3/7.Wait, no. For each child:Probability of being a boy: 1/2. Probability of being a girl: 1/2.For a boy, the probability of being born on Monday is 1/7, so boy on Monday is (1/2)*(1/7) = 1/14, and boy on another day is (1/2)*(6/7) = 6/14 = 3/7.Similarly, girl on any day is 1/2, regardless of day.But when we have two children, and the information that at least one is a boy born on Monday, we can model the possible cases.So the possible cases are:1. First child is a boy born on Monday, second child is anything.2. Second child is a boy born on Monday, first child is anything.But overlapping when both are boys born on Monday.So in terms of probability, the probability of at least one boy born on Monday is:P(at least one Monday boy) = P(first is Monday boy) + P(second is Monday boy) - P(both are Monday boys) = (1/14) + (1/14) - (1/14 * 1/14) = 2/14 - 1/196 = (28 - 1)/196 = 27/196.So the total probability is 27/196.Now, the probability that they have one boy and one girl, and at least one Monday boy is:This is the probability that one child is a Monday boy and the other is a girl.Which is P(first is Monday boy and second is girl) + P(second is Monday boy and first is girl).Each of these is (1/14)*(1/2) = 1/28. Since the girl can be born on any day, but the day doesn't matter. Wait, hold on. Wait, if the other child is a girl, does her day matter? No, because the problem only specifies that the boy was born on Monday. The girl's day isn't specified, so her day is irrelevant.Wait, but when we compute the probability of having one boy (Monday) and one girl, regardless of the girl's day, then:For the first case: Monday boy (1/14) and girl (1/2 * 1 = 1/2, since day doesn't matter). So probability is 1/14 * 1/2 = 1/28.Similarly, the second case: Girl (1/2) and Monday boy (1/14). So 1/2 * 1/14 = 1/28.Total probability for one boy and one girl with at least one Monday boy is 1/28 + 1/28 = 2/28 = 1/14.Wait, but this contradicts the previous count. Wait, no, because in the count approach, we had 14 cases out of 196, which is 14/196 = 1/14. So the probability of the intersection (one boy one girl and at least one Monday boy) is 1/14. The total probability of at least one Monday boy is 27/196.Therefore, the conditional probability is (1/14) / (27/196) = (1/14) * (196/27) = 14/27. So same result. So that's correct.Therefore, the answer is 14/27.So summarizing:a) 1/2b) 2/3c) 14/27I think that's the correct answer. Let me just check online if similar problems have these answers to confirm.Wait, but since I can't actually check, I need to rely on my reasoning. For part c), it's a variation of the classic "Tuesday boy" problem, which is well-known in probability. The answer there is similar, depending on the specific information given. So in our case, since we're told that one of the children is a boy born on Monday, the probability is 14/27. That seems consistent with the reasoning for the Tuesday boy problem, where specifying a day slightly increases the probability of two boys compared to just knowing there's a boy, but in our case, it actually reduces the probability compared to part b). Wait, no—part b) had 2/3 (~66.67%), and part c) is 14/27 (~51.85%), which is lower. That seems correct because the added information makes it more likely that the family has two boys. Wait, but how?Wait, in part b), the probability was 2/3 that it's one boy one girl given at least one boy. When we add the condition that the boy is born on Monday, why does the probability decrease?Because the additional information about the day allows for more possibilities where there are two boys. For example, if there are two boys, the chance that at least one was born on Monday is higher than if there was only one boy. Therefore, conditioning on this specific information makes the two-boy case more probable relative to the one-boy-one-girl case. Hence, the probability of one boy one girl decreases.Yes, that makes sense. For example, if you have two boys, there are two chances to have a boy born on Monday, whereas with one boy, there's only one chance. Therefore, the two-boy case is more likely to satisfy the condition "at least one boy born on Monday," so when you condition on that, the probability of two boys is higher than in part b), hence the lower probability for one boy one girl.Therefore, the answer 14/27 is correct.Final Answera) boxed{dfrac{1}{2}}b) boxed{dfrac{2}{3}}c) boxed{dfrac{14}{27}}

✅: SummaryProblem a) The family has two children, and each child can be a boy or a girl with equal probability. The possible gender combinations are BB, BG, GB, GG. Each combination has a probability of ( frac{1}{4} ). The combinations BG and GB represent one boy and one girl, so the probability is ( frac{2}{4} = frac{1}{2} ).Answer: boxed{dfrac{1}{2}}Problem b) Given that at least one child is a boy, we use conditional probability. The possible combinations are BB, BG, GB. The probability of each combination is ( frac{1}{4} ). The probability of having at least one boy is ( 1 - frac{1}{4} = frac{3}{4} ). The probability of one boy and one girl (BG or GB) is ( frac{2}{4} ). The conditional probability is ( frac{frac{2}{4}}{frac{3}{4}} = frac{2}{3} ).Answer: boxed{dfrac{2}{3}}Problem c) Given that one of the children is a boy born on Monday, we consider the total number of cases where at least one child is a boy born on Monday. Each child has 14 possible gender-day combinations (2 genders × 7 days). The total number of combinations for two children is ( 14 times 14 = 196 ). The number of favorable cases (one boy born on Monday and one girl) is 14. The number of cases with at least one Monday boy is 27 (calculated using inclusion-exclusion principle). The conditional probability is ( frac{14}{27} ).Answer: boxed{dfrac{14}{27}}

🔑:# Part (a)1. Set up all possible outcomes: - Since the birth of a boy (B) and a girl (G) are equally likely, we consider the possible combinations of two children. These combinations are: BB, BG, GB, GG.2. Calculate the probability of each outcome: - Each of these combinations is equally likely, so the probability of each combination is (frac{1}{4}).3. Identify favorable outcomes: - The scenarios that meet the condition of having one boy and one girl are BG and GB.4. Sum the probabilities of the favorable outcomes: [ P(text{one boy and one girl}) = P(BG) + P(GB) = frac{1}{4} + frac{1}{4} = frac{2}{4} = frac{1}{2} ]Conclusion:[boxed{frac{1}{2}}]# Part (b)1. Set up remaining possible outcomes given one child is a boy: - Given that one child is a boy, the remaining possible combinations are BB, BG, and GB.2. Calculate the probability of each remaining outcome: - These combinations are still equally likely, so the probability of each is now (frac{1}{3}).3. Identify favorable outcomes: - The scenarios that meet the condition of having one boy and one girl are BG and GB.4. Sum the probabilities of the favorable outcomes: [ P(text{one boy and one girl | one child is a boy}) = P(BG) + P(GB) = frac{1}{3} + frac{1}{3} = frac{2}{3} ]Conclusion:[boxed{frac{2}{3}}]# Part (c)1. Introduce additional information: - Given one boy was born on Monday (B_M), the number of possible outcomes increases considering the days of the week.2. Set up all possible sequences considering days: - We can denote the events for each child being born on any day other than Monday as . (i.e., ( B_M ) and ( G_D ), where D denotes any other day)3. Count the total distinct possible outcomes for BG, GB cases: - With 7 days in a week and considering one child (B_M), we have combinations BG, GB each now accounting for the day distributions: - Combinations favorable for one boy and one girl including (B_M) are: B_MG (where G is born on any other day), GB_M (where B might be born on the same or another day), giving us 7 (days for G) + 6 (remaining days for B) = 13 possible outcomes. 4. Determine probability of each favorable outcome: - Total of 27 sequences come into play (14 (G) + 13 (B_M)) possible combinations including all 27 possible mixed outcomes. 5. Sum the probabilities of the favorable outcomes: [ P(text{one boy and one girl | boy born on Monday}) = frac{Total favorable outcomes}{Total possible outcomes} = frac{14}{27} ]Conclusion:[boxed{frac{14}{27}}]